I have that (X, Y) are uniformly distributed in the triangle defined by the vertices $(0, 0)$, $(1, 0)$, and $(0, 1)$. I am trying to find the following expectation value:
$$ E\left( (X-Y)^2 | X \right) $$
I am able to get the joint density, marginal densities, and conditional densities fairly easily. However, I am a little confused on how to go about solving this?
Best Answer
$$ \mathbb E\left( (X-Y)^2 | X \right)= \mathbb E\left( X^2+Y^2 -2XY | X \right)=X^2+E(Y^2|X)-2XE(Y|X)$$
Since $E(Yf(X)|X)=f(X)E(Y|X)$ Basic_properties.