{"version":1,"lectureId":"01M14V07J37P7MMQN9J7QVK3JS","attempt":0,"publication":{"slug":"poisson-processes-and-waiting-times","title":"Rare Events in Continuous Time: From Poisson Arrivals to Exponential Waiting Times","subject":"statistics","summary":"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.","metaDescription":"Derive Poisson counts and exponential waiting times from tiny independent intervals, then test their predictions with arrival data.","transcript":"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. 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. 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. 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. 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. 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.","watch":{"version":1,"scenes":[{"title":"Random Marks on Time","start":0,"end":115.49329166666668,"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":[{"start":0,"say":"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.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[16.324,"card is hidden from the screen — left the board."]]},{"start":17.524,"say":"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.","live":null,"does":[[17.524,"question is shown on the screen, written out."],[29.3195,"question moves to a new place on the board."]]},{"start":29.9195,"say":"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.","live":["question"],"does":[[29.9195,"timeline is shown on the screen, written out."],[32.59,"clicks is shown on the screen, written out."],[32.77,"clicks_2 is shown on the screen, written out."],[32.95,"clicks_3 is shown on the screen, written out."],[33.13,"clicks_4 is shown on the screen, written out."]]},{"start":43.1515,"say":"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.","live":["question","timeline","clicks","clicks_2","clicks_3","clicks_4"],"does":[[48.039,"one_interval is shown on the screen, written out."],[48.794000000000004,"delta is shown on the screen, written out."],[57.675000000000004,"assumptions is shown on the screen, written out."]]},{"start":58.971999999999994,"say":"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.","live":["question","timeline","assumptions","delta","clicks","clicks_2","clicks_3","clicks_4","one_interval"],"does":[[64.37,"assumptions (the \"independently\" part) is indicated — a transient flash."],[71.11500000000001,"probability is shown on the screen, written out."]]},{"start":75.5465,"say":"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.","live":["question","timeline","assumptions","delta","probability","clicks","clicks_2","clicks_3","clicks_4","one_interval"],"does":[[76.638,"probability (the \"lambda\" part) is emphasized."],[81.201,"probability (the \"Delta t\" part) is emphasized."],[81.201,"probability (the \"lambda\" part) is no longer emphasized."],[92.46249999999999,"probability (the \"Delta t\" part) is no longer emphasized."]]},{"start":93.0625,"say":"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.","live":null,"does":[[100.74800000000002,"binomial is shown on the screen, written out."],[113.51900000000002,"A box is drawn around binomial."],[114.451625,"assumptions is hidden from the screen — left the board."],[114.451625,"binomial is hidden from the screen — left the board."],[114.451625,"delta is hidden from the screen — left the board."],[114.451625,"probability is hidden from the screen — left the board."],[114.451625,"question is hidden from the screen — left the board."],[114.451625,"timeline is hidden from the screen — left the board."],[114.451625,"clicks is hidden from the screen — timeline left the board."],[114.451625,"clicks_2 is hidden from the screen — timeline left the board."],[114.451625,"clicks_3 is hidden from the screen — timeline left the board."],[114.451625,"clicks_4 is hidden from the screen — timeline left the board."],[114.451625,"one_interval is hidden from the screen — timeline left the board."]]}]},{"title":"The Poisson Limit","start":115.49329166666668,"end":235.76700000000002,"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":[{"start":115.49329166666668,"say":"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.","live":[],"does":[[115.49329166666668,"heading_scale is shown on the screen, written out."],[119.89329166666668,"table is shown on the screen, written out."],[121.83229166666668,"table is shown on the screen, written out."],[125.84929166666667,"table is shown on the screen, written out."],[129.3672916666667,"table is shown on the screen, written out."]]},{"start":132.9742916666667,"say":"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.","live":["heading_scale"],"does":[[132.9742916666667,"mu is shown on the screen, written out."],[137.54929166666668,"table (the \"column=3\" part) is emphasized."],[143.36529166666668,"goal is shown on the screen, written out."],[147.1617916666667,"goal is hidden from the screen — left the board."],[147.1617916666667,"heading_scale is hidden from the screen — left the board."],[147.1617916666667,"mu is hidden from the screen — left the board."],[147.1617916666667,"table is hidden from the screen — left the board."],[147.1617916666667,"table (the \"column=3\" part) is no longer emphasized."]]