Krdytkyu

the question asks for the expected value as well as variance of the above question.

I've attached a picture of my answer for the mean, but i can't use similar reasoning for the variance.

$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.

$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$

$ = sum_{n=1}^infty n(1-p)^{n-1}p$

Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.

$S = 1 + 2q + 3 q^2 + ;...$

$qS = ;;;;;q + 2q^2+ ;...$

$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$

implying $S = frac{1}{(1-q)^2}$.

Using this result, we get:

$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.

$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $

$= sum_{n=1}^infty n^2(1-p)^{n-1}p $

Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.

$S = 1 + 4q + 9q^2 + ; ... $

$qS = ;;;;; q+4q^2+;... $

$(1-q)S = 1 + 3q + 5q^2 + ;...$

$q(1-q)S = ;;;; q +3q^2 + ;...$

$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$

$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $

implying $S = frac{1+q}{(1-q)^3}$.

Using this result, we get:

$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.

Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.