Finding the variance of $0.2(0.8)^x$, where $x = 0, 1, 2, 3 .......$

Question

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.

Answer 1

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