WEBVTT

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Rare events happen at particular moments:

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a customer reaches a desk, a detector
records a click, or a component fails.

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We are going to build a continuous-time
model from ordinary probability, beginning

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with nothing more mysterious
than many small independent trials.

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Here is the question
that will organize the first half.

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During a stretch of time
of length T, how many events arrive?

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We want the complete distribution
of that count, not merely its average.

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Picture time as a line.

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The red marks are arrivals
from one possible run of the process.

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Another run would put them elsewhere

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and might contain a different number, but
every arrival is attached to some moment.

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Now chop the observation
time into n equal intervals.

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One interval has length delta
t, equal to T divided by n.

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We will eventually make these intervals

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so short that an event inside
one of them is genuinely rare.

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The model makes two
substantive assumptions.

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First, separate intervals
contribute independently.

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Second, the event rate
is constant, so an interval

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of length delta t has event probability
approximately lambda times delta t.

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Lambda is a rate, measured
in events per unit time.

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Multiplying it by the interval length
gives a dimensionless probability.

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As delta t shrinks,
this probability shrinks too,

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but the total expected count across
all n intervals remains lambda T.

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Each tiny interval is therefore
a Bernoulli trial: event or no event.

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Adding n independent trials
gives a binomial count X sub n,

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with n trials and success
probability lambda T over n.

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Continuous time has not appeared by magic.

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We have built an approximation
that can be pushed to a limit.

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Before taking a limit,
hold the central idea still.

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Suppose the expected count is two.

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Ten intervals can each
have probability zero point two.

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One hundred intervals
use zero point zero two,

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and one thousand use
zero point zero zero two.

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The individual trials become
rarer while their number grows.

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In every row, n times p sub n remains two.

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More generally, call the fixed
expected count mu, equal to lambda T.

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Now fix a possible count k.

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The binomial probability

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of exactly k events is the number
of ways to choose their intervals,

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times the probability
of k occupied intervals,

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times the probability
that all the others are empty.

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Substitute p sub n equal
to mu over n and rearrange.

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The probability is mu
to the k over k factorial,

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multiplied by three factors whose
limits we can read separately.

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The first factor contains k terms:

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n over n, then n minus
one over n, and so on.

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For a fixed k, every one
of those terms approaches one.

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Therefore A sub n approaches one.

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The second factor
is the classical exponential limit.

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One minus mu over n, raised to n,
approaches e to the minus mu.

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This is the surviving probability

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of seeing no event across
the complete interval.

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The last factor raises something

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approaching one to the fixed power
minus k, so it also approaches one.

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Multiplying the three limits leaves
e to the minus mu times mu to the k

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over k factorial.

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Finally replace mu by lambda T.

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This is the Poisson distribution,
obtained rather than announced.

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Its probabilities really do sum to one,

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because the remaining series
is the exponential series for e to the mu.

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Here is a Poisson distribution
with mu equal to three.

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The bars give the probabilities of zero
events, one event, two events, and so on.

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The distribution is not symmetric, but it
is concentrated around counts near three.

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Its average can be calculated exactly.

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Start with the definition of expectation:

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sum k times the probability of k.

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The factor k cancels one
factor from k factorial.

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Pull one factor of mu outside,
then re-index with j equal to k minus one.

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What remains is the sum of every
Poisson probability, so that sum is one.

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The expected count is therefore mu.

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For variance, the useful quantity
is the second factorial moment,

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N times N minus one.

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The same cancellation,
now performed twice, gives mu squared.

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Ordinary variance can be written as
the second factorial moment plus the mean,

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minus the square of the mean.

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Substitute mu squared, mu, and mu squared.

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The two squares cancel, leaving mu.

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So a Poisson count
has mean mu and variance mu.

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The standard deviation
is therefore the square root of mu.

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Relative fluctuations become smaller
when the expected count becomes large,

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even though the absolute
fluctuations grow.

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This equality has an intuitive
origin in the small-interval model.

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Independent intervals each
contribute a tiny yes-or-no uncertainty.

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Adding them gives variance
n p times one minus p,

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which approaches n p as p becomes tiny.

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More importantly, equal mean and variance
is a prediction that data can challenge.

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If repeated equal windows
have variance far above their mean,

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arrivals may be clustered
or the rate may be changing.

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If variance is far below the mean,

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some regular spacing
or dead time may be present.

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The count question
has a natural companion.

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Instead of fixing a time interval
and asking how many events occur,

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start the clock now and ask how
long it runs before the next event.

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Call that random waiting time

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W.

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On one realized timeline,

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now is the green mark and the next
arrival is the first red mark.

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The yellow span between them
is the observed value of W.

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Another run would give a different span.

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To find the distribution, ask
when W exceeds a chosen time t.

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That happens exactly when the interval
from now to t contains zero events.

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Waiting longer than t and counting
zero by t are the same event.

