# The variance of the geometric distribution

Last updated: Jan. 3, 2019
Let $X$ be a discrete random variable with the geometric distribution with parameter p ($0 \lt p \lt 1$). , where $k=1,2,\cdots.$ The variance of $X$ is .

### Proof

Let $X$ be a discrete random variable with the geometric distribution with parameter p ($0 \lt p \lt 1$). . The expectation value of $X$ is . In general, the variance is the difference between the expectation value of the square and the square of the expectation value, i.e., . Therefore, we have . To obtain the variance, we thus need to find the expectation of $X^2$.
The expectation of $X^2$ with the geometric distribution $(1)$ is .
In order to obtain the sum in the right-hand side, we focus on the Taylor expansion of $1/(1-x)$, , and differentiate the both sides, . We multiply the both sides by x, , and differentiate the both sides again, . By variable transformation, $x=1−p$, this equation becomes . We substitute this equation into $(3)$ .
From this expression and (2), the variance is obtained. Example :
Example 1:
The graph below shows the geometric distribution for $p=1/2$. The variance is .
Example 2:
The graph below shows the geometric distribution for $p=1/4$. The variance is .
Example 3:
The graph below shows the geometric distribution for $p=1/8$. The variance is .
The larger $p$ is, the more flat the graph is, and the more the variance is.