# Conditional Expectation E[X] = E[X|Y<=a] + E[X|Y>a]

Is it generally true that for random variables $X$ and $Y$, regardless of being dependent or independent, that $E[X] = E[X \mid Y \le a] + E[X \mid Y>a]$ ?

• Assume $X$ and $Y$ are highly correlated standard normal variables. Then $E[X\mid Y\le L] + E[X \mid Y > L] \approx 0 + L = L$ when $L$ is a very large number. If you assume that $L$ is a large negative number you get a contradiction. – Hunaphu May 7 '15 at 13:00
• Consider a homely analogue: is the average height equal to the average height for those below average weight PLUS the average height for those above average weight? Answer: No. – Nick Cox May 7 '15 at 13:03

Not quite, if we use the law of total expectation we would have that

$$E(X) = E(X| Y \le a)P(Y \le a) + E(X|Y > a) P(Y>a)$$

Exactly Chappers, and I can just add if $X$ and $Y$ are independent continuous random variables, $E[X|Y]=E[X]$.

So Mauro, your formula would be false because in this case, you'll have

$E[X]=E[X|Y\le a]+E[X|Y>a]=E[X]+E[X]=2*E[X]$

Finally, regardless the dependence between $X$ and $Y$, we have always:

$E[X]=E[E[X|Y]]$

• Ah yeah sorry I feel dumb now ;_; I was trying to prove it with like integrals. Just to verify though is: f(x, y<=a) +f(x,y>a) = f(x,y) Where f is the joint distribution of X,Y. If so I think I got it from the integrals too. – Mauro Augusto May 7 '15 at 13:15