# Rare Events in Continuous Time: From Poisson Arrivals to Exponential Waiting Times > Build a continuous-time rare-event model from many independent tiny intervals, derive the Poisson count distribution as a binomial limit, and verify why its variance equals its mean. Then turn from counts to waiting times, derive the exponential distribution from the chance of seeing zero events, and prove its memoryless property. Service-desk arrivals and detector clicks show how the shared rate connects both viewpoints and how the model's mean-variance prediction can be tested against observations. - Canonical watch page: [Rare Events in Continuous Time: From Poisson Arrivals to Exponential Waiting Times](https://academa.ai/lectures/poisson-processes-and-waiting-times) - Publisher: [Academa, Inc.](https://academa.ai) - Subject: Statistics - Published: 2026-08-28T22:52:00.000Z - Updated: 2026-08-28T22:52:00.000Z - Duration: PT788S (13 minutes 8 seconds) - Chapters: 6 - Views: 0 - Language: en-US - Access: Free - Video stream: [HLS content](https://academa.ai/media/l/01M14V07J37P7MMQN9J7QVK3JS/0/dark/master.m3u8) - Audiovisual record: [Semantic JSON](https://academa.ai/media/l/01M14V07J37P7MMQN9J7QVK3JS/0/semantic.json) - Thumbnail: [Image](https://academa.ai/media/l/01M14V07J37P7MMQN9J7QVK3JS/0/dark/poster.jpg) ## Description Derive Poisson counts and exponential waiting times from tiny independent intervals, then test their predictions with arrival data. ## Chapters - [00:00–01:55.493 · Random Marks on Time](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=0) - [01:55.493–03:55.767 · The Poisson Limit](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=115.49329166666668) - [03:55.767–05:57.713 · The Signature of Poisson Counts](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002) - [05:57.713–07:41.45 · From Counts to Waiting](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=357.7125416666667) - [07:41.45–09:58.205 · Memory Without Age](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=461.4499166666667) - [09:58.205–13:08 · Models Meet Data](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=598.20525) ## Transcript ### [00:00 · Random Marks on Time](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=0) Rare events happen at particular moments: a customer reaches a desk, a detector records a click, or a component fails. We are going to build a continuous-time model from ordinary probability, beginning with nothing more mysterious than many small independent trials. Here is the question that will organize the first half. During a stretch of time of length T, how many events arrive? We want the complete distribution of that count, not merely its average. Picture time as a line. The red marks are arrivals from one possible run of the process. Another run would put them elsewhere and might contain a different number, but every arrival is attached to some moment. Now chop the observation time into n equal intervals. One interval has length delta t, equal to T divided by n. We will eventually make these intervals so short that an event inside one of them is genuinely rare. The model makes two substantive assumptions. First, separate intervals contribute independently. Second, the event rate is constant, so an interval of length delta t has event probability approximately lambda times delta t. Lambda is a rate, measured in events per unit time. Multiplying it by the interval length gives a dimensionless probability. As delta t shrinks, this probability shrinks too, but the total expected count across all n intervals remains lambda T. Each tiny interval is therefore a Bernoulli trial: event or no event. Adding n independent trials gives a binomial count X sub n, with n trials and success probability lambda T over n. Continuous time has not appeared by magic. We have built an approximation that can be pushed to a limit. ### [01:55.493 · The Poisson Limit](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=115.49329166666668) Before taking a limit, hold the central idea still. Suppose the expected count is two. Ten intervals can each have probability zero point two. One hundred intervals use zero point zero two, and one thousand use zero point zero zero two. The individual trials become rarer while their number grows. In every row, n times p sub n remains two. More generally, call the fixed expected count mu, equal to lambda T. Now fix a possible count k. The binomial probability of exactly k events is the number of ways to choose their intervals, times the probability of k occupied intervals, times the probability that all the others are empty. Substitute p sub n equal to mu over n and rearrange. The probability is mu to the k over k factorial, multiplied by three factors whose limits we can read separately. The first factor contains k terms: n over n, then n minus one over n, and so on. For a fixed k, every one of those terms approaches one. Therefore A sub n approaches one. The second factor is the classical exponential limit. One minus mu over n, raised to n, approaches e to the minus mu. This is the surviving probability of seeing no event across the complete interval. The last factor raises something approaching one to the fixed power minus k, so it also approaches one. Multiplying the three limits leaves e to the minus mu times mu to the k over k factorial. Finally replace mu by lambda T. This is the Poisson distribution, obtained rather than announced. Its probabilities really do sum to one, because the remaining series is the exponential series for e to the mu. ### [03:55.767 · The Signature of Poisson Counts](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002) Here is a Poisson distribution with mu equal to three. The bars give the probabilities of zero events, one event, two events, and so on. The distribution is not symmetric, but it is concentrated around counts near three. Its average can be calculated exactly. Start with the definition of expectation: sum k times the probability of k. The factor k cancels one factor from k factorial. Pull one factor of mu outside, then re-index with j equal to k minus one. What remains is the sum of every Poisson probability, so that sum is one. The expected count is therefore mu. For variance, the useful quantity is the second factorial moment, N times N minus one. The same cancellation, now performed twice, gives mu squared. Ordinary variance can be written as the second factorial moment plus the mean, minus the square of the mean. Substitute mu squared, mu, and mu squared. The two squares cancel, leaving mu. So a Poisson count has mean mu and variance mu. The standard deviation is therefore the square root of mu. Relative fluctuations become smaller when the expected count becomes large, even though the absolute fluctuations grow. This equality has an intuitive origin in the small-interval model. Independent intervals each contribute a tiny yes-or-no uncertainty. Adding them gives variance n p times one minus p, which approaches n p as p becomes tiny. More importantly, equal mean and variance is a prediction that data can challenge. If repeated equal windows have variance far above their mean, arrivals may be clustered or the rate may be changing. If variance is far below the mean, some regular spacing or dead time may be present. ### [05:57.713 · From Counts to Waiting](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=357.7125416666667) The count question has a natural companion. Instead of fixing a time interval and asking how many events occur, start the clock now and ask how long it runs before the next event. Call that random waiting time W. On one realized timeline, now is the green mark and the next arrival is the first red mark. The yellow span between them is the observed value of W. Another run would give a different span. To find the distribution, ask when W exceeds a chosen time t. That happens exactly when the interval from now to t contains zero events. Waiting longer than t and counting zero by t are the same event. But the count by time t is Poisson with mean lambda t. Put k equal to zero in the Poisson formula. The factorial and power disappear, leaving e to the minus lambda t. That gives the survival probability, the chance the wait is still not over. Its complement gives the cumulative distribution. Differentiating gives the exponential density, lambda e to the minus lambda t. For a unit rate, the survival curve begins at one. A very short threshold is likely to be crossed without an event, while a long event-free stretch becomes progressively less likely. Move the threshold to the right. The point rides down the curve because the probability of surviving without an event decays exponentially. The mean waiting time is the area under the survival curve. Integrating e to the minus lambda t from zero to infinity gives one over lambda. Double the event rate and the average wait is cut in half. ### [07:41.45 · Memory Without Age](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=461.4499166666667) This is the claim people rightly distrust. Suppose we have already waited s units of time with no arrival. Surely, it feels, we must now be closer to the next event. The exponential model says no. Let us prove exactly what that sentence means. Begin at zero and suppose no event occurs before s. The yellow span is the waiting time we have already endured. Reaching s tells us that W is greater than s. Now ask for no event during an additional duration t. That means the total wait exceeds s plus t. The green span is the extra future time whose probability we want. Conditional probability divides the chance of surviving all the way to s plus t by the chance of having reached s. The later event is contained inside the earlier condition, so this ratio is exact. Insert the exponential survival probabilities. The numerator contains e to the minus lambda times s plus t. The denominator contains e to the minus lambda s. The factors involving s cancel. What remains is e to the minus lambda t, exactly the probability of waiting at least t from a fresh start. The elapsed wait has disappeared from the answer. These two curves make the result visible. On the left is survival for an additional time starting now. On the right is survival for the same additional time after we have already waited s. The curves are identical. Why does this feel absurd? We often imagine an