If the sum of two independent random variables is in $L^2$, is it true that both of them are in $L^1$?

Question

Let $X$ and $Y$ be two independent random variables. If $\mathbb E(X+Y)^2 < \infty$, do we have $\mathbb E |X| < \infty$ and $\mathbb E |Y| < \infty$?

What I actually want is that $X$ and $Y$ are both in $L^2$, i.e., $\mathbb E X^2 < \infty$ and $\mathbb E Y^2 < \infty$. But this can be reduced to $\mathbb E |X|\mathbb E |X| < \infty$. Thus it suffices to show that $X$ and $Y$ are in $L^1$.

It might be helpful to see that since $|X| < |Y| + |X+Y|$, (and by symmetry), either both of $X$ and $Y$ or none of them are in $L^1$. So we may assume $\mathbb E |X| = \infty$ and $\mathbb E |Y| = \infty$ and try to find a contradiction. But here is where I got stuck.

Answer 1

Notice that $$\mathbb E(X+Y)^2=\int_{\mathbb R^2}(x+y)^2\mathrm d\mathbb P_X\otimes \mathbb P_Y(x,y)=\int_{\mathbb R}\left(\int_{\mathbb R}(x+y)^2 \mathrm d \mathbb P_X(x)\right)\mathrm d\mathbb P_Y(y)=\\=\int_{\mathbb R}\mathbb E[(X+y)^2]\mathrm d\mathbb P_Y(y).$$ Since $y\mapsto \mathbb E[(X+y)^2]$ is integrable over $\mathbb R$ for some probability measure, it's in particular finite for some $y$, hence $\mathbb E(X^2)$ is finite. Similarly, one can show that $\mathbb E(Y^2)$ is finite.

Answer 2

If they have the finite mean x=$E X < \infty$ and $y= E Y <\infty$ then yes $E(X+Y)^2 = E(X^2)+E(Y^2)+2 E(X)*E(Y) = E(X^2)+E(Y^2)+2xy < \infty$. If no, I am affraid one can found a pathological case when this is not true.

Let me update myself. I think the previous answer given by David Giraudo is correct. Let me just a little expand it. $E(X+y)^2<\infty$ for some $y$ means that $E(X+y)=E(X)+y<\infty$ for this $y$. This means that $E(X)<\infty$. This means that $E(X+y)^2=E(X^2)+2E(X)y+y^2 <\infty$. So $E(X^2)<\infty$.