},{"start":147.76179166666668,"say":"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.","live":[],"does":[[147.76179166666668,"heading_limit is shown on the screen, written out."],[151.0472916666667,"p0 is shown on the screen, written out."]]},{"start":163.2347916666667,"say":"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.","live":["p0","heading_limit"],"does":[[167.04229166666667,"p1 is shown on the screen, written out."]]},{"start":175.96679166666667,"say":"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.","live":["p0","p1","heading_limit"],"does":[[176.53529166666667,"p2 is shown on the screen, written out."],[187.89029166666666,"p2 (the \"A_n\" part) is emphasized."],[190.33379166666668,"p2 moves to a new place on the board."],[190.33379166666668,"p0 is hidden from the screen — left the board."],[190.33379166666668,"p1 is hidden from the screen — left the board."],[190.33379166666668,"p2 (the \"A_n\" part) is no longer emphasized."]]},{"start":190.93379166666668,"say":"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.","live":["p2","heading_limit"],"does":[[191.28829166666668,"p3 is shown on the screen, written out."],[198.01029166666666,"p3 (the \"e^(-mu)\" part) is emphasized."],[205.7657916666667,"p3 (the \"e^(-mu)\" part) is no longer emphasized."]]},{"start":206.36579166666667,"say":"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.","live":["p2","heading_limit","p3"],"does":[[206.76029166666666,"p4 is shown on the screen, written out."],[214.42329166666667,"poisson is shown on the screen, written out."]]},{"start":220.6192916666667,"say":"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.","live":["p2","heading_limit","p3","p4","poisson"],"does":[[223.98629166666666,"A box is drawn around poisson."],[229.11829166666666,"normalization is shown on the screen, written out."],[234.72533333333337,"heading_limit is hidden from the screen — left the board."],[234.72533333333337,"normalization is hidden from the screen — left the board."],[234.72533333333337,"p2 is hidden from the screen — left the board."],[234.72533333333337,"p3 is hidden from the screen — left the board."],[234.72533333333337,"p4 is hidden from the screen — left the board."],[234.72533333333337,"poisson is hidden from the screen — left the board."]]}]},{"title":"The Signature of Poisson Counts","start":235.76700000000002,"end":357.7125416666667,"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":[{"start":235.76700000000002,"say":"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.","live":[],"does":[[235.76700000000002,"heading is shown on the screen, written out."],[235.76700000000002,"plot is shown on the screen, written out."],[240.086,"bars is shown on the screen, written out."],[240.14600000000002,"bars_2 is shown on the screen, written out."],[240.20600000000002,"bars_3 is shown on the screen, written out."],[240.26600000000002,"bars_4 is shown on the screen, written out."],[240.32600000000002,"bars_5 is shown on the screen, written out."],[240.38600000000002,"bars_6 is shown on the screen, written out."],[240.44600000000003,"bars_7 is shown on the screen, written out."],[240.50600000000003,"bars_8 is shown on the screen, written out."],[240.56600000000003,"bars_9 is shown on the screen, written out."]]},{"start":251.80850000000004,"say":"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.","live":["plot","heading","bars","bars_2","bars_3","bars_4","bars_5","bars_6","bars_7","bars_8","bars_9"],"does":[[256.406,"plot moves to a new place on the board."],[256.406,"moments is shown on the screen, written out."],[261.108,"moments (the \"k\" part) is emphasized."],[264.5445,"moments (the \"k\" part) is no longer emphasized."]]},{"start":265.1445,"say":"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.","live":null,"does":[[266.08500000000004,"moments is shown on the screen, written out."],[278.321,"moments is shown on the screen, written out."]]},{"start":280.14000000000004,"say":"For variance, the useful quantity is the second factorial moment, N times N minus one. The same cancellation, now performed twice, gives mu squared.","live":null,"does":[[283.26300000000003,"moments is shown on the screen, written out."],[290.589,"moments (the \"mu^2\" part) is emphasized."],[291.48350000000005,"moments (the \"mu^2\" part) is no longer emphasized."]]},{"start":292.0835,"say":"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.","live":null,"does":[[292.942,"moments is shown on the screen, written out."],[304.26200000000006,"moments is shown on the screen, written out."]]},{"start":306.0575,"say":"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.","live":null,"does":[[307.985,"moments is emphasized."],[308.96000000000004,"moments is emphasized."],[310.446,"moments is no longer emphasized."],[321.08050000000003,"moments is no longer emphasized."]]},{"start":321.68050000000005,"say":"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.","live":null,"does":[[335.369,"moments is indicated — a transient flash."]]