Showing $E[x]=\frac{1}{p}$ for a Geometric Random Variable

Question

Using the fact that $E(x)=\sum_{k=1}^\infty P(X \geq k)$, which I previously proved, I have to show that $E[x]=\frac{1}{p}$ for Geometric Random Variable.

I started by saying that $$P(X \leq k)=\sum_{j=1}^k (1-p)^{j-1}p=1-(1-p)^k$. Since $P(X \leq k)=1-P(X \geq k),$$ then $P(X \geq k)=1-(1-(1-p)^k)=(1-p)^k.$

Then $E(x)=\sum_{k=1}^\infty P(X \geq k) = \sum_{k=1}^\infty (1-p)^k$, but I don't see how that would equal $1/p$. Did I do something wrong? Thank you in advanced.

Answer 1

Your answer is correct.

The sum of a geometric series $\sum^\infty_{k=1}r^k=\frac1{1-r}.$

In your example, $r = 1-p.$ Thus, $\sum^\infty_{k=1}(1-p)^k=\frac1{1-(1-p)}=\frac1p.$

Answer 2

Note: $\mathsf P(X\geq k)= (1-p)^{k-1}$ because $\mathsf P(X\leq k)=1-\mathsf P(X>k)$

$\begin{align}\mathsf E(X)&=\sum_{k=1}^\infty (1-p)^{k-1} \\&= 1+\sum_{k=2}^\infty (1-p)^{k-1} \\&= 1+ (1-p)\sum_{k=1}^\infty (1-p)^{k-1} \\&= 1+(1-p)\mathsf E(X) \\[2ex]\therefore~ \mathsf E(X) &= \frac 1 p\end{align}$