Waiting-Time Distributions: Geometric and Poisson

"On average a customer files 3 claims a year." Is that the mean of a geometric count or a Poisson rate? The geometric counts trials until a success; the Poisson counts events in a fixed window. They share the actuary's favorite shortcut and one fatal trap each.

Repeat an independent trial with success probability $p$ until the first success. Two conventions appear, and the exam uses both.

Trials convention, $N$ = the trial on which the first success occurs:

$$P(N=n) = (1-p)^{n-1}p, \quad n = 1, 2, 3, \ldots$$

$E[N] = 1/p$, $\text{Var}(N) = (1-p)/p^2$. Failures convention, $Y = N-1$ = the number of failures before the first success: $E[Y] = (1-p)/p$, with the same variance. The survival function is clean:

$$P(N>n) = (1-p)^n$$

HIGH-FREQUENCY: The geometric is the only discrete memoryless distribution. Given no success in the first $k$ trials, the additional wait has the same distribution as a fresh start, so $E[N\mid N>k] = k + 1/p$.

Common mistakes

• Mixing the two geometric conventions. $E = 1/p$ counts trials; $E = (1-p)/p$ counts failures. A "mean of 5 visits" is ambiguous until you decide which. Read whether the count starts at 1 (trials) or 0 (failures).
• Forgetting the geometric is memoryless. $E[N\mid N>k] = k + 1/p$, not $1/p$. Past failures do not shorten the remaining wait.
• Integrating the Poisson. It is discrete. $P(N\le 2) = P(0) + P(1) + P(2)$, summed, never integrated.

Bottom line

• Geometric($p$): trials until the first success. Trials $N\in\{1,2,\ldots\}$: $E = 1/p$, $\text{Var} = (1-p)/p^2$. Failures $Y=N-1$: $E = (1-p)/p$, same variance.
• Geometric survival: $P(N>n) = (1-p)^n$. It is memoryless: $E[N\mid N>k] = k + 1/p$.
• Geometric variance exceeds its mean whenever $p<1$, since $(1-p)/p^2 > 1/p$; only the Poisson has $E = \text{Var}$.
• Poisson($\lambda$): events in a fixed window. $P(N=k)=e^{-\lambda}\lambda^k/k!$, and $E = \text{Var} = \lambda$.

Exam shortcut

Decide the geometric convention first (trials start at 1, failures at 0), then the mean follows. For a Poisson given a moment, write $E[N^2] = \lambda + \lambda^2$ and solve the quadratic. For combined Poisson streams, "add the lambdas, then compute one probability." "Geometric forgets, Poisson adds." Memorylessness for the geometric; rate-addition for independent Poissons.

The full lesson (about 2,125 words, 14 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 2a
• 2b
• 2c
• 2d
• 2e
• 2f

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