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Let $X,Y$ be independent r.v.s on $(\Omega, \mathscr{F}, P)$ having the same distribution i.e. the measures $P\circ X^{-1}= P\circ Y^{-1} : \mathscr{B}(\mathbb{R}) \rightarrow [0,1]$ are equal. Also, $E(|X|), E(|Y|)<\infty$. Then $E[X|X+Y]=E[Y|X+Y]$.

In the proof given here,$\mu_X$, $\mu_Y$ and $\mu_{X,Y}$ denote the distribution measures. I do not understand why

$$\int 1_B(X+Y)X \, dP = \int 1_B(x+y)x\,d\mu_{XY}(x,y)$$

A more general question would be,

if $f= \hat {f} (X,Y):\Omega \rightarrow \mathbb{R}$ where $\hat{f} :\mathbb{R}^2 \rightarrow \mathbb{R}$ is Borel measurable, then does it hold that,$$\int \hat{f}(X,Y) \, dP = \int \hat{f} \, d\mu_{X,Y} $$

Did
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Bryan Shih
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  • Switched to a more relevant title (the previous one was mentioning a result which is not the one you want to see explained). – Did Nov 12 '17 at 16:08

2 Answers2

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Let $(\Omega,\mathcal A,\mathsf P)$ be a probability space.

Every random variable vector $X\to\mathbb R^n$ induces a probability measure $\mathsf P_X$ on $(\mathbb R^n,\mathcal B^n)$.

This by $B\mapsto\mathsf P(X\in B)$, or equivalently $\mathsf P_X(B)=\mathsf P(X\in B)$.

We could also write this equality as:$$\int 1_B(x)\mathsf P_X(dx)=\int 1_B(X(\omega))\mathsf P(d\omega)\tag1$$ In $(1)$ we use indicator function $1_B$ but it can be extended to all Borel measurable functions:$$\int f(x)\mathsf P_X(dx)=\int f(X(\omega))\mathsf P(d\omega)$$

drhab
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  • the problem with this proof is that i don't see how this extends to all measurable functions. We can show it for simple functions by linearity and (1). All nonnegative measurable functions by monotone convergence. This only integrable $f$ if we were to split it into $f^+-f^-$. – Bryan Shih Nov 12 '17 at 15:49
  • @CWL what do you suppose the meaning of the integral is if $f$ is not integrable? – Jason Nov 12 '17 at 16:16
  • Ah, but how do we know $1_B(x+y)x$ is integrable? – Bryan Shih Nov 12 '17 at 16:38
  • That is a different question. Essential is that the integrands on both sides are both integrable or are both not integrable. – drhab Nov 12 '17 at 17:38
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Consider the function $(X,Y) : \Omega\to\mathbb R^2$. This defines the pushforward measure $\mu_{X,Y} = P\circ (X,Y)^{-1}$ on $\mathbb R^2$. It is well known that you integrate a measurable function $f : \mathbb R^2\to\mathbb R$ over $A\subset\mathbb R^2$ with respect to that measure as follows: $$ \int_A f\,d\mu_{X,Y} = \int_{(X,Y)^{-1}(A)}f\circ (X,Y)\,dP. $$ Here, $A = \mathbb R^2$ and $f(x,y) = 1_B(x+y)\cdot x$. So, $(X,Y)^{-1}(A)= \Omega$ and $f\circ (X,Y) = 1_B(X+Y)X$.

Friedrich Philipp
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  • I think I have the same problem with your proof and drhab. I could only show the results for all integrable functions $f$. There is one condition that I forgot to type up, that $E(|X|)<\infty$. Maybe this would help? – Bryan Shih Nov 12 '17 at 15:52
  • Sure that helps, because $E(|X|) < \infty$ means that $X$ is integrable. Consequently, $f$ is integrable. – Friedrich Philipp Nov 12 '17 at 16:04
  • Sorry, why does this imply $f$ is integrable? I could prove "If $f$ is integrable, the equality holds." But not "If $f \circ (X,Y)$ is integrable then equality holds." – Bryan Shih Nov 12 '17 at 16:27
  • See here: https://en.wikipedia.org/wiki/Pushforward_measure#Main_property:change-of-variables_formula. We know that $f\circ (X,Y) = 1_B(X+Y)X$ is integrable since $X$ is. Thus, $f$ is $\mu{X,Y}$-integrable. – Friedrich Philipp Nov 12 '17 at 16:55