Section 2.5 - Sum Expected Values

What is the expected value of the sum $X + Y$ given $X$ and $Y$ are (not necessarily independent) random variables?

From the definition of expected value, $$E(X + Y) = \sum_x\sum_y (x + y)\,p(X = x \wedge Y = y)$$ Evaluating this sum as given requires work proportional to $n^2$; can it be evaluated more efficiently? Because $X$ and $Y$ aren’t necessarily independent, the probability $p(X = x \wedge Y = y)$ can’t be simplified in the same way it was when computing the expected value of random-value products.^?That is, without independence $p(X = x \wedge Y = y)$ is not necessarily equal to $p_X(X=x)p_Y(Y=y)$. Are there some other tricks that can simplify evaluation?

Multiply $p$ through the sum $(x + y)$ and separate the summations to get $$E(X + Y) = \sum_x\sum_y x\,p(X = x \wedge Y = y)\ + \sum_x\sum_y y\,p(X = x \wedge Y = y)$$ Taking the left summation, $x$ is constant in the inner-most summation and can be brought out $$\sum_x x \sum_y \,p(X = x \wedge Y = y)$$ The sum of the probabilities for all possible values of $Y$ is one, but it’s conditioned by the probability that $X$ is equal to some value $x$, $p_X(x)$: $$\sum_x x\,p_X(x)$$ which is $E(X)$.

A similar trick applies to the right summation after the two summations are switched, and, $$E(X + Y) = E(X) + E(Y)$$ A little more algebra shows the expected value is a linear operator: $$E(aX + bY) = aE(X) + bE(Y)$$

Because random variables are closed under addition and multiplication^?That is, if $X$ and $Y)\ are random variables, then so is \(X+Y$ and \(XY)\. and addition and multiplication are associative,^?That is, $X + Y + Z = X + (Y + Z)$ and so on. the sum and product results generalize to any number of terms: $$\begin{array}{lcr} E(\sum_i X_i) &=& \sum_i E(X_i) \\ E(\prod_i X_i) &=& \prod_i E(X_i) \end{array}$$