Sum of Independent Random Variables is in L1 if and only if Random Variables are in L1

Last updated Nov 1, 2022

Suppose $(\Omega, \mathcal{M}, \mathbb{P})$ is a Probability Space and $X, Y$ are independent Random Variables on $\Omega$. Then $X+Y \in L^{1}(\Omega)$ If and Only If $X, Y \in L^{1}(\Omega)$.

($\Leftarrow$): This is the easy direction. Suppose $X \in L^{1}$ and $Y \in L^{1}$. Since Lp Spaces are Vector Spaces, we have that $X + Y \in L^{1}$ $\checkmark$

($\Rightarrow$): Suppose $X+Y \in L^{1}$. Recall from the definition of L1 Integrable Functions, this means $$\int\limits_{\Omega} |X(\omega)+Y(\omega)| d \mathbb{P}(\omega) < \infty$$ Since $(X, Y)$ are independent, they induce a product measure $F_{X,Y}$ on $\mathbb{R}^{2}$ where for $A,B \in \mathbb{R}$, $F_{X,Y}(A \times B) = F_{X}(A)F_{Y}(B)$ ($F_{X}$ and $F_{Y}$ are the Induced Probability Measure of $X$ and $Y$ respectively). Furthermore

$$\begin{align*} \int\limits_{\Omega} |X(\omega)+Y(\omega)| d \mathbb{P}(\omega) &= \int\limits_{\mathbb{R}^{2}} |x+y| F_{X,Y}(d(x,y))\\ &=\int\limits_{\mathbb{R}} \left[\int\limits_{\mathbb{R}} |x+y| F_{X}(dx)\right] F_{Y}(dy) \end{align*}$$ and $y \mapsto \int\limits_{\mathbb{R}} |x+y| F_{X}(dx) \in L^{1}(\Omega)$ by Fubini’s Theorem. Thus, for $y \in \mathbb{R}$ Almost Everywhere we have that $\int\limits_{\mathbb{R}} |x+y| F_{X}(dx) < \infty$. Pick some such $y \in \mathbb{R}$. Then

$$\begin{align*} \int\limits_{\Omega} |X(\omega)| d \mathbb{P}(\omega) &= \int\limits_\mathbb{R} |x| F_{X}(dx)\\ &=\int\limits_\mathbb{R} |x+y-y| F_{X}(dx)\\ &\leq \int\limits_\mathbb{R} |x+y| F_{X}(dx) + \int\limits_\mathbb{R} |y| F_{X}(dx)\\ &=\int\limits_\mathbb{R} |x+y| F_{X}(dx) + |y|\\ &< \infty \end{align*}$$ and $X \in L^{1}(\Omega)$. Swap $X$ and $Y$ to get the same result for $Y$. $\checkmark$ $\blacksquare$

• Expectation
• Fubini’s Theorem