Conditional Expectation $E(X\mid X+Y=k)$ when Joint pdf is given.

Question

$$f(x,y)=4 e^{-2(x+y)}$$ with $x,y > 0$.

What would be $$E(X\mid X+Y= 4), ?$$ I tried by putting Y=$4$-X and then using usual formula of conditional expectation but that doesn't seem to be correct. Do we have to use transformation? I know here $X$ and $Y$ are independent so $X\mid X+Y$ would follow $Beta(1,1)$ or uniform $(0,1)$, so do we need to use this?

In general how do we proceed in these type of questions. It's simple in discrete cases but what about continuous ones?

Yes. but that's not the case if X+Y is given to be equal to some constant right? — Ankita Goyal, Dec 22 '22 at 09:11
I meant to say $E(X|X+Y)=\frac {X+Y} 2$ so $E(X|X+Y=k)=\frac k 2$. — geetha290krm, Dec 22 '22 at 09:28
X and Y are independent exponential with parameter 2 . So, X|X+Y follows Beta of first kind with parameters 1 and 1, implies Expectation is 1/2. Am I wrong here? Also The answer to this question is 4. — Ankita Goyal, Dec 22 '22 at 09:38

score 1 · Accepted Answer · answered Dec 22 '22 at 09:15

1

It is easy to see $X$ and $Y$ are IID, and so by symmetry :

$\mathbb{E}(X|X + Y) = \mathbb{E}(Y|X+Y) = \frac{1}{2}\mathbb{E}(X+Y|X+Y) \\ = \frac{1}{2} (X + Y)$

It follows $\mathbb{E}(X|X + Y = 4) = 2$

answered Dec 22 '22 at 09:15

Aditya Dhawan

987

Conditional Expectation $E(X\mid X+Y=k)$ when Joint pdf is given.

1 Answers1