arrival that is already on its way, a bus following a timetable, or a component accumulating age. Those mechanisms carry information from the past. A constant-rate Poisson process does not: future tiny intervals are independent of past ones. Memorylessness is therefore not a universal law of waiting. It is a sharp consequence of the assumptions we chose. If the rate changes with time, arrivals repel one another, or a schedule is operating, the remaining wait can depend strongly on how long we have already waited. ### [09:58.205 · Models Meet Data](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=598.20525) Put the model to work at a service desk. Customers arrive at an average rate of six per hour. We want the probability of exactly three arrivals in half an hour, and the average time from now until the next arrival. For the count, match the units first. Thirty minutes is one half hour, so the expected count is lambda T, equal to six times one half, which is three. Use the Poisson probability with k equal to three and mu equal to three. The result is e to the minus three times three cubed over three factorial, about zero point two two four. For waiting, use the exponential mean. One over six hour is one sixth of sixty minutes, so the mean wait is ten minutes. The count calculation and the waiting calculation use the same rate lambda from two different perspectives. Now suppose a detector is observed in twelve equal one-minute windows. These are the recorded click counts. Equal windows matter because the Poisson prediction compares counts that share the same expected value. The twelve counts add to twenty-four, so their empirical mean is exactly two clicks per window. This estimates lambda times one minute. Next measure their spread around two. Using the average squared deviation, the empirical variance is twenty-six over twelve, about two point one seven. It is not exactly two, because a finite random sample does not reproduce a population identity perfectly. The mean two point zero zero and variance two point one seven are close. That does not prove the Poisson model, but it passes this first check. With many more windows, we could simulate or calculate the variation expected in these estimates and judge whether the discrepancy is unusual. The comparison becomes diagnostic when it is repeated over substantial data. Variance near the mean is compatible with independent arrivals at a stable rate. Variance much larger than the mean suggests extra variability: customers arriving in groups, bursts of detector noise, or a rate that changes through the day. Variance much smaller than the mean suggests excessive regularity: arrivals may inhibit one another, or a detector may briefly stop registering after each click. Either pattern points back to an assumption that should be reconsidered. So the complete connection is this. Chopping time into rare independent trials produces Poisson counts in the limit. Asking for zero counts up to time t produces exponential waiting. Independence gives memorylessness, and the equal mean and variance of the counts gives the model an observable signature. The formulas are useful because their assumptions are clear. Use the same rate lambda to move between counts and waits, then return to data and ask whether independence, a stable rate, and equal mean and variance are actually visible. A probability model earns its place by surviving that return. ## About Academa, Inc. Academa makes technical knowledge easier to understand through visual lectures and lets learners request new lecture videos on the topics they need. ## Complete audiovisual record Immutable source: [semantic.json](https://academa.ai/media/l/01M14V07J37P7MMQN9J7QVK3JS/0/semantic.json) Record version: 1. Render attempt: 0. ### How to read this timeline Each scene owns its object identifiers. A beat's board is the complete board when listed, empty when marked empty, and unchanged from the nearest earlier listed board in the same scene when marked unchanged. Action times are absolute positions in the published video. ### Scene 1: [Random Marks on Time](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=0) Span: 00:00–01:55.493 (0s–115.49329166666668s). #### Objects - assumptions: a Text \[text\] that says "A tiny interval has a small chance of one event, a negligible chance of two or more, and contributes independently of disjoint intervals." - binomial: a Math \[text\] that says "$X\_n tilde upright("Binomial")(n, frac(lambda T, n))$" - card: a Title that says "Introductory Probability — Rare Events in Continuous Time: From Poisson Arrivals to Exponential Waiting Times" - clicks: a Point \[red\] drawn in timeline (location=(0.8, 0.0)) - clicks\_2: a Point \[red\] drawn in timeline (location=(3.2, 0.0)) - clicks\_3: a Point \[red\] drawn in timeline (location=(5.7, 0.0)) - clicks\_4: a Point \[red\] drawn in timeline (location=(8.8, 0.0)) - delta: a Math \[text\] that says "$Delta t = frac(T, n)$" - one\_interval: a Brace \[yellow\] labelled "Delta t" drawn in timeline (x\_start=2.0, x\_end=3.0) - probability: a Math \[text\] that says "$p\_n = lambda Delta t = frac(lambda T, n)$" - question: a Panel that says "During a time interval of length $T$, how many randomly arriving events should we expect to count?" - timeline: a NumberLine labelled "t" (x\_range=(0.0, 10.0), include\_numbers=True) #### Beats ##### [00:00](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=0) Narration: Rare events happen at particular moments: a customer reaches a desk, a detector records a click, or a component fails. We are going to build a continuous-time model from ordinary probability, beginning with nothing more mysterious than many small independent trials. Board: Empty. Actions: - [00:00](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=0): card is shown on the screen, written out. - [00:1.5](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=1.5): card: enter:write-left-to-right. - [00:16.324](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=16.324): card is hidden from the screen — left the board. ##### [00:17.524](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=17.524) Narration: Here is the question that will organize the first half. During a stretch of time of length T, how many events arrive? We want the complete distribution of that count, not merely its average. Board: Unchanged from the preceding beat in this scene. Actions: - [00:17.524](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=17.524): question is shown on the screen, written out. - [00:29.32](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=29.3195): question moves to a new place on the board. ##### [00:29.919](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=29.9195) Narration: Picture time as a line. The red marks are arrivals from one possible run of the process. Another run would put them elsewhere and might contain a different number, but every arrival is attached to some moment. Board: question — a Panel that says "During a time interval of length $T$, how many randomly arriving events should we expect to count?" Actions: - [00:29.919](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=29.9195): timeline is shown on the screen, written out. - [00:32.59](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=32.59): clicks is shown on the screen, written out. - [00:32.77](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=32.77): clicks\_2 is shown on the screen, written out. - [00:32.95](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=32.95): clicks\_3 is shown on the screen, written out. - [00:33.13](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=33.13): clicks\_4 is shown on the screen, written out. ##### [00:43.151](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=43.1515) Narration: Now chop the observation time into n equal intervals. One interval has length delta t, equal to T divided by n. We will eventually make these intervals so short that an event inside one of them is genuinely rare. Board: question — a Panel that says "During a time interval of length $T$, how many randomly arriving events should we expect to count?"; timeline — a NumberLine labelled "t" (x\_range=(0.0, 10.0), include\_numbers=True); clicks — a Point \[red\] drawn in timeline (location=(0.8, 0.0)); clicks\_2 — a Point \[red\] drawn in timeline (location=(3.2, 0.0)); clicks\_3 — a Point \[red\] drawn in timeline (location=(5.7, 0.0)); clicks\_4 — a Point \[red\] drawn in timeline (location=(8.8, 0.0)) Actions: - [00:48.039](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=48.039): one\_interval is shown on the screen, written out. - [00:48.794](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=48.794000000000004): delta is shown on the screen, written out. - [00:57.675](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=57.675000000000004): assumptions is shown on the screen, written out. ##### [00:58.972](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=58.971999999999994) Narration: The model makes two substantive assumptions. First, separate intervals contribute independently. Second, the event rate is constant, so an interval of length delta t has event probability approximately lambda times delta t. Board: question — a Panel that says "During a time interval of length $T$, how many randomly arriving events should we expect to count?"; timeline — a NumberLine labelled "t" (x\_range=(0.0, 10.0), include\_numbers=True); assumptions — a Text \[text\] that says "A tiny interval has a small chance of one event, a negligible chance of two or more, and contributes independently of disjoint intervals."; delta — a Math \[text\] that says "$Delta t = frac(T, n)$"; clicks — a Point \[red\] drawn in timeline (location=(0.8, 0.0)); clicks\_2 — a Point \[red\] drawn in timeline (location=(3.2, 0.0)); clicks\_3 — a Point \[red\] drawn in timeline (location=(5.7, 0.0)); clicks\_4 — a Point \[red\] drawn in timeline (location=(8.8, 0.0)); one\_interval — a Brace \[yellow\] labelled "Delta t" drawn in timeline (x\_start=2.0, x\_end=3.0) Actions: - [01:4.37](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=64.37): assumptions (the "independently" part) is indicated — a transient flash. - [01:11.115](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=71.11500000000001): probability is shown on the screen, written