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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


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    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!!
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