How to Prove the Variance of a Geometric Distribution

A geometric random variable models the number of independent trials needed to get the first success. Its variance depends on the success probability, but the distribution is defined using two common counting conventions. This proof starts with the convention that counts the successful trial, then shows why the other convention has the same variance.

Define the geometric distribution

Let each trial succeed with probability $p$ , where $0 < p \leq 1$ , and write $q = 1-p$ . Define $X$ to be the number of trials up to and including the first success. Then $X$ takes values $1,2,3,\ldots$ , and

\Pr(X=k)=q^{k-1}p.

For example, $X=3$ means the first two trials fail and the third succeeds. The probability-generating function (PGF) is

G(s)=\mathbb{E}[s^X]=\sum_{k=1}^{\infty}q^{k-1}p s^k=\frac{ps}{1-qs}.

The geometric series converges when $|qs|<1$ . Since $q<1$ when $p>0$ , the function and its derivatives are well-defined at $s=1$ .

Derive the mean and second factorial moment

For a PGF, the first derivative at 1 gives the mean, and the second derivative at 1 gives the second factorial moment, $\mathbb{E}[X(X-1)]$ . Differentiate:

\begin{aligned} G'(s) &= \frac{p}{(1-qs)^2},\\ G''(s) &= \frac{2pq}{(1-qs)^3}. \end{aligned}

At $s=1$ , the denominator $1-q$ equals $p$ . Therefore,

\mathbb{E}[X]=G'(1)=\frac{1}{p}, \qquad \mathbb{E}[X(X-1)]=G''(1)=\frac{2q}{p^2}.

The second derivative does not directly give $\mathbb{E}[X^2]$ . Instead, use $X^2=X(X-1)+X$ , so

\mathbb{E}[X^2]=\frac{2q}{p^2}+\frac{1}{p}.

Now apply $\operatorname{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ :

\begin{aligned} \operatorname{Var}(X) &=\frac{2q}{p^2}+\frac{1}{p}-\frac{1}{p^2}\\ &=\frac{2q+p-1}{p^2}\\ &=\frac{q}{p^2}\\ &=\frac{1-p}{p^2}. \end{aligned}

Thus, for the convention that counts the successful trial, the mean is $1/p$ and the variance is $(1-p)/p^2$ .

Worked example

Suppose a trial succeeds with probability $p=0.2$ . Then $q=0.8$ . The expected number of trials is

\mathbb{E}[X]=\frac{1}{0.2}=5,

and the variance is

\operatorname{Var}(X)=\frac{0.8}{(0.2)^2}=20.

The standard deviation is therefore $\sqrt{20}\approx 4.47$ trials. A variance larger than the mean is possible here: many trials may be needed when success is relatively unlikely, creating a broad distribution.

As a boundary case, if $p=1$ , the first trial always succeeds. The formula gives variance $0$ , as expected for a constant random variable.

If you count failures instead

Some texts define a geometric random variable $Y$ as the number of failures before the first success. Its possible values are $0,1,2,\ldots$ , with $Y=X-1$ . Subtracting a constant changes the mean but not the variance, so

\mathbb{E}[Y]=\frac{1-p}{p}, \qquad \operatorname{Var}(Y)=\frac{1-p}{p^2}.

Check which convention a problem uses before quoting the mean. The variance formula is the same for both conventions, provided $p$ means the probability of success on each trial.

Common mistakes

Mixing the two conventions: The means $1/p$ and $(1-p)/p$ correspond to different definitions. State whether the variable counts trials or failures.
Treating $G''(1)$ as the second moment: It equals $\mathbb{E}[X(X-1)]$ , not $\mathbb{E}[X^2]$ . Add $\mathbb{E}[X]$ to obtain the second moment.
Using the failure probability as $p$ without saying so: If a source uses $p$ for the probability of failure, rewrite the formula in terms of the success probability before comparing results.
Forgetting the squared mean: Variance is $\mathbb{E}[X^2]-(\mathbb{E}[X])^2$ , not just the second moment.

References

A First Course in Probability — Sheldon Ross

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