out. ##### [01:15.546](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=75.5465) Narration: Lambda is a rate, measured in events per unit time. Multiplying it by the interval length gives a dimensionless probability. As delta t shrinks, this probability shrinks too, but the total expected count across all n intervals remains lambda T. Board: question — a Panel that says "During a time interval of length $T$, how many randomly arriving events should we expect to count?"; timeline — a NumberLine labelled "t" (x\_range=(0.0, 10.0), include\_numbers=True); assumptions — a Text \[text\] that says "A tiny interval has a small chance of one event, a negligible chance of two or more, and contributes independently of disjoint intervals."; delta — a Math \[text\] that says "$Delta t = frac(T, n)$"; probability — a Math \[text\] that says "$p\_n = lambda Delta t = frac(lambda T, n)$"; clicks — a Point \[red\] drawn in timeline (location=(0.8, 0.0)); clicks\_2 — a Point \[red\] drawn in timeline (location=(3.2, 0.0)); clicks\_3 — a Point \[red\] drawn in timeline (location=(5.7, 0.0)); clicks\_4 — a Point \[red\] drawn in timeline (location=(8.8, 0.0)); one\_interval — a Brace \[yellow\] labelled "Delta t" drawn in timeline (x\_start=2.0, x\_end=3.0) Actions: - [01:16.638](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=76.638): probability (the "lambda" part) is emphasized. - [01:21.201](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=81.201): probability (the "Delta t" part) is emphasized. - [01:21.201](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=81.201): probability (the "lambda" part) is no longer emphasized. - [01:32.462](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=92.46249999999999): probability (the "Delta t" part) is no longer emphasized. ##### [01:33.063](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=93.0625) Narration: Each tiny interval is therefore a Bernoulli trial: event or no event. Adding n independent trials gives a binomial count X sub n, with n trials and success probability lambda T over n. Continuous time has not appeared by magic. We have built an approximation that can be pushed to a limit. Board: Unchanged from the preceding beat in this scene. Actions: - [01:40.748](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=100.74800000000002): binomial is shown on the screen, written out. - [01:53.519](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=113.51900000000002): A box is drawn around binomial. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): assumptions is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): binomial is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): delta is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): probability is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): question is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): timeline is hidden from the screen — left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): clicks is hidden from the screen — timeline left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): clicks\_2 is hidden from the screen — timeline left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): clicks\_3 is hidden from the screen — timeline left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): clicks\_4 is hidden from the screen — timeline left the board. - [01:54.452](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=114.451625): one\_interval is hidden from the screen — timeline left the board. ### Scene 2: [The Poisson Limit](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=115.49329166666668) Span: 01:55.493–03:55.767 (115.49329166666668s–235.76700000000002s). #### Objects - goal: a Math \[text\] that says "$n p\_n = mu$" - heading\_limit: a Heading that says "Let the Number of Intervals Grow" - heading\_scale: a Heading that says "Rarer Trials, Same Expected Count" - mu: a Math \[text\] that says "$mu = lambda T$" - normalization: a Math \[text\] that says "$sum\_(k=0)^infinity e^(-mu) frac(mu^k, k!) = e^(-mu)e^mu = 1$" - p0: a Math \[text\] that says "$P(X\_n=k) = frac(n!, k! (n-k)!) (frac(mu, n))^k (1-frac(mu, n))^(n-k)$" - p1: a Math \[text\] that says "$P(X\_n=k) = frac(mu^k, k!) A\_n B\_n C\_n$" - p2: a Math \[text\] that says "$A\_n = frac(n(n-1) dots (n-k+1), n^k) arrow.r 1$" - p3: a Math \[text\] that says "$B\_n = (1-frac(mu, n))^n arrow.r e^(-mu)$" - p4: a Math \[text\] that says "$C\_n = (1-frac(mu, n))^(-k) arrow.r 1$" - poisson: a Math \[text\] that says "$P(N(T)=k) = e^(-lambda T) frac((lambda T)^k, k!), quad k=0,1,2,dots$" - table: a Table \[text\] that says "Intervals $n$ Chance $p\_n$ Expected count $n p\_n$ 10 0.2 2 100 0.02 2 1000 0.002 2" (rows=(('Intervals $n$', 'Chance $p\_n$', 'Expected count $n p\_n$'), (…, header=True) #### Beats ##### [01:55.493](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=115.49329166666668) Narration: Before taking a limit, hold the central idea still. Suppose the expected count is two. Ten intervals can each have probability zero point two. One hundred intervals use zero point zero two, and one thousand use zero point zero zero two. Board: Empty. Actions: - [01:55.493](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=115.49329166666668): heading\_scale is shown on the screen, written out. - [01:59.893](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=119.89329166666668): table is shown on the screen, written out. - [02:1.832](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=121.83229166666668): table is shown on the screen, written out. - [02:5.849](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=125.84929166666667): table is shown on the screen, written out. - [02:9.367](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=129.3672916666667): table is shown on the screen, written out. ##### [02:12.974](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=132.9742916666667) Narration: The individual trials become rarer while their number grows. In every row, n times p sub n remains two. More generally, call the fixed expected count mu, equal to lambda T. Board: heading\_scale — a Heading that says "Rarer Trials, Same Expected Count" Actions: - [02:12.974](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=132.9742916666667): mu is shown on the screen, written out. - [02:17.549](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=137.54929166666668): table (the "column=3" part) is emphasized. - [02:23.365](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=143.36529166666668): goal is shown on the screen, written out. - [02:27.162](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.1617916666667): goal is hidden from the screen — left the board. - [02:27.162](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.1617916666667): heading\_scale is hidden from the screen — left the board. - [02:27.162](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.1617916666667): mu is hidden from the screen — left the board. - [02:27.162](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.1617916666667): table is hidden from the screen — left the board. - [02:27.162](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.1617916666667): table (the "column=3" part) is no longer emphasized. ##### [02:27.762](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.76179166666668) Narration: Now fix a possible count k. The binomial probability of exactly k events is the number of ways to choose their intervals, times the probability of k occupied intervals, times the probability that all the others are empty. Board: Empty. Actions: - [02:27.762](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=147.76179166666668): heading\_limit is shown on the screen, written out. - [02:31.047](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=151.0472916666667): p0 is shown on the screen, written out. ##### [02:43.235](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=163.2347916666667) Narration: Substitute p sub n equal to mu over n and rearrange. The probability is mu to the k over k factorial, multiplied by three factors whose limits we can read separately. Board: p0 — a Math \[text\] that says "$P(X\_n=k) = frac(n!, k! (n-k)!) (frac(mu, n))^k (1-frac(mu, n))^(n-k)$"; heading\_limit — a Heading that says "Let the Number of Intervals Grow" Actions: - [02:47.042](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=167.04229166666667): p1 is shown on the screen, written out. ##### [02:55.967](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=175.96679166666667) Narration: The first factor contains k terms: n over n, then n minus one over n, and so on. For a fixed k, every one of those terms approaches one. Therefore A sub n approaches one. Board: p0 — a Math \[text\] that says "$P(X\_n=k) = frac(n!, k! (n-k)!) (frac(mu, n))^k (1-frac(mu, n))^(n-k)$"; p1 — a Math \[text\] that says "$P(X\_n=k) = frac(mu^k, k!) A\_n B\_n C\_n$"; heading\_limit — a Heading that says "Let the Number of Intervals Grow" Actions: - [02:56.535](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=176.53529166666667): p2 is shown on the screen, written out. - [03:7.89](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=187.89029166666666): p2 (the "A\_n" part) is emphasized. - [03:10.334](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=190.33379166666668): p2 moves to a new place on the board. - [03:10.334](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=190.33379166666668): p0 is hidden from the screen — left the board. - [03:10.334](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=190.33379166666668): p1 is hidden from the screen — left the board. - [03:10.334](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=190.33379166666668): p2 (the "A\_n" part) is no longer emphasized. ##### [03:10.934](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=190.93379166666668) Narration: The second factor is the classical exponential limit. One minus mu over n, raised to n, approaches e to the minus mu. This is the surviving probability of seeing no event across the complete interval. Board: p2 — a Math \[text\] that says "$A\_n = frac(n(n-1) dots (n-k+1), n^k) arrow.r 1$"; heading\_limit — a Heading that says "Let the Number of Intervals Grow" Actions: - [03:11.288](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=191.28829166666668): p3 is shown on the screen, written