Interchanging random variables with the same distribution

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?

• 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 $$P[(X,Y,Z) \in A] = P[(X,Y,W) \in A]$$ Aug 14, 2022 at 15:05

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}

• 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 $$\mathbb{P}(f(X, Y) \in B, f(X, Z) \in B) = \mathbb{P}((X, Y) \in f^{-1}(B), (X, Z) \in f^{-1}(B)).$$ Aug 14, 2022 at 11:02
• @Harry Good question. The argument is too long for a comment. See my edited answer. Aug 14, 2022 at 14:40