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Let $X, W, Y, Z$ be $\mathbb{R}$-valued random variables on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that $Y, Z, W$ are i.i.d. and $\sigma(X)$ is independent of $\sigma(W, Y, Z)$. Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a measurable function and $B \in \mathcal{B}(\mathbb{R})$.

Is it true that $$ \mathbb{P} (f(X, Y) \in B, f(X, Z) \in B) = \mathbb{P}(f(X, Y) \in B, f(X, W) \in B). $$

The probability on the left-hand side depends on the joint law of $X$, $Y$, $Z$, which is given by the joint law of $X$ and $(Y, Z)$. Since $X$ is independent of $(Y, Z)$, the joint law is given by the product of the two laws. But $(Y, Z)$ has the same law as $(Y, W)$, and so the joint law of $X, Y, Z$ is the same as that of $X, Y, W$. Can this be used to show the equality?

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  • $\begingroup$ Yes, since $(X,Y,Z)$ has the same joint CDF as $(X,Y,W)$, we know $(X,Y,Z)$ has the same distribution as $(X,Y,W)$ and so for all Borel measurable sets $A \subseteq \mathbb{R}^3$ we have $$P[(X,Y,Z) \in A] = P[(X,Y,W) \in A]$$ $\endgroup$
    – Michael
    Aug 14, 2022 at 15:05

1 Answer 1

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The statement is true. By Theorem $20.3$ of Probability and Measure and the independence assumption, \begin{align} & P[f(X, Y) \in B, f(X, Z) \in B] = \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, Z) \in B]\mu(dx), \\ & P[f(X, Y) \in B, f(X, W) \in B] = \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, W) \in B]\mu(dx), \end{align}

where $\mu$ is the distribution of $X$.

It is therefore sufficient to show $P[f(x, Y) \in B, f(x, Z) \in B] = P[f(x, Y) \in B, f(x, W) \in B]$ for any fixed $x \in \mathbb{R}^1$, which is the consequence of (use the i.i.d. assumption):
\begin{align*} & P[f(x, Y) \in B, f(x, Z) \in B] = P[f(x, Y) \in B]P[f(x, Z) \in B] \\ =& P[f(x, Y) \in B]^2 \\ =& P[f(x, Y) \in B]P[f(x, W) \in B] = P[f(x, Y) \in B, f(x, W) \in B]. \end{align*}


More details of applying Theorem $20.3$: basically identify the $X$ in the question as the $X$ in the theorem, and identify $(Y, Z)$ (or $(Y, W)$) in the question as the $Y$ in the theorem. Argue as follows:

Define the mapping $g: \mathbb{R}^3 \to \mathbb{R}^2$ by $g(x, y, z) = (f(x, y), f(x, z))$. Clearly $g$ is measurable $\mathscr{R}^3/\mathscr{R}^2$ and $[f(X, Y) \in B, f(X, Z) \in B] = [g(X, Y, Z) \in B \times B]$. Therefore,

\begin{align} & P[f(X, Y) \in B, f(X, Z) \in B] = P[g(X, Y, Z) \in B \times B] \\ =& P[(X, Y, Z) \in g^{-1}(B \times B)] \\ =& \int_{-\infty}^\infty P[(x, Y, Z) \in g^{-1}(B \times B)]\mu(dx) \\ =& \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, Z) \in B]\mu(dx). \\[2em] & P[f(X, Y) \in B, f(X, W) \in B] = P[g(X, Y, W) \in B \times B] \\ =& P[(X, Y, W) \in g^{-1}(B \times B)] \\ =& \int_{-\infty}^\infty P[(x, Y, W) \in g^{-1}(B \times B)]\mu(dx) \\ =& \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, W) \in B]\mu(dx). \end{align}

P.S. Using the notation I introduced above, I think your (verbal) argument in the question actually works too (and gives an even shorter proof), as it fully takes advantage of the fact $(X, Y, Z) \overset{d}{=} (X, Y, W)$ under given assumptions. If we denote their common distribution on $(\mathbb{R}^3, \mathscr{R}^3)$ as $\eta$ (it can be easily seen that $\eta = \mu \times \nu \times \nu$, which is a product measure of the distribution $\mu$ of $X$ and the distribution of $\nu \times \nu$ of $(Y, Z)$ (or $(Y, W)$)), and note that $g^{-1}(B \times B)$ is a Borel set in $\mathscr{R}^3$, it follows that

\begin{align} & P[f(X, Y) \in B, f(X, Z) \in B] = P[(X, Y, Z) \in g^{-1}(B \times B)] \\ =& \eta(g^{-1}(B \times B)) \\ =& P[(X, Y, W) \in g^{-1}(B \times B)] = P[f(X, Y) \in B, f(X, W) \in B]. \end{align}

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  • $\begingroup$ The theorem states that if $X$ and $Y$ are independent, then for $A \in \mathcal{B}(\mathbb{R})$ and $B \in \mathcal{B}(\mathbb{R})^2$ $$ \mathbb{P}((X, Y) \in B) = \int_{\mathbb{R}} (\mathbb{P} (x, Y) \in B) \, \mu (dx), $$ $$ \mathbb{P}(X \in A, (X, Y) \in B) = \int_{A} (\mathbb{P} (x, Y) \in B) \, \mu (dx). $$ Could you clarify how you apply this exactly? We can of course write $$ \mathbb{P}(f(X, Y) \in B, f(X, Z) \in B) = \mathbb{P}((X, Y) \in f^{-1}(B), (X, Z) \in f^{-1}(B)). $$ $\endgroup$
    – Harry
    Aug 14, 2022 at 11:02
  • $\begingroup$ @Harry Good question. The argument is too long for a comment. See my edited answer. $\endgroup$
    – Zhanxiong
    Aug 14, 2022 at 14:40

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