out. - [03:18.01](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=198.01029166666666): p3 (the "e^(-mu)" part) is emphasized. - [03:25.766](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=205.7657916666667): p3 (the "e^(-mu)" part) is no longer emphasized. ##### [03:26.366](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=206.36579166666667) Narration: The last factor raises something approaching one to the fixed power minus k, so it also approaches one. Multiplying the three limits leaves e to the minus mu times mu to the k over k factorial. Board: p2 — a Math \[text\] that says "$A\_n = frac(n(n-1) dots (n-k+1), n^k) arrow.r 1$"; heading\_limit — a Heading that says "Let the Number of Intervals Grow"; p3 — a Math \[text\] that says "$B\_n = (1-frac(mu, n))^n arrow.r e^(-mu)$" Actions: - [03:26.76](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=206.76029166666666): p4 is shown on the screen, written out. - [03:34.423](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=214.42329166666667): poisson is shown on the screen, written out. ##### [03:40.619](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=220.6192916666667) Narration: Finally replace mu by lambda T. This is the Poisson distribution, obtained rather than announced. Its probabilities really do sum to one, because the remaining series is the exponential series for e to the mu. Board: p2 — a Math \[text\] that says "$A\_n = frac(n(n-1) dots (n-k+1), n^k) arrow.r 1$"; heading\_limit — a Heading that says "Let the Number of Intervals Grow"; p3 — a Math \[text\] that says "$B\_n = (1-frac(mu, n))^n arrow.r e^(-mu)$"; p4 — a Math \[text\] that says "$C\_n = (1-frac(mu, n))^(-k) arrow.r 1$"; poisson — a Math \[text\] that says "$P(N(T)=k) = e^(-lambda T) frac((lambda T)^k, k!), quad k=0,1,2,dots$" Actions: - [03:43.986](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=223.98629166666666): A box is drawn around poisson. - [03:49.118](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=229.11829166666666): normalization is shown on the screen, written out. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): heading\_limit is hidden from the screen — left the board. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): normalization is hidden from the screen — left the board. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): p2 is hidden from the screen — left the board. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): p3 is hidden from the screen — left the board. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): p4 is hidden from the screen — left the board. - [03:54.725](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=234.72533333333337): poisson is hidden from the screen — left the board. ### Scene 3: [The Signature of Poisson Counts](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002) Span: 03:55.767–05:57.713 (235.76700000000002s–357.7125416666667s). #### Objects - bars: a Polygon \[blue\] drawn in plot (vertices=((-0.4, 0.0), (0.4, 0.0), (0.4, 0.0498), (-0.4, 0.0498)), fill\_opacity=0.5) - bars\_2: a Polygon \[blue\] drawn in plot (vertices=((0.6, 0.0), (1.4, 0.0), (1.4, 0.1494), (0.6, 0.1494)), fill\_opacity=0.5) - bars\_3: a Polygon \[blue\] drawn in plot (vertices=((1.6, 0.0), (2.4, 0.0), (2.4, 0.224), (1.6, 0.224)), fill\_opacity=0.5) - bars\_4: a Polygon \[blue\] drawn in plot (vertices=((2.6, 0.0), (3.4, 0.0), (3.4, 0.224), (2.6, 0.224)), fill\_opacity=0.5) - bars\_5: a Polygon \[blue\] drawn in plot (vertices=((3.6, 0.0), (4.4, 0.0), (4.4, 0.168), (3.6, 0.168)), fill\_opacity=0.5) - bars\_6: a Polygon \[blue\] drawn in plot (vertices=((4.6, 0.0), (5.4, 0.0), (5.4, 0.1008), (4.6, 0.1008)), fill\_opacity=0.5) - bars\_7: a Polygon \[blue\] drawn in plot (vertices=((5.6, 0.0), (6.4, 0.0), (6.4, 0.0504), (5.6, 0.0504)), fill\_opacity=0.5) - bars\_8: a Polygon \[blue\] drawn in plot (vertices=((6.6, 0.0), (7.4, 0.0), (7.4, 0.0216), (6.6, 0.0216)), fill\_opacity=0.5) - bars\_9: a Polygon \[blue\] drawn in plot (vertices=((7.6, 0.0), (8.4, 0.0), (8.4, 0.0081), (7.6, 0.0081)), fill\_opacity=0.5) - heading: a Heading that says "A Poisson Count With Mean Three" - moments: a Derivation \[text\] that says "$E\[N\] &= sum\_(k=0)^infinity k e^(-mu) frac(mu^k, k!) \\ &= mu sum\_(j=0)^infinity e^(-mu) frac(mu^j, j!) \\ &= mu \\ E\[N(N-1)\] &= mu^2 \\ op("Var")(N) &= E\[N(N-1)\] + E\[N\] - E\[N\]^2 \\ &= mu$" - plot: an Axes (x\_range=(-0.5, 8.5), y\_range=(0.0, 0.26), x\_ticks\_every=1.0) - prediction: a Text \[text\] that says "Across many equal time windows, the sample mean and sample variance should be close if a constant-rate Poisson model is appropriate." #### Beats ##### [03:55.767](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002) Narration: Here is a Poisson distribution with mu equal to three. The bars give the probabilities of zero events, one event, two events, and so on. The distribution is not symmetric, but it is concentrated around counts near three. Board: Empty. Actions: - [03:55.767](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002): heading is shown on the screen, written out. - [03:55.767](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=235.76700000000002): plot is shown on the screen, written out. - [04:0.086](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.086): bars is shown on the screen, written out. - [04:0.146](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.14600000000002): bars\_2 is shown on the screen, written out. - [04:0.206](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.20600000000002): bars\_3 is shown on the screen, written out. - [04:0.266](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.26600000000002): bars\_4 is shown on the screen, written out. - [04:0.326](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.32600000000002): bars\_5 is shown on the screen, written out. - [04:0.386](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.38600000000002): bars\_6 is shown on the screen, written out. - [04:0.446](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.44600000000003): bars\_7 is shown on the screen, written out. - [04:0.506](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.50600000000003): bars\_8 is shown on the screen, written out. - [04:0.566](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=240.56600000000003): bars\_9 is shown on the screen, written out. ##### [04:11.809](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=251.80850000000004) Narration: Its average can be calculated exactly. Start with the definition of expectation: sum k times the probability of k. The factor k cancels one factor from k factorial. Board: plot — an Axes (x\_range=(-0.5, 8.5), y\_range=(0.0, 0.26), x\_ticks\_every=1.0); heading — a Heading that says "A Poisson Count With Mean Three"; bars — a Polygon \[blue\] drawn in plot (vertices=((-0.4, 0.0), (0.4, 0.0), (0.4, 0.0498), (-0.4, 0.0498)), fill\_opacity=0.5); bars\_2 — a Polygon \[blue\] drawn in plot (vertices=((0.6, 0.0), (1.4, 0.0), (1.4, 0.1494), (0.6, 0.1494)), fill\_opacity=0.5); bars\_3 — a Polygon \[blue\] drawn in plot (vertices=((1.6, 0.0), (2.4, 0.0), (2.4, 0.224), (1.6, 0.224)), fill\_opacity=0.5); bars\_4 — a Polygon \[blue\] drawn in plot (vertices=((2.6, 0.0), (3.4, 0.0), (3.4, 0.224), (2.6, 0.224)), fill\_opacity=0.5); bars\_5 — a Polygon \[blue\] drawn in plot (vertices=((3.6, 0.0), (4.4, 0.0), (4.4, 0.168), (3.6, 0.168)), fill\_opacity=0.5); bars\_6 — a Polygon \[blue\] drawn in plot (vertices=((4.6, 0.0), (5.4, 0.0), (5.4, 0.1008), (4.6, 0.1008)), fill\_opacity=0.5); bars\_7 — a Polygon \[blue\] drawn in plot (vertices=((5.6, 0.0), (6.4, 0.0), (6.4, 0.0504), (5.6, 0.0504)), fill\_opacity=0.5); bars\_8 — a Polygon \[blue\] drawn in plot (vertices=((6.6, 0.0), (7.4, 0.0), (7.4, 0.0216), (6.6, 0.0216)), fill\_opacity=0.5); bars\_9 — a Polygon \[blue\] drawn in plot (vertices=((7.6, 0.0), (8.4, 0.0), (8.4, 0.0081), (7.6, 0.0081)), fill\_opacity=0.5) Actions: - [04:16.406](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=256.406): plot moves to a new place on the board. - [04:16.406](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=256.406): moments is shown on the screen, written out. - [04:21.108](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=261.108): moments (the "k" part) is emphasized. - [04:24.545](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=264.5445): moments (the "k" part) is no longer emphasized. ##### [04:25.144](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=265.1445) Narration: Pull one factor of mu outside, then re-index with j equal to k minus one. What remains is the sum of every Poisson probability, so that sum is one. The expected count is therefore mu. Board: Unchanged from the preceding beat in this scene. Actions: - [04:26.085](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=266.08500000000004): moments is shown on the screen, written out. - [04:38.321](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=278.321): moments is shown on the screen, written out. ##### [04:40.14](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=280.14000000000004) Narration: For variance, the useful quantity is the second factorial moment, N times N minus one. The same cancellation, now performed twice, gives mu squared. Board: Unchanged from the preceding beat in this scene. Actions: - [04:43.263](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=283.26300000000003): moments is shown on the screen, written out. - [04:50.589](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=290.589): moments (the "mu^2" part) is emphasized. - [04:51.484](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=291.48350000000005): moments (the "mu^2" part) is no longer emphasized. ##### [04:52.084](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=292.0835) Narration: Ordinary variance can be written as the second factorial moment plus the mean, minus the square of the mean. Substitute mu squared, mu, and mu squared. The two squares cancel, leaving mu. Board: Unchanged from the preceding beat in this scene. Actions: - [04:52.942](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=292.942): moments is shown on the screen, written out. - [05:4.262](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=304.26200000000006): moments is shown on the screen, written out. ##### [05:6.058](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=306.0575) Narration: So a Poisson