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?