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But the count by time t
is Poisson with mean lambda t.

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Put k equal to zero
in the Poisson formula.

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The factorial and power disappear,
leaving e to the minus lambda t.

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That gives the survival probability,
the chance the wait is still not over.

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Its complement gives
the cumulative distribution.

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Differentiating gives the exponential
density, lambda e to the minus lambda t.

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For a unit rate,
the survival curve begins at one.

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A very short threshold is likely
to be crossed without an event,

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while a long event-free stretch
becomes progressively less likely.

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Move the threshold to the right.

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The point rides down the curve

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because the probability of surviving
without an event decays exponentially.

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The mean waiting time is the area
under the survival curve.

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Integrating e to the minus lambda t from
zero to infinity gives one over lambda.

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Double the event rate
and the average wait is cut in half.

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This is the claim people rightly distrust.

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Suppose we have already waited
s units of time with no arrival.

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Surely, it feels, we must now
be closer to the next event.

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The exponential model says no. Let us
prove exactly what that sentence means.

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Begin at zero and suppose
no event occurs before s.

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The yellow span is the waiting
time we have already endured.

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Reaching s tells us
that W is greater than s.

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Now ask for no event
during an additional duration

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t. That means the total
wait exceeds s plus t.

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The green span is the extra future
time whose probability we want.

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Conditional probability divides

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the chance of surviving all the way to s
plus t by the chance of having reached s.

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The later event is contained inside the
earlier condition, so this ratio is exact.

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Insert the exponential
survival probabilities.

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The numerator contains e
to the minus lambda times s plus t.

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The denominator contains
e to the minus lambda s.

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The factors involving s cancel.

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What remains is e to the minus lambda t,

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exactly the probability of waiting
at least t from a fresh start.

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The elapsed wait
has disappeared from the answer.

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These two curves make the result visible.

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On the left is survival
for an additional time starting now.

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On the right is survival for the same
additional time after we have already

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waited s. The curves are identical.

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Why does this feel absurd?

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We often imagine an arrival
that is already on its way,

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a bus following a timetable,
or a component accumulating age.

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Those mechanisms carry
information from the past.

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A constant-rate Poisson process does not:

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future tiny intervals
are independent of past ones.

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Memorylessness is therefore
not a universal law of waiting.

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It is a sharp consequence
of the assumptions we chose.

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If the rate changes with time,

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arrivals repel one another,
or a schedule is operating,

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the remaining wait can depend strongly
on how long we have already waited.

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Put the model to work at a service desk.

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Customers arrive
at an average rate of six per hour.

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We want the probability of exactly
three arrivals in half an hour,

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and the average time
from now until the next arrival.

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For the count, match the units first.

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Thirty minutes is one half hour,

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so the expected count is lambda T, equal
to six times one half, which is three.

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Use the Poisson probability with k
equal to three and mu equal to three.

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The result is e to the minus three
times three cubed over three factorial,

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about zero point two two four.

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For waiting, use the exponential mean.

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One over six hour is one sixth of sixty
minutes, so the mean wait is ten minutes.

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The count calculation and the waiting
calculation use the same rate lambda

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from two different perspectives.

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Now suppose a detector is observed
in twelve equal one-minute windows.

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These are the recorded click counts.

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Equal windows matter

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because the Poisson prediction compares
counts that share the same expected value.

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The twelve counts add to twenty-four,

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so their empirical mean
is exactly two clicks per window.

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This estimates lambda times one minute.

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Next measure their spread around two.

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Using the average squared deviation,

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the empirical variance is twenty-six
over twelve, about two point one seven.

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It is not exactly two,

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because a finite random sample does not
reproduce a population identity perfectly.

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The mean two point zero zero and
variance two point one seven are close.

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That does not prove the Poisson
model, but it passes this first check.

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With many more windows, we could simulate
or calculate the variation expected

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in these estimates and judge
whether the discrepancy is unusual.

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The comparison becomes diagnostic
when it is repeated over substantial data.

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Variance near the mean is compatible with
independent arrivals at a stable rate.

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Variance much larger than
the mean suggests extra variability:

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customers arriving in groups,

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bursts of detector noise,
or a rate that changes through the day.

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Variance much smaller than the mean
suggests excessive regularity:

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arrivals may inhibit one another,

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or a detector may briefly stop
registering after each click.

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Either pattern points back to an
assumption that should be reconsidered.

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So the complete connection is this.

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Chopping time into rare independent trials
produces Poisson counts in the limit.

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Asking for zero counts up to time
t produces exponential waiting.

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Independence gives memorylessness, and
the equal mean and variance of the counts

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gives the model an observable signature.

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The formulas are useful
because their assumptions are clear.

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Use the same rate lambda to move
between counts and waits, then return

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to data and ask whether independence,

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a stable rate, and equal mean
and variance are actually visible.

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A probability model earns
its place by surviving that return.