If $X_1$ and $X_2$ are i.i.d., integrable, nonnegative, then $\min(X_1,X_2)^2$ is integrable

Question

Let $X_1$, $X_2$ be i.i.d., integrable, nonnegative real random variables, and $Y=\min(X_1,X_2)$. Show that $Y^2$ is integrable.

Does anyone have some hints or tips on how to show this?

Answer 1

Solution 1

Since $Y^2 \leq X_1X_2$, we have $\mathbb{E}(Y^2) \leq \mathbb{E}(X_1X_2)=\mathbb{E}(X_1)\mathbb{E}(X_2) < +\infty$.

Solution 2

Let $g(t)=\mathbb{P}(X_1>t)$.

• Show that $\mathbb{E}(X_1)=\int_0^\infty g(t) \,\mathrm{d}t$.
• Show that $\mathbb{E}(Y^2)=\int_0^\infty 2tg(t)^2 \,\mathrm{d}t$.
• Show that when $t$ tends to $+\infty$, $g(t)=o(1/t)$. For instance, you could show that $xg(x) \leq 2\int_{x/2}^x g(t) \,\mathrm{d}t$ for any $x>0$.
• Show that $tg(t)^2=o(g(t))$.
• Conclude.

Edit : as pointed out by Did in the comments, we need another argument to conclude.

Edit 2 : I added two steps. This should work now.

Edit 3 : That was simpler than I thought. Thanks to D. Thomine.

Edit 4 : added Vim's solution since it is much simpler.