count has mean mu and variance mu. The standard deviation is therefore the square root of mu. Relative fluctuations become smaller when the expected count becomes large, even though the absolute fluctuations grow. Board: Unchanged from the preceding beat in this scene. Actions: - [05:7.985](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=307.985): moments is emphasized. - [05:8.96](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=308.96000000000004): moments is emphasized. - [05:10.446](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=310.446): moments is no longer emphasized. - [05:21.081](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=321.08050000000003): moments is no longer emphasized. ##### [05:21.681](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=321.68050000000005) Narration: This equality has an intuitive origin in the small-interval model. Independent intervals each contribute a tiny yes-or-no uncertainty. Adding them gives variance n p times one minus p, which approaches n p as p becomes tiny. Board: Unchanged from the preceding beat in this scene. Actions: - [05:35.369](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=335.369): moments is indicated — a transient flash. ##### [05:38.663](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=338.6625) Narration: More importantly, equal mean and variance is a prediction that data can challenge. If repeated equal windows have variance far above their mean, arrivals may be clustered or the rate may be changing. If variance is far below the mean, some regular spacing or dead time may be present. Board: Unchanged from the preceding beat in this scene. Actions: - [05:41.53](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=341.53000000000003): prediction is shown on the screen, written out. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): heading is hidden from the screen — left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): moments is hidden from the screen — left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): plot is hidden from the screen — left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_2 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_3 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_4 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_5 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_6 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_7 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_8 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): bars\_9 is hidden from the screen — plot left the board. - [05:56.671](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=356.670875): prediction is hidden from the screen — left the board. ### Scene 4: [From Counts to Waiting](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=357.7125416666667) Span: 05:57.713–07:41.45 (357.7125416666667s–461.4499166666667s). #### Objects - bridge: a Math \[text\] that says "$W \> t quad \<=\> quad N(t)=0$" - cdf: a Math \[text\] that says "$P(W\<=t)=1-e^(-lambda t)$" - density: a Math \[text\] that says "$f\_W(t)=lambda e^(-lambda t)$" - later\_click: a Point \[red\] drawn in waiting\_line (location=(4.6, 0.0)) - mean\_wait: a Math \[text\] that says "$E\[W\]=integral\_0^infinity P(W\>t) dif t=frac(1, lambda)$" - next\_click: a Point \[red\] labelled "upright("next event")" drawn in waiting\_line (location=(2.4, 0.0)) - question: a Panel that says "Starting now, how long must we wait until the next event?" - start: a Point \[green\] labelled "upright("now")" drawn in waiting\_line - survival: a Math \[text\] that says "$P(W\>t)=P(N(t)=0)=e^(-lambda t)$" - survival\_curve: a FunctionPlot \[blue\] drawn in survival\_plot (function=\, x\_range=(0.0, 5.0)) - survival\_plot: an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0) - survival\_point: a PlotPoint \[yellow\] labelled "0.5" drawn in survival\_plot (target='survival\_curve', x=\) - threshold: a VariableNumber (initial\_value=0.5, format\_spec='.1f') - wait\_brace: a Brace \[yellow\] labelled "W" drawn in waiting\_line (x\_start=0.0, x\_end=2.4) - waiting\_line: a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True) #### Beats ##### [05:57.713](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=357.7125416666667) Narration: The count question has a natural companion. Instead of fixing a time interval and asking how many events occur, start the clock now and ask how long it runs before the next event. Call that random waiting time W. Board: Empty. Actions: - [05:57.713](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=357.7125416666667): question is shown on the screen, written out. - [06:10.507](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=370.5065416666667): question moves to a new place on the board. ##### [06:11.107](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=371.1065416666667) Narration: On one realized timeline, now is the green mark and the next arrival is the first red mark. The yellow span between them is the observed value of W. Another run would give a different span. Board: question — a Panel that says "Starting now, how long must we wait until the next event?" Actions: - [06:11.107](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=371.1065416666667): waiting\_line is shown on the screen, written out. - [06:13.893](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=373.8925416666667): start is shown on the screen, written out. - [06:14.915](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=374.9145416666667): next\_click is shown on the screen, written out. - [06:18.096](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=378.0955416666667): wait\_brace is shown on the screen, written out. - [06:21.335](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=381.3345416666667): later\_click is shown on the screen, written out. ##### [06:24.269](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=384.2685416666667) Narration: To find the distribution, ask when W exceeds a chosen time t. That happens exactly when the interval from now to t contains zero events. Waiting longer than t and counting zero by t are the same event. Board: question — a Panel that says "Starting now, how long must we wait until the next event?"; waiting\_line — a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True); start — a Point \[green\] labelled "upright("now")" drawn in waiting\_line; next\_click — a Point \[red\] labelled "upright("next event")" drawn in waiting\_line (location=(2.4, 0.0)); later\_click — a Point \[red\] drawn in waiting\_line (location=(4.6, 0.0)); wait\_brace — a Brace \[yellow\] labelled "W" drawn in waiting\_line (x\_start=0.0, x\_end=2.4) Actions: - [06:30.294](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=390.2935416666667): waiting\_line moves to a new place on the board. - [06:30.294](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=390.2935416666667): bridge is shown on the screen, written out. ##### [06:38.58](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=398.5795416666667) Narration: But the count by time t is Poisson with mean lambda t. Put k equal to zero in the Poisson formula. The factorial and power disappear, leaving e to the minus lambda t. Board: question — a Panel that says "Starting now, how long must we wait until the next event?"; waiting\_line — a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True); bridge — a Math \[text\] that says "$W \> t quad \<=\> quad N(t)=0$"; start — a Point \[green\] labelled "upright("now")" drawn in waiting\_line; next\_click — a Point \[red\] labelled "upright("next event")" drawn in waiting\_line (location=(2.4, 0.0)); later\_click — a Point \[red\] drawn in waiting\_line (location=(4.6, 0.0)); wait\_brace — a Brace \[yellow\] labelled "W" drawn in waiting\_line (x\_start=0.0, x\_end=2.4) Actions: - [06:44.165](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=404.1645416666667): survival is shown on the screen, written out. - [06:48.797](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=408.7965416666667): survival (the "e^(-lambda t)" part) is emphasized. - [06:51.107](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=411.1065416666667): survival (the "e^(-lambda t)" part) is no longer emphasized. ##### [06:51.707](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=411.7065416666667) Narration: That gives the survival probability, the chance the wait is still not over. Its complement gives the cumulative distribution. Differentiating gives the exponential density, lambda e to the minus lambda t. Board: question — a Panel that says "Starting now, how long must we wait until the next event?"; waiting\_line — a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True); bridge — a Math \[text\] that says "$W \> t quad \<=\> quad N(t)=0$"; survival — a Math \[text\] that says "$P(W\>t)=P(N(t)=0)=e^(-lambda t)$"; start — a Point \[green\] labelled "upright("now")" drawn in waiting\_line; next\_click — a Point \[red\] labelled "upright("next event")" drawn in waiting\_line (location=(2.4, 0.0)); later\_click — a Point \[red\] drawn in waiting\_line (location=(4.6, 0.0)); wait\_brace — a Brace \[yellow\] labelled "W" drawn in waiting\_line (x\_start=0.0, x\_end=2.4) Actions: - [06:57.025](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=417.0245416666667): cdf is shown on the screen, written out. - [07:2.121](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=422.1205416666667): density is shown on the screen, written out. