If $X$ and $Y$ are iid, then $f(X,Y)$ and $f(Y,X)$ are identitically distributed

Question

Let $X$ and $Y$ be iid. Then $(X,Y) \sim (Y,X)$. Proof:

$$\mathbb P((X,Y)\in B)=\int 1_{B}(X,Y) d \mathbb P=\int_{\mathbb R^2} 1_B(x,y) \mathbb P_{X,Y}(d(x,y))$$ now use independent and identitically distributed \begin{align*} &=\int \int 1_B(x,y) \mathbb P_X(dx)\mathbb P_Y(dy)\\ &=\int \int 1_B (y,x) \mathbb P_X(dy)\mathbb P_Y(dx) \\ &= \int \int 1_B (y,x) \mathbb P_Y(dy)\mathbb P_X(dx)\\ &= \int_{\mathbb R^2} 1_B (y,x) \mathbb P_{Y,X}(d(y,x))\\ &=\mathbb P((Y,X)\in B) \end{align*} Now say $f$ is measurable on $\mathbb R^2$. Then $$\mathbb P(f(X,Y)\in B)=\mathbb P((X,Y) \in f^{-1} B )=\mathbb P((Y,X) \in f^{-1} B )=\mathbb P(f(Y,X)\in B)$$

Is the last step correct?

Answer 1

Your steps are correct.

Alternate solution for the first part:

I assume that $X,Y$ are real valued. Note that $\mathcal B (\Bbb R ^2) = \sigma (\mathcal E)$ with $\mathcal E := \{A \times B : A,B\in \mathcal B (\Bbb R)\}$. Note that $\mathcal E$ is closed under $\cap $-operations. Let $A\times B \in \mathcal E$, then due to independence

$$\Bbb P ((X,Y) \in A\times B) = \Bbb P (X\in A , Y\in B) = \Bbb P (X\in A ) \Bbb P( Y\in B)$$ Further, by the fact that $X$ and $Y$ have the same distribution and are independent we have that $$\Bbb P (X\in A ) \Bbb P( Y\in B) = \Bbb P (Y\in A ) \Bbb P( X\in B) = \Bbb P (Y\in A , X\in B) = \Bbb P ((Y,X) \in A\times B)$$ Therfore we have that $$\Bbb P \circ (X,Y)^{-1} = \Bbb P \circ (Y,X)^{-1}$$ on $\mathcal E$. By the uniqueness of measures theorem (or here) we have that this equality holds on $\sigma (\mathcal E)= \mathcal B (\Bbb R ^2)$.

Answer 2

Your proof is correct.

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Here is a one-line solution using image measures (please, read this carefully, such arguments can buy you lots of time!)

$$\mathbb{P}_{f(X,Y)} = (\mathbb{P}_{(X,Y)})_f = (\mathbb{P}_X \otimes \mathbb{P}_Y)_f = (\mathbb{P}_Y \otimes \mathbb{P}_X)_f = (\mathbb{P}_{(Y,X)})_f = \mathbb{P}_{f(Y,X)}$$

Second equality follows because for independent variables the joint distribution of $(X,Y)$ is equal to the product measure of the distributions. The third equality follows because $X,Y$ have the same distribution.