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?

Section 2.5 - Product Expected Value

Given independent random variables \(X\) and \(Y\),^?That is, knowning \(X = x\) provides no help in determining what value \(Y\) might be, and vice versa (knowing \(Y = y\) says nothing about \(X\)’s value. what is the expected value of the product \(XY\)?

From the definition of expected vlaue, \[ E(XY) = \sum_x\sum_y xy\,p(XY = xy)\] Because \(X\) and \(Y\) are independent, the probability of their product is the product of their probabilities \[p(XY = xy) = p_X(X = x)\,p_Y(Y = y)\] and the expected value becomes \[ E(XY) = \sum_x\sum_y xy\,p_X(X = x)\,p_Y(Y = y)\] where \(p_X\) is the probability distribution for \(X\) and similarly for \(p_Y\).^?Independence includes not only values, but probability distributions too.

Evaluating the expected value as given takes work proportional to \(n^2\). Can the sum be evaluated with less work?