When can I say "WLOG that $E(X)=E(Y)=0$" in the proof of probability inequalities In Grimmett's 《Probability and Random Processes》, there is an exercise in Section 4.6 :
Let $X$ and $Y$ be random variables with correlation $\rho$. Show that $E(var(Y|X))\leq(1-\rho^2)var(Y)$
And in the solution manual, it is claimed in the first line that "We may assume without loss of generality that $EX=EY=0$", and I can't figure out under what circumstances can I apply the technique.
Best regards!
 A: Define variables $Y’ = Y - \mathbb{E}[Y]$ and similarly for $X’$. Then $X’$ and $Y’$ have correlation $\rho$, and $\mathbb{E}[var(X’ \mid Y’)] = \mathbb{E}[var(X \mid Y)]$. Furthermore, $\mathbb{E}[X’] = \mathbb{E}[Y’] = 0$. This is why we can assume WLOG that the expected values of $X$ and $Y$ are zero - we can reduce to this case using $X’$ and $Y’$.
A: I don't have a intuitive explanation why we the expected values do not affect $E\left( Var(Y|X)\right)$. So I show the term when the random variables $X,Y$ do have an expected value and are normal distributed with the correlation coefficient $\rho$. From the law of total variance we know that
$${Var}(Y) = E[{Var}(Y |X)] + Var(E[Y |X])\Rightarrow E[{Var}(Y |X)]={Var}(Y)-Var(E[Y |X])  $$
And you probably know that
$$E[Y |X]=\mu_y+\rho \frac{\sigma_y}{\sigma_x}(X-\mu_x)$$
If we now calculate the variance of it, the expected values disappear. Thus $E\left( Var(Y|X)\right)$ has no expected value at all. I agree to the comments that the w.l.o.g assumption should have been explained. Maybe it is explained in the corresponding chapter of the book.
