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Iirc, one of my professors offered the following proof for total expectation:

If $X \in L^1 (\Omega, \mathscr{F}, \mathbb{P})$ has pdf $f_X$ and $Y > \in L^1 (\Omega, \mathscr{F}, \mathbb{P})$ has pdf $f_Y$, then $E[E[X|Y]] = E[X]$, assuming $f_{X|Y=y}$ exists.

Pf: $LHS = \int_\mathbb{R} (E[X|Y=y]) f_Y(y) dy$

$=\int_\mathbb{R} (\int_\mathbb{R} x f_{X|Y=y} (x)) f_Y(y) dy$

$=\int_\mathbb{R}\int_\mathbb{R} x f_{X|Y=y} (x) dx f_Y(y) dy$

$=\int_\mathbb{R}\int_\mathbb{R} x f_{XY} (x,y) dx dy$

$=\int_\mathbb{R}\int_\mathbb{R} x f_{XY}(x,y) dy dx$ (*)

$=\int_\mathbb{R} x \int_\mathbb{R} f_{XY}(x,y) dy dx$

$=\int_\mathbb{R} x f_X dx$

$=RHS$

QED

Is that right?

If it is wrong, why?

If it is right, what is the justification for (*)?

I thought the justification is that

$$\int_\mathbb{R}\int_\mathbb{R} f(x,y) dx dy = \int_\mathbb{R}\int_\mathbb{R} f(x,y) dy dx$$

$\because f(x,y) \geq 0$ and by Tonelli's theorem for non-negative functions

but actually it may not extend to saying that

$$\int_\mathbb{R}\int_\mathbb{R} \color{red}{x} f(x,y) dx dy = \int_\mathbb{R}\int_\mathbb{R} \color{red}{x} f(x,y) dy dx$$

since $X \geq 0$ is not assumed.

My professor said that it has something to do with uniform or Lipschitz continuity. Does it?

BCLC
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  • If the random variables are $L^1$, you can use Fubini's theorem to switch the order of integration. – yarmenti Sep 21 '15 at 10:42
  • @yarmenti How? $ $ – BCLC Sep 21 '15 at 10:49
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    FYI, in general, $$E[E[X|Y]] \ne \int_\mathbb{R} E[X|Y] f_Y(y) dy=E[X|Y] .$$ – Did Sep 21 '15 at 10:58
  • @Did Why? It looks to be the same as in the discrete case here. – BCLC Sep 21 '15 at 11:02
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    Yeah--except that since you copy things without knowing their definitions, slips do occur, which make that what you write and what you are copying from are two different matters. – Did Sep 21 '15 at 11:04
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    Maybe a problem of notation, I guess Did wants to say that: $E(E(X|Y)) = \int_{\mathbb{R}} E(X|Y=y) f_Y(y) dy$ ... – yarmenti Sep 21 '15 at 11:05
  • @Did I didn't copy from the discrete case :P I tried to rewrite the proof from memory. So is it just the $= y$ ? – BCLC Sep 21 '15 at 11:06
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    Second problem (more serious, and already explained in details multiple times to you): in your setting, the object $f_{X\mid Y=y}$ might not exist at all. – Did Sep 21 '15 at 11:06
  • @yarmenti Thanks! Again how can we use Fubini's? – BCLC Sep 21 '15 at 11:07
  • @Did Thanks! Edited – BCLC Sep 21 '15 at 11:08

1 Answers1

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Since $f_{X,Y}(x,y)$ is non-negative when $x$ is non-negative, and $-x\;f_{X,Y}(x,y)$ is non-negative when $x$ is negative, we can apply Tonelli's Theorem for non negative functions by separating the integral into two.

$$\begin{align}\int_\Bbb R\int_\Bbb R x\;f_{X,Y}(x,y)\operatorname d x\operatorname d y & =\int_\Bbb R\left(\int_{\Bbb R^+} x f_{X,Y}(x,y) \operatorname d x-\int_{\Bbb R^-} (-x)f_{X,Y}(x,y)\operatorname d x\right)\operatorname d y \\[1ex] ~ & = \int_{\Bbb R^+} x\int_{\Bbb R}f_{X,Y}(x,y)\operatorname d y\operatorname d x-\int_{\Bbb R^-}(-x)\int_\Bbb R f_{X,Y}(x,y)\operatorname d y\operatorname d x \\[1ex] ~ & = \int_\Bbb R x\int_\Bbb R f_{X,Y}(x,y)\operatorname d y\operatorname d x\end{align}$$

Graham Kemp
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