> [!NOTE] Theorem > Let $p\in(0,1].$ Let $X$ be a [[Discrete random variables|discrete real-valued random]] that has a [[Geometric Distribution|geometric distribution]] with parameter $p.$ Then $X$ is [[Square-Integrable Discrete Real-Valued Random Variable|square integrable]] and its [[Variance of Square-Integrable Discrete Real-Valued Random Variable|variance]] is $\frac{(1-p)}{p^{2}}.$ **Proof**: By [[Expectation of Real-Valued Function of Discrete Real-Valued Random Variable]], we have $\begin{aligned} \mathbb{E}[X^{2}]& =\sum_{n=1}^\infty n^2\cdot p\cdot x^{n-1} \\ &=p\sum_{n=1}^\infty n\cdot x^{n-1}+p\sum_{n=1}^\infty n\cdot(n-1)\cdot x^{n-1} \\ &=p\sum_{n=0}^{\infty}n\cdot x^{n-1}+px\sum_{n=0}^{\infty}n\cdot(n-1)\cdot x^{n-2} \\ &=p\sum_{n=0}^\infty\frac{\mathrm{d}}{\mathrm{d}x}[x^n]+px\sum_{n=0}^\infty\frac{\mathrm{d}^2}{\mathrm{d}x^2}[x^n] \\ &=p\cdot\frac{\mathrm{d}}{\mathrm{d}x}\Big[\sum_{n=0}^\infty x^n\Big]+px\cdot\frac{\mathrm{d}^2}{\mathrm{d}x^2}\Big[\sum_{n=0}^\infty x^n\Big] \\ &=p\frac1{(1-x)^2}+px\frac2{(1-x)^3} \\ &=\frac1p+2\cdot\frac{1-p}{p^2} \\ &=\frac{2-p}{p^2}. \end{aligned}$using [[Geometric Series]], thus $X$ is indeed square-integrable. Now by [[Expectation of Geometric Distribution]], $\mathbb{E}[X]= \frac{1}{p}$ so $\text{Var}(X)=\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2} = \frac{2-p}{p^{2}}-\frac{1}{p^{2}}=\frac{1-p}{p^{2}}.$