Question

Conditional expectation "substitution rule" \[E[g(X,Y)|Y=y] = E[g(X, y)|Y=y]\]. How can I prove this?

Answer 1

@reemii

Answer 2

I tried writing the LHS as:

Answer 3

\[\int_{-\infty}^{\infty}g(x,y)~f_{X|Y}(x|y)~dx\]

Answer 4

that is correctly done at the moment.

Answer 5

and I ended up writing the same thing for the RHS ... so, that means both are true? It seems a bit trivial to me :S

Answer 6

sorry, what you wrote is actually the RHS.

Answer 7

This is an intuitive result:
If P(Y=y)>0 (denominator issue), then \[E[G|Y=y] = \frac{E[G 1\!\!\!1(Y=y)]}{P(Y=y)}\]
Take your time to consider this. It is quite intuitive.

Answer 8

\(1\!\!\!1(A)\)) is the indicator variable of the set A.

Answer 9

what it says is that the average value of G given that Y=y, is the average value of G, restricted to the set of possibilities for which Y=y, then scaled by the right factor. (to understand the scaling, consider a constant function for G)

Answer 10

With this, the LHS is \(\frac{E[g(X,Y)1(Y=y)]}{P(Y=y)}\).
The RHS is \(\frac{E[g(X,y)1(Y=y)]}{P(Y=y)}\).
And now you notice that if \(Y\neq y\), then the indicator variable \(1(Y=y)\) is equal to zero. Therefore, only the \(Y=y\) is counted in this expectation, and so you can replace \(g(X,Y)\) by \(g(X,y)\).

Answer 11

Sorry for the delay. I didn't log in for a while. Omg thank you sooo much though :D

Answer 12

:)