5
$\begingroup$

This problem can be found here, which is a previous prelim exam problem of UT Austin.

Let $X$ and $Y$ be two independent random variables with $X+Y \in L^1$. Show that $X\in L^1$.

Generally, in real analysis, $f+g\in L^1$ does not imply $f\in L^1$, so I guess this must have something to do with their independence.

I guess it might be something like $EX=E(X+Y|Y=y)-y$, but I'm not sure whether I can write like that without knowing $X\in L^1$ or $Y\in L^1$ first.

Or, it should be proved in another way?

Could you please help? Thanks.

  • 1
    @tomasz they do not necessarily have densities, but you can do something similar.2012-12-23

1 Answers 1

3

Recall that $X$ and $Y$ are independent if and only if the joint measure $P_{X,Y}$ is the product measure $P_X \otimes P_Y$. The statement $X + Y \in L^1$ implies that we can apply Fubini's theorem to the integrand $\int |X + Y |dP_{X,Y} = \int |X + Y| dP_X \otimes dP_Y = \int \int |X + Y| dP_X dP_Y$ to find that for almost every $Y$ slice of $X + Y$ we have that $|X + y|$ is in $L^1$, and it is not hard to see that this implies that $X$ itself is in $L^1$.

  • 0
    @ChrisJanjigian: Thank you, I got it. :)2012-12-24