# Conditional Expectation

1. Jan 2, 2008

### jimmy1

[SOLVED] Conditional Expectation

I'm trying to understand the following proof I saw in a book. It says that:
$$E[Xg(Y)|Y] = g(Y)E[X|Y]$$ where X and Y are discrete random variables and g(Y) is a function of the random variable Y.

Now they give the following proof:

$$E[Xg(Y)|Y] = \sum_{x}x g(Y) f_{x|y}(x|y)$$
$$= g(Y)\sum_{x}x f_{x|y}(x|y)$$
$$= g(Y)E[X|Y]$$

Now, the proof is very simple as they are just using the definition of conditional expectation (ie. $$E[X|Y]= \sum_{x}x f_{x|y}(x|y)$$).

But, would this formula also work for say 3 variables? That is, $$E[Xg(Z)|Y] = g(Z)E[X|Y]$$, where Z is another discrete random variable.

It probably isn't right, but from the above proof I can't immediatley see what's wrong with it, as I'll be just switching the g(Y) with a g(Z), and as the summation in the proof is over x, I can take g(Z) out of the summation and similarly get the result $$E[Xg(Z)|Y] = g(Z)E[X|Y]$$ ??

Last edited: Jan 2, 2008
2. Jan 2, 2008

### EnumaElish

No, the proof works because the event {Xg(Y)|Y=y} = {Xg(y)|y} = g(y){X|y}; but the event {Xg(Z)|y} cannot be written similarly, in general.

In general, V = Xg(Z) is a new random variable and has its own distribution and conditional distribution; so the sum is over v.

If X and Z are independent then E[Xg(Z)|y] = E[X|y]E[g(Z)|y].

If Z = h(Y) then E[Xg(h(y))|y] = g(h(y))E[X|y].

3. Jan 2, 2008

### jimmy1

Ah yes, that clears things up. Cheers, you've been great help once again!!