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.? Are there some other tricks that can simplify evaluation?