},{"start":338.6625,"say":"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.","live":null,"does":[[341.53000000000003,"prediction is shown on the screen, written out."],[356.670875,"heading is hidden from the screen — left the board."],[356.670875,"moments is hidden from the screen — left the board."],[356.670875,"plot is hidden from the screen — left the board."],[356.670875,"bars is hidden from the screen — plot left the board."],[356.670875,"bars_2 is hidden from the screen — plot left the board."],[356.670875,"bars_3 is hidden from the screen — plot left the board."],[356.670875,"bars_4 is hidden from the screen — plot left the board."],[356.670875,"bars_5 is hidden from the screen — plot left the board."],[356.670875,"bars_6 is hidden from the screen — plot left the board."],[356.670875,"bars_7 is hidden from the screen — plot left the board."],[356.670875,"bars_8 is hidden from the screen — plot left the board."],[356.670875,"bars_9 is hidden from the screen — plot left the board."],[356.670875,"prediction is hidden from the screen — left the board."]]}]},{"title":"From Counts to Waiting","start":357.7125416666667,"end":461.4499166666667,"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=<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=<VariableNumber threshold = 4.0>)","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":[{"start":357.7125416666667,"say":"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.","live":[],"does":[[357.7125416666667,"question is shown on the screen, written out."],[370.5065416666667,"question moves to a new place on the board."]]},{"start":371.1065416666667,"say":"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.","live":["question"],"does":[[371.1065416666667,"waiting_line is shown on the screen, written out."],[373.8925416666667,"start is shown on the screen, written out."],[374.9145416666667,"next_click is shown on the screen, written out."],[378.0955416666667,"wait_brace is shown on the screen, written out."],[381.3345416666667,"later_click is shown on the screen, written out."]]},{"start":384.2685416666667,"say":"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.","live":["question","waiting_line","start","next_click","later_click","wait_brace"],"does":[[390.2935416666667,"waiting_line moves to a new place on the board."],[390.2935416666667,"bridge is shown on the screen, written out."]]},{"start":398.5795416666667,"say":"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.","live":["question","waiting_line","bridge","start","next_click","later_click","wait_brace"],"does":[[404.1645416666667,"survival is shown on the screen, written out."],[408.7965416666667,"survival (the \"e^(-lambda t)\" part) is emphasized."],[411.1065416666667,"survival (the \"e^(-lambda t)\" part) is no longer emphasized."]]},{"start":411.7065416666667,"say":"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.","live":["question","waiting_line","bridge","survival","start","next_click","later_click","wait_brace"],"does":[[417.0245416666667,"cdf is shown on the screen, written out."],[422.1205416666667,"density is shown on the screen, written out."],[425.4760416666667,"cdf moves to a new place on the board."],[425.4760416666667,"density moves to a new place on the board."],[425.4760416666667,"survival moves to a new place on the board."],[425.4760416666667,"bridge is hidden from the screen — left the board."],[425.4760416666667,"waiting_line is hidden from the screen — left the board."],[425.4760416666667,"start is hidden from the screen — waiting_line left the board."],[425.4760416666667,"next_click is hidden from the screen — waiting_line left the board."],[425.4760416666667,"later_click is hidden from the screen — waiting_line left the board."],[425.4760416666667,"wait_brace is hidden from the screen — waiting_line left the board."]]},{"start":426.6760416666667,"say":"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.","live":["question","survival","cdf","density"],"does":[[426.6760416666667,"survival_plot is shown on the screen, written out."],[428.7775416666667,"survival_curve is shown on the screen, written out."],[431.3785416666667,"survival_point is shown on the screen, written out."]]},{"start":438.7115416666667,"say":"Move the threshold to the right. The point rides down the curve because the probability of surviving without an event decays exponentially.","live":["question","survival","cdf","density","survival_plot","survival_curve","survival_point"],"does":[[440.0815416666667,"survival_point is redrawn as the numbers it depends on change."],[440.0815416666667,"threshold ticks to 4.0."]]},{"start":447.5430416666667,"say":"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.","live":null,"does":[[448.08854166666674,"mean_wait is shown on the screen, written out."],[459.47854166666673,"A box is drawn around mean_wait."],[460.40825000000007,"cdf is hidden from the screen — left the board."],[460.40825000000007,"density is hidden from the screen — left the board."],[460.40825000000007,"mean_wait is hidden from the screen — left the board."],[460.40825000000007,"question is hidden from the screen — left the board."],[460.40825000000007,"survival is hidden from the screen — left the board."],[460.40825000000007,"survival_plot is hidden from the screen — left the board."],[460.40825000000007,"survival_curve is hidden from the screen — survival_plot left the board."],[460.40825000000007,"survival_point is hidden from the screen — survival_plot left the board."]]