- #1
nonequilibrium
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- 2
Homework Statement
Well, not really, but in essence that's the part I'm having trouble with. The actual question is
The equality seems obvious enough, but I'm unsure how to actually prove it...Show that for continuous random variables, [itex]E(Xh(Y)|Y) = h(Y) E(X|Y)[/itex].
Homework Equations
N/A
The Attempt at a Solution
So it seems I have to prove that [itex]P( \{ E(Xh(Y)|Y) = h(Y) E(X|Y) \} ) = 1[/itex].
Can anybody tell me how they would start?
I shall say how I would proceed if pressed:
As I see it, E(Xh(Y)|Y) is actually a function with possible values [itex]E(X h(Y) | Y = y_0)[/itex] and we can write [itex]E(X h(Y) | Y = y_0) = \iint x h(y) f_{X,Y|Y=y_0}(x,y) \mathrm d x \mathrm d x[/itex] and since [itex]f_{X,Y|Y=y_0}(x,y) = f_{X|Y=y_0} \delta(y-y_0)[/itex] (or perhaps this needs to be proven too?) we get [itex]E(X h(Y) | Y = y_0) = h(y_0) \int x f_{X|Y=y_0}(x) \mathrm d x = h(y_0) E(X|Y) [/itex] and hence E(Xh(Y)|Y) is the same random variable as h(Y) E(X|Y).
Is this acceptable?