1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional Expectation

  1. Jan 2, 2008 #1
    [SOLVED] Conditional Expectation

    I'm trying to understand the following proof I saw in a book. It says that:
    [tex]E[Xg(Y)|Y] = g(Y)E[X|Y][/tex] 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:

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

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

    But, would this formula also work for say 3 variables? That is, [tex]E[Xg(Z)|Y] = g(Z)E[X|Y][/tex], 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 [tex]E[Xg(Z)|Y] = g(Z)E[X|Y][/tex] ??
     
    Last edited: Jan 2, 2008
  2. jcsd
  3. Jan 2, 2008 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    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].
     
  4. Jan 2, 2008 #3
    Ah yes, that clears things up. Cheers, you've been great help once again!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Conditional Expectation
Loading...