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Let $X$ a random variable on $(\Omega,\mathcal{F},P),\mathcal{G}$ a sub $\sigma$-algebra. Prove that, almost surely,

$$E[X^2|\mathcal{G}]=2\int_0^{\infty}xP(X>x|\mathcal{G})dx$$ The above formula was true for simple function $X,$ if we pick an increasing function $X_n$ to $X,$ there is a problem in applying the conditional monotone theorem in $\int_0^{\infty}xP(X_n>x)dx$ because $P(X_n>x|\mathcal{G}) \to P(X>x | \mathcal{G})$ on an event $E_x$ (which depends of $x>0$)such that $P(E_x)=1,$ so we can't change interchange limit-integral.

What do you suggest to solve this problem?

john
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    There is a big issue: For each $x$, $P(X>x \mid \mathcal{G})$ is only determined up to a set of measure zero. If for each $x$, we fix a choice for the random variable $P(X>x \mid \mathcal{G})$, how to guarantee the joint measurability of the map $(x,\omega)\mapsto P(X>x \mid \mathcal{G})(\omega)$? – Danny Pak-Keung Chan Mar 05 '21 at 23:14
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    You need regular conditional probability or its alike in order to even make sense of the integral in the right-hand side, precisely because of the reason pointed out by other users. – Sangchul Lee Mar 06 '21 at 05:00

2 Answers2

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I have got a slightly different answer. Let $P_{\mathcal{G}}$ be your conditional measure (that is $P_{\mathcal{G}}(A)=P(A|\mathcal{G})$), then $$ E[X^2|\mathcal{G}]= \int X^2 \,dP_{\mathcal{G}}= \int \int_{0}^{\infty}1(X^2>l)\,dl \,dP_{\mathcal{G}} = \int \int_{0}^{\infty}1(|X|>\sqrt{l})\,dl \,dP_{\mathcal{G}}. $$ Now let $x=\sqrt{l}$ then $$ E[X^2|\mathcal{G}]=2\int \int_{0}^{\infty}x1(|X|>x)\,dx \,dP_{\mathcal{G}}. $$ Now use Tonelli's theorem and interchange the integral to get

$$ E[X^2|\mathcal{G}]=2\int_{0}^{\infty}x\int 1(|X|>x) \,dP_{\mathcal{G}}\,dx = 2 \int_{0}^{\infty}xP(|X|>x|\mathcal{G})\,dx. $$

Suman Chakraborty
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Let $\nu$ denote the regular conditional distribution of $X$ given $\mathcal{G}$. Then for any suitable function $g$, \begin{align} \mathsf{E}[g(X)\mid \mathcal{G}](\omega)&=\int g(x)\nu(\omega,dx) \\ &=g(0)+\int_0^\infty g'(x)\nu(\omega,(x,\infty))\, dx-\int_{-\infty}^0 g'(x)\nu(\omega,(-\infty,x])\, dx \quad\text{a.s.} \end{align} (See, e.g., this question.) Now, plug $g(x)=x^2$.

  • The first line is this a version of Fubini theorem? Also, can't see in the second equality how did you obtain $\nu(w,(x,\infty))$? – john Mar 06 '21 at 00:06
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    Look at Proposition 10.16 on page 12 here. Also, have you looked at the referenced post? Just replace $\mathsf{P}(X>x)$ with $\nu(\cdot, (x,\infty))$ there. –  Mar 06 '21 at 00:20