# Showing that the product of two integrable complex valued random variables is integrable

Let $$S_1 = A_1 + iB_1, S_2 = C_2 + iD_2$$ be two integrable complex valued random variables on the probability space $$(\Omega, F, P)$$, where $$A_1, B_1, C_2, D_2$$ are real valued r.v.s on the same space. Then, in order to show that the product $$S_1S_2$$ is integrable, I have to show that the products of the real and imaginary parts of $$S_1, S_2$$ are integrable as well, i.e. $$\mathbb{E}(A_1B_1) < +\infty$$ etc. I do know that the product of two independent integrable r.v.s is integrable as well, but evidently we do not have that luxury here. Therefore, I was wondering whether the following applicaiton of the Fubini's theorem would help: Define $$T_1, T_2 \in \{A_1, B_1, C_2, D_2\}, T_1 \neq T_2$$. Then clearly $$\left|T_1\cdot T_2\right|$$ is measurable w.r.t. the product sigma-algebra $$F \otimes F$$ (isn't it true that this would hold event if $$S_1$$ and $$S_2$$ would live in different probability spaces?). So now the expected value of the absolute value of the component product is: $$\mathbb{E}(|T_1\cdot T_2|) = \int_{\Omega^2}|T_1\cdot T_2|d(P\otimes P) = \int_{\Omega_1}\left(\int_{\Omega_2}|T_1(\omega_1)\cdot T_2(\omega_2)|dP(\omega_2)\right)dP(\omega_1)$$ from which we obtain that $$\int_{\Omega_1}|T_1(\omega_1)|\left(\int_{\Omega_2}|T_2(\omega_2)|dP(\omega_2)\right)dP(\omega_1) = \int_{\Omega_1}|T_1(\omega_1)|\mathbb{E}(T_2)dP(\omega_1) = \mathbb{E}(T_1)\mathbb{E}(T_2) < +\infty$$ by assumption, as $$S_1, S_2 \in \mathcal{L}^1(P)$$.

If the above reasoning was valid, then this would show that the product of $$S_1$$ and $$S_2$$ is integrable as a linear combination of integrable functions.

Note that by definition, $$X \in L^1$$ if and only if $$|X| \in L^1$$. Since $$|S_1S_2| = |S_1||S_2|$$, this means that you can WLOG assume that $$S_1, S_2 \geq 0$$. Still, this reduction doesn't help since the proposition is false. The product of two $$L^1$$ variables is in general not $$L^1$$. For example, pick $$p \in \mathbb{R}$$ such that $$x^p \in L^1([0, 1])$$ but $$x^p \cdot x^p = x^{2p} \notin L^1([0, 1])$$.

• Hmm, I see. I suppose if we make the assumptions more stringent by requiring that $S_1$ and $S_2$ are independent, then this proof is quite straightforward: By assumption the components of $S_1$ and $S_2$ must be mutually independent since otherwise $S_1$ and $S_2$ would not be independent. Commented Feb 17, 2022 at 20:36