Proof

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Discrete Case

Let f(x) be the probability function such that \(f(x_{i})=P(X=x_{i})\).

\(E[aX+b]=\displaystyle\sum_{i}\ (ax_{i}+b)\cdot f(x_{i})=a\displaystyle\sum_{i}\ x_{i}f(x_{i})+b\displaystyle\sum_{i}f(x_{i})=aE[X]+b\ \ \ (\ \text{∵}\ \sum_{i} f(x_{i})=1)\)

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Continuous Case

Let f(x) be the probability function such that \(f(x)=P(X=x)\).

\(E[aX+b]=\int_{-\infty}^{\infty}\ (ax+b)\cdot f(x)\ dx=a\int_{-\infty}^{\infty}\ xf(x)\ dx+b\int_{-\infty}^{\infty}\ f(x)\ dx=aE[X]+b\ \ \ (\ \text{∵}\ \int_{-\infty}^{\infty}\ f(x)\ dx=1)\)