> [!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 the [[Expectation of Discrete Real-Valued Random Variable|expectation]] of $X$ is $\frac{1}{p}.$ ###### Proof by definition of expectation We have $\begin{align} \mathbb{E}[X] &= \sum_{n=1}^{\infty} p_{X}(n)\cdot n \\ &= \sum _{n=1}^{\infty} np\cdot (1-p)^{n-1} = p \sum_{n=0}^{\infty} n (1-p)^{n-1} \tag{1} \end{align}$Let $x=1-p.$ Then $\begin{align} (1) &= p \sum_{n=0}^{\infty} \frac{d}{dx}(x^{n}) \\ & = p \frac{d}{dx}\left( \sum_{n=0}^{\infty}x^{n} \right) \\ & = p \cdot \frac{d}{dx}\left[ \frac{1}{1-x} \right] \\ &= p \cdot \left( \frac{1}{(1-x)^{2}} \right) =p \cdot \frac{1}{p^{2}} \\ &= \frac{1}{p} \end{align}$using [[Geometric Series]]. ###### Proof by tail probabilities formula for expectation By [[Expectation of Integrable Non-negative Discrete Real-Valued Random Variable Equals Sum of Tail Probabilities]], $\mathbb{E}[X]=\sum_{i=0}^{\infty} \mathbb{P}(X>i)=\sum_{i=0}^{\infty}(1-p)^{i} \frac{1}{1-(1-p)}=\frac{1}{p}.$