Question

Formula for conditional expectation?

Answer 1

If assuming the continuous case, I know that
\[E[g(X)|Y=y]=\int_{-\infty}^{\infty}g(x)~f_{X|Y} (x|y) dx\]

But if we have something more general, like
\[E[g(X,Y)|Y=y]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x,y)~f_{X|Y}(x|y)~dx~dy?\] Would that be the correct formula to compute such a conditional expectation?

Answer 2

I'm having trouble finding this online

Answer 3

http://isfaserveur.univ-lyon1.fr/~stephane.loisel/prerequis_esp_cond.pdf
\[E(Y |X = x) = Z Ω Y (ω)P(dw|X = x)\]
\[ = R X=x Y (ω)P(dw) P(X = x) \]
\[= E(Y 1(X=x)) P(X = x)\]

Answer 4

i don't think I'm quite that advanced.. I never studied sigma fields :S

Answer 5

Ok I finally found something: http://math.arizona.edu/~tgk/563/cond_exp.pdf
page 4. They used the discrete case. What's the intuition of having only one summation/integral, is it because if we're given \(Y=y\), then it's sort of considered like a constant, so we just sum/integrate over x? @ForsureYunome