Geometric Distribution

Study guide: Exam P-1

The Geometric Distribution X \sim \text{Geom}(p) counts the number of independent Bernoulli trials until the first success. It is the discrete analogue of the Exponential distribution and shares its memoryless property: P(X > m+n \mid X > m) = P(X > n).

P(X = k) = (1-p)^{k-1}p, \quad k = 1, 2, 3, \ldots

\text{where } p \in (0,1) = \text{probability of success on each trial}

E[X] = \frac{1}{p}, \qquad \text{Var}(X) = \frac{1-p}{p^2}

• Two parameterizations exist and they are not interchangeable. The form above counts trials X \in \{1,2,3,\ldots\} with E[X] = 1/p. The other counts failures before the first success, Y = X - 1 \in \{0,1,2,\ldots\}, with P(Y=k) = (1-p)^k p and E[Y] = (1-p)/p. The variance \,(1-p)/p^2 is the same for both. Decide from the wording which is being counted before touching a formula — a mismatch shifts every answer by exactly 1.
• The survival function is clean: P(X > n) = (1-p)^n — "the first n trials all failed." Use it instead of summing the PMF.

Number of Policies Until First Claim Each policy independently has a 20% chance of generating a claim. Let X = the number of policies reviewed until the first claim.

Answer X \sim \text{Geom}(0.20). The probability the first claim is on the 3rd policy: P(X=3) = (0.80)^2(0.20) = 0.128 The expected number of policies until the first claim: E[X] = 1/0.20 = 5.

Using Memorylessness After a Dry Spell With the same 20% claim probability per policy, 8 policies have been reviewed with no claim. Find the probability that the first claim comes on the 11th policy or later.

Answer Memorylessness discards the 8 wasted trials entirely:

\begin{align*}
P(X > 8 + 2 \mid X > 8) &= P(X > 2) \\
                        &= (0.80)^2 \\
                        &= 0.64
\end{align*}

The 8 claim-free reviews carry no information about when the next claim arrives — the process does not "become due."