}]},{"title":"Memory Without Age","start":461.4499166666667,"end":598.20525,"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=<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=<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":[{"start":461.4499166666667,"say":"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.","live":[],"does":[[461.4499166666667,"question is shown on the screen, written out."],[477.9014166666667,"question moves to a new place on the board."]]},{"start":478.50141666666667,"say":"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.","live":["question"],"does":[[478.50141666666667,"line is shown on the screen, written out."],[479.2099166666667,"zero is shown on the screen, written out."],[481.7989166666667,"s_mark is shown on the screen, written out."],[483.1799166666667,"s_brace is shown on the screen, written out."]]},{"start":490.7929166666667,"say":"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.","live":["question","line","zero","s_mark","s_brace"],"does":[[495.5999166666667,"sum_mark is shown on the screen, written out."],[498.5719166666667,"t_brace is shown on the screen, written out."]]},{"start":502.8639166666667,"say":"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.","live":["question","line","zero","s_mark","s_brace","sum_mark","t_brace"],"does":[[503.1659166666667,"line moves to a new place on the board."],[503.1659166666667,"memory_work is shown on the screen, written out."]]},{"start":516.0259166666667,"say":"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.","live":null,"does":[[516.3739166666667,"memory_work is shown on the screen, written out."]]},{"start":528.7699166666666,"say":"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.","live":null,"does":[[530.4999166666666,"memory_work is shown on the screen, written out."],[540.8669166666667,"A box is drawn around memory_work."],[542.7364166666666,"line is hidden from the screen — left the board."],[542.7364166666666,"zero is hidden from the screen — line left the board."],[542.7364166666666,"s_mark is hidden from the screen — line left the board."],[542.7364166666666,"s_brace is hidden from the screen — line left the board."],[542.7364166666666,"sum_mark is hidden from the screen — line left the board."],[542.7364166666666,"t_brace is hidden from the screen — line left the board."],[542.7364166666666,"memory_work is hidden from the screen — left the board."],[542.7364166666666,"question is hidden from the screen — left the board."]]},{"start":543.9364166666667,"say":"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.","live":[],"does":[[543.9364166666667,"left_axes is shown on the screen, written out."],[547.4659166666667,"left_axes moves to a new place on the board."],[547.4659166666667,"left_curve is shown on the screen, written out."],[547.4659166666667,"left_label is shown on the screen, written out."],[551.3089166666666,"right_axes is shown on the screen, written out."],[551.3089166666666,"right_curve is shown on the screen, written out."],[551.3089166666666,"right_label is shown on the screen, written out."]]},{"start":559.2234166666667,"say":"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.","live":["left_label","left_axes","right_label","right_axes","left_curve","right_curve"],"does":[[577.3699166666667,"right_curve is indicated — a transient flash."]]},{"start":579.8974166666667,"say":"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.","live":null,"does":[[585.5749166666667,"left_curve is indicated — a transient flash."],[585.5749166666667,"right_curve is indicated — a transient flash."],[597.1635833333333,"left_axes is hidden from the screen — left the board."],[597.1635833333333,"left_curve is hidden from the screen — left_axes left the board."],[597.1635833333333,"left_label is hidden from the screen — left the board."],[597.1635833333333,"right_axes is hidden from the screen — left the board."],[597.1635833333333,"right_curve is hidden from the screen — right_axes left the board."],[597.1635833333333,"right_label is hidden from the screen — left the board."]]}]},{"title":"Models Meet Data","start":598.20525,"end":788.3507916666666,"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":[{"start":598.20525,"say":"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.","live":[],"does":[[598.20525,"service_question is shown on the screen, written out."],[599.95825,"service_line is shown on the screen, written out."],[606.40225,"service_clicks is shown on the screen, written out."],[606.60225,"service_clicks_2 is shown on the screen, written out."],[606.80225,"service_clicks_3 is shown on the screen, written out."]]},{"start":611.64575,"say":"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.","live":["service_line","service_question","service_clicks","service_clicks_2","service_clicks_3"],"does":[[617.11425,"service_line moves to a new place on the board."],[617.11425,"service_mu is shown on the screen, written out."],[621.3512499999999,"service_mu (the \"3\" part) is emphasized."]]},{"start":622.69475,"say":"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.","live":["service_line","service_mu","service_question","service_clicks","service_clicks_2","service_clicks_3"],"does":[[622.69475,"service_mu (the \"3\" part) is no longer emphasized."],[623.46125,"service_count is shown on the screen, written out."],[633.70125,"service_count (the \"0.224\" part) is indicated — a transient flash."]]