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): cdf moves to a new place on the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): density moves to a new place on the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): survival moves to a new place on the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): bridge is hidden from the screen — left the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): waiting\_line is hidden from the screen — left the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): start is hidden from the screen — waiting\_line left the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): next\_click is hidden from the screen — waiting\_line left the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): later\_click is hidden from the screen — waiting\_line left the board. - [07:5.476](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=425.4760416666667): wait\_brace is hidden from the screen — waiting\_line left the board. ##### [07:6.676](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=426.6760416666667) Narration: For a unit rate, the survival curve begins at one. A very short threshold is likely to be crossed without an event, while a long event-free stretch becomes progressively less likely. Board: question — a Panel that says "Starting now, how long must we wait until the next event?"; survival — a Math \[text\] that says "$P(W\>t)=P(N(t)=0)=e^(-lambda t)$"; cdf — a Math \[text\] that says "$P(W\<=t)=1-e^(-lambda t)$"; density — a Math \[text\] that says "$f\_W(t)=lambda e^(-lambda t)$" Actions: - [07:6.676](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=426.6760416666667): survival\_plot is shown on the screen, written out. - [07:8.778](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=428.7775416666667): survival\_curve is shown on the screen, written out. - [07:11.379](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=431.3785416666667): survival\_point is shown on the screen, written out. ##### [07:18.712](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=438.7115416666667) Narration: Move the threshold to the right. The point rides down the curve because the probability of surviving without an event decays exponentially. Board: question — a Panel that says "Starting now, how long must we wait until the next event?"; survival — a Math \[text\] that says "$P(W\>t)=P(N(t)=0)=e^(-lambda t)$"; cdf — a Math \[text\] that says "$P(W\<=t)=1-e^(-lambda t)$"; density — a Math \[text\] that says "$f\_W(t)=lambda e^(-lambda t)$"; survival\_plot — an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0); survival\_curve — a FunctionPlot \[blue\] drawn in survival\_plot (function=\, x\_range=(0.0, 5.0)); survival\_point — a PlotPoint \[yellow\] labelled "0.5" drawn in survival\_plot (target='survival\_curve', x=\) Actions: - [07:20.082](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=440.0815416666667): survival\_point is redrawn as the numbers it depends on change. - [07:20.082](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=440.0815416666667): threshold ticks to 4.0. ##### [07:27.543](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=447.5430416666667) Narration: The mean waiting time is the area under the survival curve. Integrating e to the minus lambda t from zero to infinity gives one over lambda. Double the event rate and the average wait is cut in half. Board: Unchanged from the preceding beat in this scene. Actions: - [07:28.089](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=448.08854166666674): mean\_wait is shown on the screen, written out. - [07:39.479](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=459.47854166666673): A box is drawn around mean\_wait. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): cdf is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): density is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): mean\_wait is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): question is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): survival is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): survival\_plot is hidden from the screen — left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): survival\_curve is hidden from the screen — survival\_plot left the board. - [07:40.408](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=460.40825000000007): survival\_point is hidden from the screen — survival\_plot left the board. ### Scene 5: [Memory Without Age](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=461.4499166666667) Span: 07:41.45–09:58.205 (461.4499166666667s–598.20525s). #### Objects - heading: a Heading that says "The Remaining Wait Has the Same Law" - left\_axes: an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0) - left\_curve: a FunctionPlot \[blue\] drawn in left\_axes (function=\, x\_range=(0.0, 5.0)) - left\_label: a Tex \[text\] that says "Starting now" - line: a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True) - memory\_work: a Derivation \[text\] that says "$P(W\>s+t \| W\>s) &= frac(P(W\>s+t), P(W\>s)) \\ &= frac(e^(-lambda(s+t)), e^(-lambda s)) \\ &= e^(-lambda t) = P(W\>t)$" - question: a Panel that says "If no event has arrived after waiting $s$, should the remaining wait now be shorter?" - right\_axes: an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0) - right\_curve: a FunctionPlot \[blue\] drawn in right\_axes (function=\, x\_range=(0.0, 5.0)) - right\_label: a Tex \[text\] that says "After surviving $s$" - s\_brace: a Brace \[yellow\] labelled "s" drawn in line (x\_start=0.0, x\_end=2.0) - s\_mark: a Point \[yellow\] labelled "s" drawn in line (location=(2.0, 0.0)) - sum\_mark: a Point \[red\] labelled "s+t" drawn in line (location=(4.0, 0.0)) - t\_brace: a Brace \[green\] labelled "t" drawn in line (x\_start=2.0, x\_end=4.0) - zero: a Point \[green\] labelled "0" drawn in line #### Beats ##### [07:41.45](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=461.4499166666667) Narration: This is the claim people rightly distrust. Suppose we have already waited s units of time with no arrival. Surely, it feels, we must now be closer to the next event. The exponential model says no. Let us prove exactly what that sentence means. Board: Empty. Actions: - [07:41.45](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=461.4499166666667): question is shown on the screen, written out. - [07:57.901](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=477.9014166666667): question moves to a new place on the board. ##### [07:58.501](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=478.50141666666667) Narration: Begin at zero and suppose no event occurs before s. The yellow span is the waiting time we have already endured. Reaching s tells us that W is greater than s. Board: question — a Panel that says "If no event has arrived after waiting $s$, should the remaining wait now be shorter?" Actions: - [07:58.501](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=478.50141666666667): line is shown on the screen, written out. - [07:59.21](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=479.2099166666667): zero is shown on the screen, written out. - [08:1.799](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=481.7989166666667): s\_mark is shown on the screen, written out. - [08:3.18](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=483.1799166666667): s\_brace is shown on the screen, written out. ##### [08:10.793](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=490.7929166666667) Narration: Now ask for no event during an additional duration t. That means the total wait exceeds s plus t. The green span is the extra future time whose probability we want. Board: question — a Panel that says "If no event has arrived after waiting $s$, should the remaining wait now be shorter?"; line — a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True); zero — a Point \[green\] labelled "0" drawn in line; s\_mark — a Point \[yellow\] labelled "s" drawn in line (location=(2.0, 0.0)); s\_brace — a Brace \[yellow\] labelled "s" drawn in line (x\_start=0.0, x\_end=2.0) Actions: - [08:15.6](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=495.5999166666667): sum\_mark is shown on the screen, written out. - [08:18.572](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=498.5719166666667): t\_brace is shown on the screen, written out. ##### [08:22.864](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=502.8639166666667) Narration: Conditional probability divides the chance of surviving all the way to s plus t by the chance of having reached s. The later event is contained inside the earlier condition, so this ratio is exact. Board: question — a Panel that says "If no event has arrived after waiting $s$, should the remaining wait now be shorter?"; line — a NumberLine labelled "t" (x\_range=(0.0, 6.0), include\_numbers=True); zero — a Point \[green\] labelled "0" drawn in line; s\_mark — a Point \[yellow\] labelled "s" drawn in line (location=(2.0, 0.0)); s\_brace — a Brace \[yellow\] labelled "s" drawn in line (x\_start=0.0, x\_end=2.0); sum\_mark — a Point \[red\] labelled "s+t" drawn in line (location=(4.0, 0.0)); t\_brace — a Brace \[green\] labelled "t" drawn in line (x\_start=2.0, x\_end=4.0) Actions: - [08:23.166](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=503.1659166666667): line moves to a new place on the board. - [08:23.166](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=503.1659166666667): memory\_work is shown on the screen, written out. ##### [08:36.026](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=516.0259166666667) Narration: Insert the exponential survival probabilities. The numerator contains e to the minus lambda times s plus t. The denominator contains e to the minus lambda s. Board: Unchanged from the preceding beat in this scene. Actions: - [08:36.374](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=516.3739166666667): memory\_work is shown on the screen, written out. ##### [08:48.77](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=528.7699166666666) Narration: The factors involving s cancel. What remains is e to the minus lambda t, exactly the probability of waiting at least t from a fresh start. The elapsed wait has disappeared from the answer. Board: Unchanged from the preceding beat in this scene. Actions: - [08:50.5](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=530.4999166666666): memory\_work is shown on the screen, written out. - [09:0.867](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=540.8669166666667): A box is drawn around memory\_work. