User:Yungam99

Geometric distribution case
If X is a random variable and has Geometric distribution (the parameter is p, and the number of event is k),


 * $$ \operatorname{E}[X] = \sum_{k=1}^\infty kp(1-p)^{k-1} = \frac{1}{p}$$

because it can be derived from the equation below.

$$ \sum_{k=1}^\infty kx^{k-1} = \frac{1}{(1-x)^2}$$