},{"start":635.0677499999999,"say":"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.","live":["service_line","service_mu","service_count","service_question","service_clicks","service_clicks_2","service_clicks_3"],"does":[[635.63625,"service_wait is shown on the screen, written out."],[642.8232499999999,"A box is drawn around service_wait."],[649.97475,"service_count is hidden from the screen — left the board."],[649.97475,"service_line is hidden from the screen — left the board."],[649.97475,"service_clicks is hidden from the screen — service_line left the board."],[649.97475,"service_clicks_2 is hidden from the screen — service_line left the board."],[649.97475,"service_clicks_3 is hidden from the screen — service_line left the board."],[649.97475,"service_mu is hidden from the screen — left the board."],[649.97475,"service_question is hidden from the screen — left the board."],[649.97475,"service_wait is hidden from the screen — left the board."]]},{"start":651.17475,"say":"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.","live":[],"does":[[651.17475,"detector_heading is shown on the screen, written out."],[654.64625,"detector is shown on the screen, written out."],[656.01625,"detector is shown on the screen, written out."],[656.11625,"detector is shown on the screen, written out."],[656.21625,"detector is shown on the screen, written out."],[656.31625,"detector is shown on the screen, written out."],[656.41625,"detector is shown on the screen, written out."],[656.51625,"detector is shown on the screen, written out."]]},{"start":664.24375,"say":"The twelve counts add to twenty-four, so their empirical mean is exactly two clicks per window. This estimates lambda times one minute.","live":["detector_heading"],"does":[[665.1492499999999,"detector (the \"column=2\" part) is emphasized."],[665.1492499999999,"detector (the \"column=4\" part) is emphasized."],[667.80825,"sample_mean is shown on the screen, written out."],[673.21875,"detector (the \"column=2\" part) is no longer emphasized."],[673.21875,"detector (the \"column=4\" part) is no longer emphasized."]]},{"start":673.81875,"say":"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.","live":["sample_mean","detector_heading"],"does":[[674.8512499999999,"sample_variance is shown on the screen, written out."],[682.7692499999999,"sample_variance (the \"2.17\" part) is emphasized."],[690.06025,"sample_variance (the \"2.17\" part) is no longer emphasized."]]},{"start":690.66025,"say":"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.","live":["sample_mean","sample_variance","detector_heading"],"does":[[695.0722499999999,"verdict is shown on the screen, written out."],[699.65825,"A box is drawn around verdict."],[708.40075,"detector is hidden from the screen — left the board."],[708.40075,"detector_heading is hidden from the screen — left the board."],[708.40075,"sample_mean is hidden from the screen — left the board."],[708.40075,"sample_variance is hidden from the screen — left the board."],[708.40075,"verdict is hidden from the screen — left the board."]]},{"start":709.60075,"say":"The comparison becomes diagnostic when it is repeated over substantial data. Variance near the mean is compatible with independent arrivals at a stable rate.","live":[],"does":[[709.60075,"check_heading is shown on the screen, written out."],[715.26625,"near is shown on the screen, written out."]]},{"start":719.77825,"say":"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.","live":["near","check_heading"],"does":[[720.9392499999999,"high is shown on the screen, written out."]]},{"start":730.7692499999999,"say":"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.","live":["near","high","check_heading"],"does":[[731.8952499999999,"low is shown on the screen, written out."],[745.79275,"check_heading is hidden from the screen — left the board."],[745.79275,"high is hidden from the screen — left the board."],[745.79275,"low is hidden from the screen — left the board."],[745.79275,"near is hidden from the screen — left the board."]]},{"start":746.39275,"say":"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.","live":[],"does":[[747.50725,"final_formula is shown on the screen, written out."]]},{"start":768.2742499999999,"say":"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.","live":["final_formula"],"does":[[785.64325,"A box is drawn around final_formula."],[787.309125,"final_formula is hidden from the screen — left the board."]]}]}]},"durationSeconds":788,"chapters":[{"title":"Random Marks on Time","startSeconds":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. 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."},{"title":"The Poisson Limit","startSeconds":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. 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."},{"title":"The Signature of Poisson Counts","startSeconds":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. 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."},{"title":"From Counts to Waiting","startSeconds":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. 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."},{"title":"Memory Without Age","startSeconds":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. 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."},{"title":"Models Meet Data","startSeconds":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. 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."}]}}