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): line is hidden from the screen — left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): zero is hidden from the screen — line left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): s\_mark is hidden from the screen — line left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): s\_brace is hidden from the screen — line left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): sum\_mark is hidden from the screen — line left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): t\_brace is hidden from the screen — line left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): memory\_work is hidden from the screen — left the board. - [09:2.736](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=542.7364166666666): question is hidden from the screen — left the board. ##### [09:3.936](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=543.9364166666667) Narration: These two curves make the result visible. On the left is survival for an additional time starting now. On the right is survival for the same additional time after we have already waited s. The curves are identical. Board: Empty. Actions: - [09:3.936](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=543.9364166666667): left\_axes is shown on the screen, written out. - [09:7.466](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=547.4659166666667): left\_axes moves to a new place on the board. - [09:7.466](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=547.4659166666667): left\_curve is shown on the screen, written out. - [09:7.466](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=547.4659166666667): left\_label is shown on the screen, written out. - [09:11.309](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=551.3089166666666): right\_axes is shown on the screen, written out. - [09:11.309](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=551.3089166666666): right\_curve is shown on the screen, written out. - [09:11.309](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=551.3089166666666): right\_label is shown on the screen, written out. ##### [09:19.223](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=559.2234166666667) Narration: Why does this feel absurd? We often imagine an arrival that is already on its way, a bus following a timetable, or a component accumulating age. Those mechanisms carry information from the past. A constant-rate Poisson process does not: future tiny intervals are independent of past ones. Board: left\_label — a Tex \[text\] that says "Starting now"; left\_axes — an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0); right\_label — a Tex \[text\] that says "After surviving $s$"; right\_axes — an Axes (x\_range=(0.0, 5.0), y\_range=(0.0, 1.05), x\_ticks\_every=1.0); left\_curve — a FunctionPlot \[blue\] drawn in left\_axes (function=\, x\_range=(0.0, 5.0)); right\_curve — a FunctionPlot \[blue\] drawn in right\_axes (function=\, x\_range=(0.0, 5.0)) Actions: - [09:37.37](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=577.3699166666667): right\_curve is indicated — a transient flash. ##### [09:39.897](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=579.8974166666667) Narration: Memorylessness is therefore not a universal law of waiting. It is a sharp consequence of the assumptions we chose. If the rate changes with time, arrivals repel one another, or a schedule is operating, the remaining wait can depend strongly on how long we have already waited. Board: Unchanged from the preceding beat in this scene. Actions: - [09:45.575](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=585.5749166666667): left\_curve is indicated — a transient flash. - [09:45.575](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=585.5749166666667): right\_curve is indicated — a transient flash. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): left\_axes is hidden from the screen — left the board. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): left\_curve is hidden from the screen — left\_axes left the board. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): left\_label is hidden from the screen — left the board. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): right\_axes is hidden from the screen — left the board. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): right\_curve is hidden from the screen — right\_axes left the board. - [09:57.164](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=597.1635833333333): right\_label is hidden from the screen — left the board. ### Scene 6: [Models Meet Data](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=598.20525) Span: 09:58.205–13:8.351 (598.20525s–788.3507916666666s). #### Objects - check\_heading: a Heading that says "What the Mean-Variance Comparison Can Reveal" - detector: a Table \[text\] that says "Window Clicks Window Clicks 1 0 7 4 2 1 8 1 3 2 9 0 4 1 10 2 5 3 11 3 6 2 12 5" (rows=(('Window', 'Clicks', 'Window', 'Clicks'), ('1', '0', '7', '4')…, header=True) - detector\_heading: a Heading that says "Twelve One-Minute Detector Windows" - final\_formula: a Math \[text\] that says "$upright("counts"): thin N(T) tilde upright("Poisson")(lambda T) quad arrow.r quad upright("waits"): thin W tilde upright("Exponential")(lambda)$" - heading: a Heading that says "One Rate, Two Questions" - high: a Text \[text\] that says "Variance much larger than the mean: look for clustering, changing rates, or dependence." - low: a Text \[text\] that says "Variance much smaller than the mean: look for regular spacing, inhibition, or detector dead time." - near: a Text \[text\] that says "Variance near the mean: a constant-rate Poisson model remains plausible." - sample\_mean: a Math \[text\] that says "$overline(x)=frac(24, 12)=2.00$" - sample\_variance: a Math \[text\] that says "$hat(v)=frac(1, 12) sum\_(i=1)^12 (x\_i-2)^2 =frac(26, 12) approx 2.17$" - service\_clicks: a Point \[red\] drawn in service\_line (location=(4.0, 0.0)) - service\_clicks\_2: a Point \[red\] drawn in service\_line (location=(9.0, 0.0)) - service\_clicks\_3: a Point \[red\] drawn in service\_line (location=(23.0, 0.0)) - service\_count: a Math \[text\] that says "$P(N=3)=e^(-3) frac(3^3, 3!) approx 0.224$" - service\_line: a NumberLine labelled "t thin (upright("minutes"))" (x\_range=(0.0, 30.0), include\_numbers=True, ticks\_every=10.0) - service\_mu: a Math \[text\] that says "$mu=lambda T=6(0.5)=3$" - service\_question: a Panel that says "Customers arrive at a desk at an average rate of 6 per hour. What is the chance of exactly 3 arrivals in 30 minutes, and what is the mean wait for the next customer?" - service\_wait: a Math \[text\] that says "$E\[W\]=frac(1, lambda)=frac(1, 6) upright("hour") =10 upright(" minutes")$" - text: a Text \[text\] that says "Counts in a fixed interval and waits to the next event are two descriptions of the same constant-rate independent-arrival model." - verdict: a Math \[text\] that says "$overline(x) approx hat(v)$" #### Beats ##### [09:58.205](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=598.20525) Narration: Put the model to work at a service desk. Customers arrive at an average rate of six per hour. We want the probability of exactly three arrivals in half an hour, and the average time from now until the next arrival. Board: Empty. Actions: - [09:58.205](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=598.20525): service\_question is shown on the screen, written out. - [09:59.958](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=599.95825): service\_line is shown on the screen, written out. - [10:6.402](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=606.40225): service\_clicks is shown on the screen, written out. - [10:6.602](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=606.60225): service\_clicks\_2 is shown on the screen, written out. - [10:6.802](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=606.80225): service\_clicks\_3 is shown on the screen, written out. ##### [10:11.646](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=611.64575) Narration: For the count, match the units first. Thirty minutes is one half hour, so the expected count is lambda T, equal to six times one half, which is three. Board: service\_line — a NumberLine labelled "t thin (upright("minutes"))" (x\_range=(0.0, 30.0), include\_numbers=True, ticks\_every=10.0); service\_question — a Panel that says "Customers arrive at a desk at an average rate of 6 per hour. What is the chance of exactly 3 arrivals in 30 minutes, and what is the mean wait for the next customer?"; service\_clicks — a Point \[red\] drawn in service\_line (location=(4.0, 0.0)); service\_clicks\_2 — a Point \[red\] drawn in service\_line (location=(9.0, 0.0)); service\_clicks\_3 — a Point \[red\] drawn in service\_line (location=(23.0, 0.0)) Actions: - [10:17.114](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=617.11425): service\_line moves to a new place on the board. - [10:17.114](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=617.11425): service\_mu is shown on the screen, written out. - [10:21.351](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=621.3512499999999): service\_mu (the "3" part) is emphasized. ##### [10:22.695](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=622.69475) Narration: Use the Poisson probability with k equal to three and mu equal to three. The result is e to the minus three times three cubed over three factorial, about zero point two two four. Board: service\_line — a NumberLine labelled "t thin (upright("minutes"))" (x\_range=(0.0, 30.0), include\_numbers=True, ticks\_every=10.0); service\_mu — a Math \[text\] that says "$mu=lambda T=6(0.5)=3$"; service\_question — a Panel that says "Customers arrive at a desk at an average rate of 6 per hour. What is the chance of exactly 3 arrivals in 30 minutes, and what is the mean wait for the next customer?"; service\_clicks — a Point \[red\] drawn in service\_line (location=(4.0, 0.0)); service\_clicks\_2 — a Point \[red\] drawn in service\_line (location=(9.0, 0.0)); service\_clicks\_3 — a Point \[red\] drawn in service\_line (location=(23.0, 0.0)) Actions: - [10:22.695](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=622.69475): service\_mu (the "3" part) is no longer emphasized. - [10:23.461](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=623.46125): service\_count is shown on the screen, written out. - [10:33.701](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=633.70125): service\_count (the "0.224" part) is indicated — a transient flash. ##### [10:35.068](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=635.0677499999999) Narration: For waiting, use the exponential mean. One over six hour is one sixth of sixty minutes, so the mean wait is ten minutes. The count calculation and the waiting calculation use the same rate lambda from two different perspectives. Board: service\_line — a NumberLine labelled "t thin (upright("minutes"))" (x\_range=(0.0, 30.0), include\_numbers=True, ticks\_every=10.0); service\_mu — a Math \[text\] that says "$mu=lambda T=6(0.5)=3$"; service\_count — a Math \[text\] that says "$P(N=3)=e^(-3) frac(3^3, 3!) approx 0.224$"; service\_question — a Panel that says "Customers arrive at a desk at an average rate of 6 per hour. What is the chance of exactly 3 arrivals in 30 minutes, and what is the mean wait for the next customer?"; service\_clicks — a Point \[red\] drawn in service\_line (location=(4.0, 0.0)); service\_clicks\_2 — a Point \[red\] drawn in service\_line (location=(9.0, 0.0)); service\_clicks\_3 — a Point \[red\] drawn in service\_line (location=(23.0, 0.0)) Actions: - [10:35.636](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=635.63625): service\_wait is shown on the screen, written out. - [10:42.823](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=642.8232499999999): A box is drawn around service\_wait. