Geometric distribution: $$ f_X(k) = p (1 - p)^{k-1}, \quad k = 1,2,3,\cdots $$ The expectation:
$$ \begin{align} \mathbb{E}(X) &= \sum_{k=1}^\infty k p (1 - p)^{k - 1} \\ &= \sum_{k=1}^\infty p(1-p)^{k-1} + \sum_{k=2}^\infty (k - 1) p(1-p)^{k - 1} \\ &= 1 + \sum_{k=1}^\infty k p(1-p)^k \\ &= 1 + (1 - p) \sum_{k=1}^\infty k p(1-p)^{k - 1} \\ &= 1 + (1 - p) \mathbb{E}(X) \end{align} $$
Solving for the expectation, we get $\mathbb{E}(X) = 1/p$.
Another version of Geometric distribution: $$ f_X(k) = p (1 - p)^k, \quad k = 0, 1,2,3,\cdots $$
$$ \begin{align} \mathbb{E}(X) &= \sum_{k=0}^\infty k p (1 - p)^{k} \\ &= \sum_{k=1}^\infty kp(1-p)^{k}\\ &= \sum_{k=1}^\infty (k-1)p(1-p)^{k}+\sum_{k=1}^\infty p(1-p)^{k}\\ &= (1-p)\sum_{k=1}^\infty (k-1)p(1-p)^{k-1}+\sum_{k=1}^\infty p(1-p)^{k}\\ &= (1 - p)\mathbb{E}(X) +\sum_{k=1}^\infty p(1-p)^{k} \\ &= (1 - p)\mathbb{E}(X) +\sum_{k=0}^\infty p(1-p)^{k}-p \\ &= (1 - p)\mathbb{E}(X) +1-p \\ \end{align} $$
from where we get $$\mathbb{E}(X) = (1-p)/p$$
For the variance we also have:
$$ \begin{align} \mathbb{E}(X^2) &= \sum_{k=1}^\infty k^2 p (1 - p)^{k - 1} \\ &= \sum_{k=1}^\infty k p(1-p)^{k-1} + \sum_{k=2}^\infty (k^2 - k) p(1-p)^{k - 1} \\ &= \mathbb{E}(X) + (1 - p) \sum_{k=1}^\infty (k^2 + k) p(1-p)^{k - 1} \\ &= \mathbb{E}(X) + (1 - p) \mathbb{E}(X) + (1 - p) \sum_{k=1}^\infty k^2 p(1-p)^{k-1} \\ &= \frac{2 - p}{p} + (1 - p) \mathbb{E}(X^2) \end{align} $$
Solving for the expectation, we get $\mathbb{E}(X^2) = (2 - p) / p^2$.
Finally,
$$ \mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = \frac{2 - p}{p^2} - \frac{1}{p^2} = \frac{1 - p}{p^2} $$
For the variant version of Geometric distribution: $$ f_X(k) = p (1 - p)^k, \quad k = 0, 1,2,3,\cdots $$ $$ \begin{align} \mathbb{E}(X^2) &= \sum_{k=0}^\infty k^2 p (1 - p)^{k} \\ &= \sum_{k=1}^\infty k^2 p(1-p)^{k}\\ &= \sum_{k=0}^\infty (k+1)^2 p(1-p)^{k+1}\\ &= \sum_{k=0}^\infty k^2 p(1-p)^{k+1}+2\sum_{k=0}^\infty k p(1-p)^{k+1}+\sum_{k=0}^\infty p(1-p)^{k+1}\\ &= (1-p)\mathbb{E}(X^2)+2(1-p)\mathbb{E}(X)+(1-p)\\ &= (1-p)\mathbb{E}(X^2)+2(1-p)^2/p+(1-p)\\ &= (1-p)\mathbb{E}(X^2)+\frac{2(1-p)^2}{p}+\frac{p(1-p)}{p}\\ &= (1-p)\mathbb{E}(X^2)+\frac{2-3p+p^2}{p}\\ \end{align} $$ We get $$\mathbb{E}(X^2) = \frac{2-3p+p^2}{p^2}$$ And then the variance: $$ \mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = \frac{2-3p+p^2}{p^2} - \frac{(1-p)^2}{p^2} = \frac{1 - p}{p^2} $$
Why Geometric distribution has different expectations but the same variance?
