The Geometric Distribution

Definition Consider a sequence of independent trials, each a success with the same probability $p$, where $0 < p \le 1$. Let $T$ be the number of trials up to and including the first success. Counting the successful trial is the trial-count convention, and in it the support of $T$ is $\{1, 2, 3, \dots\}$ when $p < 1$. The event $\{T = k\}$ means that the first $k - 1$ trials fail and trial $k$ succeeds, so independence gives

for every positive integer $k$, and $P(T = x) = 0$ for every other real $x$. We say that $T$ has the geometric distribution with parameter $p$. When $p = 1$ the first trial succeeds with certainty, so $T = 1$ and the support is $\{1\}$. The formula gives this with the usual convention $(1 - p)^0 = 1$. The value $p = 0$ is excluded, because then the probability of ever succeeding is $0$.

Definition Under the same independent trials with $0 < p \le 1$, let $F$ be the number of failures before the first success. Then $F = T - 1$, the support of $F$ is $\{0, 1, 2, \dots\}$ when $p < 1$ and $\{0\}$ when $p = 1$, and

for every integer $j \ge 0$. Some references also call this distribution geometric, so a formula or a mean quoted for a geometric distribution is ambiguous until the convention is stated. Since the two variables differ by the constant $1$, their means differ by $1$ and their variances are equal.

Definition For a real number $r \ne 1$ and a positive integer $m$, the finite geometric series has the sum

To see this, call the sum $S$. Every power of $r$ except $1$ and $r^m$ cancels in $S - rS$, so $(1 - r)S = 1 - r^m$. If $\lvert r \rvert < 1$, then $r^m \to 0$ as $m \to \infty$, and the infinite series converges with

If $\lvert r \rvert \ge 1$, the terms $r^i$ do not tend to $0$ and the infinite series diverges.

Definition Let $T$ be geometric with parameter $p$ in the trial-count convention and $q = 1 - p$. The event $\{T > k\}$ means that the first $k$ trials all fail, whatever happens afterward. Hence for every integer $k \ge 0$, with $q^0 = 1$,

The second line is the CDF of $T$ at the integer $k$. For $k \ge 1$, the event $\{T \ge k\}$ is the event $\{T > k - 1\}$, so it has probability $q^{k-1}$.

Definition Let $T$ be geometric with parameter $p$, where $0 < p < 1$, and let $q = 1 - p$. For all integers $m \ge 0$ and $n \ge 0$,

The conditioning event has probability $q^m > 0$, and since $\{T > m + n\}$ is contained in $\{T > m\}$, the left side is $q^{m+n} / q^m = q^n$. Equivalently, given $T > m$, the remaining wait $T - m$ has the same geometric distribution as $T$. When $p = 1$ and $m \ge 1$, the event $\{T > m\}$ has probability $0$ and the conditional probability is not defined.

Definition If $T$ is geometric with parameter $p$, where $0 < p \le 1$, in the trial-count convention, then

Both moments are finite because the tail $q^k$ shrinks geometrically. For the mean, a variable with values in the positive integers satisfies $E[T] = \sum_{k \ge 1} P(T \ge k)$, so with $q = 1 - p$

For the second moment $s = E[T^2]$, condition on the first trial. If it succeeds, $T = 1$. If it fails, $T = 1 + T'$, where the remaining wait $T'$ does not depend on the first trial and has the same distribution as $T$. So

Solving gives $s = (2 - p)/p^2$, and subtracting $(E[T])^2 = 1/p^2$ gives the variance. The failure count $F = T - 1$ has $E[F] = (1 - p)/p$ and the same variance.