> [!NOTE] Definition 1 (Expectation of Discrete Real-Valued Random Variable) > Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a [[Probability Space|probability space]]. Let $X$ be a [[Discrete random variables|discrete real-valued random variable]] on $(\Omega, \mathcal{F}, \mathbb{P}).$ The *expectation* of $X$ is given by $\mathbb{E}[X] = \sum_{x:\mathbb{P}(X=x)>0} x \cdot \mathbb{P}(X=x)\left( = \sum_{x\in D_{X}}x\cdot p_{X}(x) \right)$whenever the sum is [[Absolutely Convergent Series|absolutely convergent]], otherwise $\mathbb{E}[X]$ is not defined. $p_{X}$ denotes the [[Probability Mass Function of Discrete Real-Valued Random Variable|PMF]] of $X$ and $D_{X}$ denote its [[Discrete Support of Distribution of Discrete Real-Valued Random Variable|support]]. **Notation**: $X=x$ denotes the set $\{ \omega: X(\omega) = x\}$ or $X^{-1}(x),$ the [[Preimage (of set under a function)|preimage]] of $x$ under $X.$ > [!Example] > Toss a fair coin $4$ times and count the number of Heads. $\begin{aligned} \mathbb{E}[X]& =0\cdot\mathbb{P}(X=0)+1\cdot\mathbb{P}(X=1)+2\cdot\mathbb{P}(X=2)+ \\ &+3\cdot\mathbb{P}(X=3)+4\cdot\mathbb{P}(X=4) \\ &=0\cdot\frac1{16}+1\cdot\frac4{16}+2\cdot\frac6{16}+3\cdot\frac4{16}+4\cdot\frac1{16} \\ &=2. \end{aligned}$ # Properties By [[Expectation of Discrete Real-Valued Random Variable is Unitary]], By [[Expectation of Real-Valued Function of Discrete Real-Valued Random Variable]], ... By [[Expectation of Real-Valued Function of Bivariate Discrete Real-Valued Random Variable]], ... This gives [[Expectation of Discrete Real-Valued Random Variable is Linear]] which asserts that .... By [[Expectation of Discrete Real-Valued Random Variable is Monotone]], ... # Applications **Integrable discrete real-valued random variable**: If the sum $\sum_{x:\mathbb{P}(X=x)} x \cdot \mathbb{P}(X=x)$ converges (that is, $\mathbb{E}[X]$ is defined) then $X$ is said to be [[Integrable Discrete Real-Valued Random Variable|integrable]]. **Examples**: By [[Expectation of Poisson Distribution]], if $X\sim \text{Poisson}(\lambda)$ then $\mathbb{E}[X]=\lambda.$ By [[Expectation of Binomial Distribution]], if $X \sim \text{Binon}(n,p)$ then $\mathbb{E}[X]=np.$ By [[Expectation of Geometric Distribution]],