Calculating Nested Expectation

Question

If $X $and $Y$ are random variables and $c$ is a constant, is it valid to say that $E(XY-2E(X)Y-c(Y-E(Y)))=E(XY)-E(XY)-c(E(Y)-E(Y))$

Answer 1

No; most of your calculation is correct, but there's one small error. You can recall that expectations are linear which means two things: first, if $X$ and $Y$ are random variables, then $$\mathbb E[X+Y]=\mathbb E[X]+\mathbb E[Y]$$ and also if $X$ is a random variable and $c$ is a constant, then $$\mathbb E[cX]=c\mathbb E[X].$$ Your manipulation is fine in breaking apart the terms except that it appears that you transformed the term $\mathbb E[2\mathbb E[X]Y]$ into $\mathbb E[XY]$, which is incorrect.

Rather, what you can do is note that $2\mathbb E[X]$ is a constant, so you can pull it out as $2\mathbb E[X]\mathbb E[Y]$. This is likely not the same as $2\mathbb E[XY]$ unless $X$ and $Y$ are independent. For instance, if $X$ was a normally distributed random variable with mean $0$ and standard deviation $1$ and $Y=X$, one would have $\mathbb E[X]\mathbb E[Y] = 0$ whereas $XY$ would always be positive and $\mathbb E[XY] = 1$.