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jimmy1
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[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] ??
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] ??
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