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_count is hidden from the screen — left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_line is hidden from the screen — left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_clicks is hidden from the screen — service\_line left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_clicks\_2 is hidden from the screen — service\_line left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_clicks\_3 is hidden from the screen — service\_line left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_mu is hidden from the screen — left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_question is hidden from the screen — left the board. - [10:49.975](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=649.97475): service\_wait is hidden from the screen — left the board. ##### [10:51.175](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=651.17475) Narration: Now suppose a detector is observed in twelve equal one-minute windows. These are the recorded click counts. Equal windows matter because the Poisson prediction compares counts that share the same expected value. Board: Empty. Actions: - [10:51.175](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=651.17475): detector\_heading is shown on the screen, written out. - [10:54.646](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=654.64625): detector is shown on the screen, written out. - [10:56.016](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.01625): detector is shown on the screen, written out. - [10:56.116](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.11625): detector is shown on the screen, written out. - [10:56.216](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.21625): detector is shown on the screen, written out. - [10:56.316](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.31625): detector is shown on the screen, written out. - [10:56.416](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.41625): detector is shown on the screen, written out. - [10:56.516](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=656.51625): detector is shown on the screen, written out. ##### [11:4.244](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=664.24375) Narration: The twelve counts add to twenty-four, so their empirical mean is exactly two clicks per window. This estimates lambda times one minute. Board: detector\_heading — a Heading that says "Twelve One-Minute Detector Windows" Actions: - [11:5.149](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=665.1492499999999): detector (the "column=2" part) is emphasized. - [11:5.149](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=665.1492499999999): detector (the "column=4" part) is emphasized. - [11:7.808](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=667.80825): sample\_mean is shown on the screen, written out. - [11:13.219](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=673.21875): detector (the "column=2" part) is no longer emphasized. - [11:13.219](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=673.21875): detector (the "column=4" part) is no longer emphasized. ##### [11:13.819](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=673.81875) Narration: Next measure their spread around two. Using the average squared deviation, the empirical variance is twenty-six over twelve, about two point one seven. It is not exactly two, because a finite random sample does not reproduce a population identity perfectly. Board: sample\_mean — a Math \[text\] that says "$overline(x)=frac(24, 12)=2.00$"; detector\_heading — a Heading that says "Twelve One-Minute Detector Windows" Actions: - [11:14.851](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=674.8512499999999): sample\_variance is shown on the screen, written out. - [11:22.769](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=682.7692499999999): sample\_variance (the "2.17" part) is emphasized. - [11:30.06](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=690.06025): sample\_variance (the "2.17" part) is no longer emphasized. ##### [11:30.66](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=690.66025) Narration: The mean two point zero zero and variance two point one seven are close. That does not prove the Poisson model, but it passes this first check. With many more windows, we could simulate or calculate the variation expected in these estimates and judge whether the discrepancy is unusual. Board: sample\_mean — a Math \[text\] that says "$overline(x)=frac(24, 12)=2.00$"; sample\_variance — a Math \[text\] that says "$hat(v)=frac(1, 12) sum\_(i=1)^12 (x\_i-2)^2 =frac(26, 12) approx 2.17$"; detector\_heading — a Heading that says "Twelve One-Minute Detector Windows" Actions: - [11:35.072](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=695.0722499999999): verdict is shown on the screen, written out. - [11:39.658](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=699.65825): A box is drawn around verdict. - [11:48.401](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=708.40075): detector is hidden from the screen — left the board. - [11:48.401](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=708.40075): detector\_heading is hidden from the screen — left the board. - [11:48.401](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=708.40075): sample\_mean is hidden from the screen — left the board. - [11:48.401](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=708.40075): sample\_variance is hidden from the screen — left the board. - [11:48.401](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=708.40075): verdict is hidden from the screen — left the board. ##### [11:49.601](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=709.60075) Narration: The comparison becomes diagnostic when it is repeated over substantial data. Variance near the mean is compatible with independent arrivals at a stable rate. Board: Empty. Actions: - [11:49.601](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=709.60075): check\_heading is shown on the screen, written out. - [11:55.266](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=715.26625): near is shown on the screen, written out. ##### [11:59.778](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=719.77825) Narration: Variance much larger than the mean suggests extra variability: customers arriving in groups, bursts of detector noise, or a rate that changes through the day. Board: near — a Text \[text\] that says "Variance near the mean: a constant-rate Poisson model remains plausible."; check\_heading — a Heading that says "What the Mean-Variance Comparison Can Reveal" Actions: - [12:0.939](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=720.9392499999999): high is shown on the screen, written out. ##### [12:10.769](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=730.7692499999999) Narration: Variance much smaller than the mean suggests excessive regularity: arrivals may inhibit one another, or a detector may briefly stop registering after each click. Either pattern points back to an assumption that should be reconsidered. Board: near — a Text \[text\] that says "Variance near the mean: a constant-rate Poisson model remains plausible."; high — a Text \[text\] that says "Variance much larger than the mean: look for clustering, changing rates, or dependence."; check\_heading — a Heading that says "What the Mean-Variance Comparison Can Reveal" Actions: - [12:11.895](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=731.8952499999999): low is shown on the screen, written out. - [12:25.793](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=745.79275): check\_heading is hidden from the screen — left the board. - [12:25.793](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=745.79275): high is hidden from the screen — left the board. - [12:25.793](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=745.79275): low is hidden from the screen — left the board. - [12:25.793](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=745.79275): near is hidden from the screen — left the board. ##### [12:26.393](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=746.39275) Narration: So the complete connection is this. Chopping time into rare independent trials produces Poisson counts in the limit. Asking for zero counts up to time t produces exponential waiting. Independence gives memorylessness, and the equal mean and variance of the counts gives the model an observable signature. Board: Empty. Actions: - [12:27.507](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=747.50725): final\_formula is shown on the screen, written out. ##### [12:48.274](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=768.2742499999999) Narration: The formulas are useful because their assumptions are clear. Use the same rate lambda to move between counts and waits, then return to data and ask whether independence, a stable rate, and equal mean and variance are actually visible. A probability model earns its place by surviving that return. Board: final\_formula — a Math \[text\] that says "$upright("counts"): thin N(T) tilde upright("Poisson")(lambda T) quad arrow.r quad upright("waits"): thin W tilde upright("Exponential")(lambda)$" Actions: - [13:5.643](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=785.64325): A box is drawn around final\_formula. - [13:7.309](https://academa.ai/lectures/poisson-processes-and-waiting-times?t=787.309125): final\_formula is hidden from the screen — left the board.