Showing that the Geometric distribution $E(X)=\frac 1p$

Question

So I have $X \sim \text{Geom}(p)$ and the probability mass function is:

$$p(1-p)^{x-1}$$

From the definition that:

$$\sum_{n=1}^\infty ns^{n-1} = \frac {1}{(1-s)^2}$$

How would I show that the $E(X)=\frac 1p$

Answer 1

What is $s$ in your case? Consider:

$$ E(X)=\sum_{x=1}^\infty x p(1-p)^{x-1}=p\sum_{x=1}^\infty x(1-p)^{x-1}=\cdots=? $$

Can you see how to take it from here?

Answer 2

\begin{eqnarray} E(X)&=&\sum_{x=1}^\infty x p(1-p)^{x-1}\\ &=&p\sum_{x=1}^\infty x(1-p)^{x-1}\\ &=&p\sum_{x=1}^\infty -\frac{d}{dp}(1-p)^{x}\\ &=&-p\left[\frac{d}{dp} \sum_{x=1}^\infty (1-p)^x\right]\\ &=&-p\cdot \frac{d}{dp}\frac{1-p}{1-(1-p)} \\ &=&-p\cdot \frac{d}{dp}\frac{1-p}{p} \\ &=&-p\cdot \frac{-1}{p^2} \\ &=& \frac{1}{p}\\ \end{eqnarray}