Rigorous proof that $E(X|X+Z) = E(Y|Y+Z)$ when $X, Y, Z$ are independent and $X\overset{d}{=} Y$

Question

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and let $X, Y, Z:\Omega\rightarrow[-\infty,\infty]$ be independent random variables, such that $X, Y$ are $\mathbb{P}$-integrable and identically distributed. By symmetry considerations, intuitively $E(X|X+Z) = E(Y|Y+Z)$. Can this intuition be given a valid formal statement? For instance, is it true that $E(X|X+Z) \overset{d}{=} E(Y|Y+Z)$? If so, how can this be proved rigorously?

Answer 1

I will show that $$ \mathbb{E}(X|X+Z) \overset{d}{=} \mathbb{E}(Y|Y+Z)\tag{*} $$ by proving the more general claim:

Claim

Let $(\Omega_i, \mathcal{F}_i, \mathbb{P}_i)$ be a probability space for $i \in \{1,2\}$, and let $U^i_1, U^i_2:\Omega_i\rightarrow[-\infty,\infty]$ be random variables. Write $U^i = (U^i_1, U^i_2)$. Assume that

1. $U^i_1$ is $\mathbb{P}_i$-integrable.
2. $U^1 \overset{d}{=} U^2$.
3. $U^1_2 \overset{d}{=} U^2_2$.

Then $\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \overset{d}{=} \mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2)$.

We can obtain $(*)$ from the claim by setting $$ \begin{align*} \Omega_1 &= \Omega_2 = \Omega,\\ \mathcal{F}_1 &= \mathcal{F}_2 = \mathcal{F},\\ \mathbb{P}_1 &= \mathbb{P}_2 = \mathbb{P},\\ U^1_1 &= X,\\ U^1_2 &= X + Z,\\ U^2_1 &= Y,\\ U^2_2 &= Y + Z. \end{align*} $$

Note that since (a) $X, Z$ are independent, (b) $Y, Z$ are independet, (c) $X \overset{d}{=} Y$, we have $(X,Z) \overset{d}{=}(Y,Z)$, and therefore $X+Z \overset{d}{=} Y+Z$ and $(X, X+Z) \overset{d}{=} (Y, Y+Z)$.

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Proof of the claim

By definition of conditional expectation, there are Borel functions $\alpha_1, \alpha_2:[-\infty,\infty]\rightarrow[-\infty,\infty]$ such that $\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2) = \alpha_i\circ U^i_2$. Hence, $\alpha_i\circ U^i_2$ is $\mathbb{P}_i$-integrable. Hence, $\alpha_i$ is $\mathbb{P}_{U^i_2}$-integrable.

For every Borel set $A \subseteq \mathbb{R}$, define $g_A:\mathbb{R}^2\rightarrow\mathbb{R}$ by $g_A(x,y) = x\mathbb{1}_A(y)$. Then $U^i_1\mathbb{1}_{\{U^i_2\in A\}} = U^i_1\big(\mathbb{1}_A\circ U^i_2\big) = g_A\circ U^i$.

Then, for every Borel set $A \subseteq \mathbb{R}$ we have $$ \begin{align*} \int_A \alpha_i\ d\mathbb{P}_{U^i_2} &= \int_{\{U^i_2 \in A\}}\alpha_i\circ U^i_2\ d\mathbb{P}_i\\ &= \int_{\{U^i_2 \in A\}}\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2)\ d\mathbb{P}_i\\ &= \int_{\{U^i_2 \in A\}}U^i_1\ d\mathbb{P}_i\\ &= \int U^i_1\mathbb{1}_{\{U^i_2 \in A\}}\ d\mathbb{P}_i\\ &= \int g_A\circ U^i\ d\mathbb{P}_i\\ &= \int g_A\ d\mathbb{P}_{U^i}. \end{align*} $$

Since $U^1 \overset{d}{=} U^2$, we have $\mathbb{P}_{U^1} = \mathbb{P}_{U^2}$, and therefore $\int_A \alpha_1\ d\mathbb{P}_{U^1_2} = \int_A \alpha_2\ d\mathbb{P}_{U^2_2}$ for every Borel set $A \subseteq \mathbb{R}$. Since $U^1_2 \overset{d}{=} U^2_2$, we have $\mathbb{P}_{U^1_2} = \mathbb{P}_{U^2_2}$, and therefore for every Borel set $A \subseteq \mathbb{R}$ we have $\int_A \alpha_1 - \alpha_2\ d\mathbb{P}_{U^i_2} = 0$, which implies that $\alpha_1 - \alpha_2 = 0$ $\mathbb{P}_{U^i_2}$-a.s., i.e. that $\alpha_1 = \alpha_2$ $\mathbb{P}_{U^i_2}$-a.s.

Now, let $A \subseteq \mathbb{R}$ be a Borel set. Then $$ \begin{align*} \mathbb{P}_i\big(\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2) \in A\big) &= \mathbb{P}_i\big(\alpha_i\circ U^i_2 \in A\big)\\ &= \mathbb{P}_{U^i_2}(\alpha_i \in A)\\ &= \mathbb{P}_{U^i_2}\big(\{\alpha_i \in A\}\cap\{\alpha_1 = \alpha_2\}\big). \end{align*} $$

It follows that $\mathbb{P}_1\big(\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \in A\big) = \mathbb{P}_2\big(\mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2) \in A\big)$ for every Borel set $A \subseteq \mathbb{R}$. In other words, $\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \overset{d}{=} \mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2)$, Q.E.D.

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Remark

The above proof shows that if $(\Omega_1, \mathcal{F}_1, \mathbb{P}_1) = (\Omega_2, \mathcal{F}_2, \mathbb{P}_2)$ and if $U^1_2 = U^2_2$ $\mathbb{P}$-a.s., then $\mathbb{E}(U^1_1|U^1_2) = \mathbb{E}(U^2_1|U^2_2)$ $\mathbb{P}$-a.s. ($\mathbb{P} = \mathbb{P}_1 = \mathbb{P}_2$.)

Indeed, under these conditions define $$ \begin{align*} N_0 &= \{U^1_2 \neq U^2_2\},\\ N_i &= \big\{U^i_2 \in \{\alpha_1 \neq \alpha_2)\}\big\},\quad i \in \{1,2\}\\ N &= N_0 \cup N_1 \cup N_2. \end{align*} $$ Then $$ \begin{align*} \mathbb{P}(N_0) &= 0,\\ \mathbb{P}(N_i) &= \mathbb{P}_{U^i_2}(\alpha_1 \neq \alpha_2) = 0,\quad i \in \{1,2\}\\ \mathbb{P}(N) &= 0, \end{align*} $$ and for every $\omega \in \Omega\setminus N$ we have $$ \big(\mathbb{E}(U^1_1|U^1_2)\big)(\omega) = (\alpha_1\circ U^1_2)(\omega) = (\alpha_2\circ U^1_2)(\omega) = \big(\mathbb{E}(U^2_1|U^2_2)\big)(\omega). $$

In particular, $\mathbb{E}(X|X+Y) = \mathbb{E}(Y|X+Y)$ $\mathbb{P}$-a.s.