Expectation is Linear/Discrete

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ and $Y$ be random variables on $\struct {\Omega, \Sigma, \Pr}$.

Let $\expect X$ denote the expectation of $X$.


Then:

$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \, \expect X + \beta \, \expect Y$


Proof

Follows directly from Expectation of Function of Joint Probability Mass Distribution, thus:


\(\ds \expect {\alpha X + \beta Y}\) \(=\) \(\ds \sum_x \sum_y \paren {\alpha x + \beta y} \, \map \Pr {X = x, Y = y}\) Expectation of Function of Joint Probability Mass Distribution
\(\ds \) \(=\) \(\ds \alpha \sum_x x \sum_y \map \Pr {X = x, Y = y}\)
\(\ds \) \(+\) \(\ds \beta \sum_y y \sum_x \map \Pr {X = x, Y = y}\)
\(\ds \) \(=\) \(\ds \alpha \sum_x x \, \map \Pr {X = x} + \beta \sum_y y \, \map \Pr {Y = y}\) Definition of Marginal Probability Mass Function
\(\ds \) \(=\) \(\ds \alpha \, \expect X + \beta \, \expect Y\) Definition of Expectation

$\blacksquare$