Translating expected values between two sets of related iid variables

Question

The setting: $\mu$ is a probability measure on $\mathbb{R}$, $f: \mathbb{R} \to [0, \infty)$ so that $0 < ||f||_{L^1(\mu)} < \infty$, and $v$ is another probability measure defined by $v(A) = \frac{1}{||f||_{L^1(\mu)}} \int_A f(x) \mu(dx)$. Also, for any function g, $\int g(x) v(dx) = \frac{1}{||f||_{L^1(\mu)}} \int_{\mathbb{R}} g(x) f(x) \mu(dx)$

The problem: Show that if $X_1, X_2, ..., X_n$ are iid with distribution $\mu$ and $Y_1, Y_2, ..., Y_n$ are iid with distribution $v$, then for any bounded Borel measurable function $F: \mathbb{R}^n \to \mathbb{R}$, $E[F(Y_1 Y_2, ..., Y_n)] = \frac{1}{||f||_{L^1(\mu)}} E[f(X_1) f(X_2) ... f(X_n) F(X_1, X_2, ..., X_n)]$.

I'm reasonably sure that since F is bounded Borel measurable and the $Y_i's$ and $X_i's$ are both random variables, $F(Y_1 Y_2, ..., Y_n)$ and $F(X_1 X_2, ..., X_n)$ are also random variables, maybe with distributions $v^n$ and $\mu^n$ respectively (not a priori necessarily related in any meaningful way to $\mu$ and $v$). If I'm reading my textbook correctly, that means that $E[F(Y_1 Y_2, ..., Y_n)] = \int \int ... \int_{\mathbb{R}^n} F(y_1, ... y_n) v^n(dy_1, ..., dy_n)$, and $\frac{1}{||f||_{L^1(\mu)}} E[f(X_1) f(X_2) ... f(X_n) F(X_1, X_2, ..., X_n)]$ $= \frac{1}{||f||_{L^1(\mu)}} \int \int ... \int_{\mathbb{R}^n} F(x_1, ..., x_n) f(x_1) ... f(x_n) \mu^n(dx_1, ..., dx_n)$, but I'm not really sure how I should interpret either of these statements to make the two integrals equal to each other. How exactly should I relate the $\mu^n$, $v^n$ to the original measures?

Answer 1

First of all,

$$\mathbb{E}(F(Y_1,\ldots,Y_n)) = \int \dots \int F(y_1,\ldots,y_n) \, d\mathbb{P}_{Y_1,\ldots,Y_n}(y_1,\ldots,y_n) \tag{1}$$

where $\mathbb{P}_{Y_1,\ldots,Y_n}$ denotes the joint distribution of the random vector $(Y_1,\ldots,Y_n)$ with respect to $\mathbb{P}$. Since the random variables are independent, the joint distribution equals the product of the marginals, i.e.

$$d\mathbb{P}_{Y_1,\ldots,Y_n}(y_1,\ldots,y_n) = d\nu(y_1) \dots d\nu(y_n) = \frac{1}{\|f\|_{L^1}^n} f(y_1) \, d\mu(y_1) \ldots f(y_n) \, d\mu_n.$$

Plugging this into $(1)$ yields

$$\mathbb{E}(F(Y_1,\ldots,Y_n)) =\frac{1}{\|f\|_{L^1}^n} \int \dots \int F(y_1,\ldots,y_n) f(y_1) \cdots f(y_n) \, d\mu(y_1) \dots d\mu_n(y_n).$$

Using again the independence, we get

$$\begin{align*} \mathbb{E}(F(Y_1,\ldots,Y_n)) &=\frac{1}{\|f\|_{L^1}^n} \int \dots \int F(y_1,\ldots,y_n) f(y_1) \cdots f(y_n) \, d\mathbb{P}_{X_1,\ldots,X_n}(y_1,\ldots,y_n) \\ &= \frac{1}{\|f\|_{L^1}^n}\mathbb{E}(F(X_1,\ldots,X_n) \cdot f(X_1) \cdots f(X_n)). \end{align*}$$

Answer 2

This is because $(Y_1,\ldots,Y_n)$ has density $$ (y_1,\ldots,y_n)\mapsto \|f\|^{-n}f(y_1)\cdots f(y_n) $$ with respect to $\mu^n$. Here $\mu^n$ is the product measure of $\mu$ by itself $n$ times, i.e. , $\mu^n=\mu\otimes\cdots\otimes\mu$. This follows immediately by the independence assumption and that $\nu$ has density $\|f\|^{-1} f$ with respect to $\mu$. For $A=A_1\times\cdots\times A_n$ with $A_1,\ldots,A_n$ being Borel sets, one has $$ \begin{align} P_{(Y_1,\ldots,Y_n)}(A)&=\prod_{i=1}^n\nu(A_i)=\prod_{i=1}\|f\|^{-n}\int_{A_i} f(y)\mu(\mathrm dy)\\ &=\int_A \|f\|^{-n}f(y_1)\cdots f(y_n)\mu^n(\mathrm dy_1,\ldots,\mathrm dy_n) \end{align} $$ where in the last step we have used Tonelli (everything is non-negative). Since this holds for all measurable rectangles, it holds for any Borel set $A\in\mathcal{B}(\mathbb{R}^n)$ showing that $(Y_1,\ldots,Y_n)$ has the wanted density.