$E\left[\prod_{i=1}^nX_i\right]=\prod_{i=1}^nE\left[X_i\right]$ for all independent and real-valued random variables

Question

Let $(\Omega,\mathcal{A},P)$ be a measurable space and $X_1,\ldots,X_n:\Omega\to\mathbb{R}$ be independent random variables with $\color{red}{\prod_{i=1}^nX_i\in\mathcal{L}^1(P)}$ $\;\Rightarrow$ $$E\left[\prod_{i=1}^nX_i\right]=\prod_{i=1}^nE\left[X_i\right]$$ Proof:

• Let $$f:\mathbb{R}^n\to\mathbb{R}\;,\;\;\;x\mapsto\prod_{i=1}^nx_i$$
• From basic facts, we know that $f$ is $\mathcal{B}(\mathbb{R}^n)$-$\mathcal{B}(\mathbb{R})$-measurable
• Let $X:=(X_1,\ldots,X_n)$ and $P_X:=P\circ X^{-1}$.
• Since $X_1,\ldots,X_n$ are independent, we've got $$P_X=\bigotimes_{i=1}^nP_{X_i}\tag{1}$$
• Thus, \begin{equation}\begin{split} E\left[\prod_{i=1}^nX_i\right]&=&\int f(x)\;P_X(dx)\\ &\stackrel{(2)}{=}&\int\cdots\int x_1\cdot\ldots\cdot x_n\;P_{X_1}(dx_1)\cdots P_{X_n}(dx_n)\\ &\stackrel{(*)}{=}&\int x_1\;P_{X_1}(dx_1)\;\cdots\int x_n P_{X_n}(dx_n) \\&=&\prod_{i=1}^nE\left[X_i\right] \end{split}\end{equation}

Question: Does $(*)$ hold without further explanations (e.g. Fubini's theorem)? Moreover: How can we show that the $\color{red}{\text{red}}$ assumption holds iff $X_1,\ldots,X_n\in\mathcal{L}^1(P)$.

Answer 1

Once you write an integral over $\mathbb R^n$ as a product/iteration of integrals, we have to justify it. In this context, note that $|x_1\dots x_n|$ is integrable with respect to the measure $P_{X_1}(dx_1)\;\cdots\int x_n P_{X_n}(dx_n)$.

For the red part, we have to use the fact that if $U$ and $V$ are independent and bounded random variables, then $\mathbb E[UV]=\mathbb E[U]\mathbb E[V]$. To this this, start from the case where $U$ and $V$ are linear combination of $\sigma(U)$ (respectively $\sigma(V)$) measurable sets, then approximate. Finally, consider $U_n:=U\chi\{ |U|\leqslant n\} $, $V_n:=V\chi\{ |V|\leqslant n\}$, use the previous case and conclude by dominated convergence.

Answer 2

There are a few issues with regards integrability that require some explanation.

1. By definition, independence of $\{X_1,\ldots,X_n\}$ means that for any Borel sets $A_1,\ldots,A_n$ if the real line, $\mathbb{P}[X_1\in A_1,\ldots,X_n\in X_n]= \mathbb{P}[X_1\in A_1]\cdot\ldots\cdot\mathbb{P}[A_n\in A_n]$. With a little more effort (starting with simple functions, induction and what not) we can see that independence of $\{X_1,\ldots,X_n\}$ is independent iff $$E[f_1(X_1)\cdot\ldots f_n(X_n)]=E[f(X_1)]\cdot\ldots\cdot E[f_n(X_n)]$$ for all bounded measurable functions $f_1,\ldots,f_n$ on the real line.

2. We now show that $Y=X_1\cdot\ldots\cdot X_n$ is integrable: Consider the function $\phi_m(x)=\max(-m,\min(x,m))$. Each $\phi_m$ is a nice (Borel) measurable bounded function. Hence $f_m(Y_j)\in \mathcal{L}_1(P)$ and and $|\phi_m(X_j)|\leq |X_j|$ for all $j=1,\ldots,n$ and $m\in\mathbb{N}$. Then, by Fatou's lemma and dominated convergence \begin{align} E[|Y|]&=\liminf_m E[|\phi_m(X_1)|\cdot\ldots\cdot |\phi_m(Y_n)|]\\ &=\liminf_mE[|\phi_m(X_1)|]\cdot\ldots\cdot E[|\phi_m(X_n)|]\\ &=E[|X_1|]\cdot\ldots\cdot E[|X_n|]<\infty \end{align} 3.Now that we have integration issues out of the way, the conclusion follows again by dominated convergence: \begin{align} E[Y]&=E[X_1\cdot\ldots\cdot X_n]=E[\lim_m\,\phi_m(X_1)\cdot\ldots\cdot\phi_m(X_n)]\\ &=\lim_mE[\phi_m(X_1)\cdot\ldots\cdot\phi_m(X_n)]=\lim_mE[\phi_m(X_1)]\cdot\ldots\cdot E[\phi_m(X_n)]\\ &=E[X_1]\cdot\ldots\cdot E[X_n] \end{align}