I am working with this problem. Can someone take a look at it?

"Suppose $X$ and $Y$ are arbitrary random variables with finite second moments. you are told $P(X+Y=0)<1$. Then is

$$ \sqrt{E[(X+Y)^2]} \leq \sqrt{E[X^2]} + \sqrt{E[Y^2]} $$ necessarily true?

To do this, I deduced it is enough to show $(X+Y)^2 \leq X^2 + Y^2$ and then take expectations and square root.

I started by $(X+Y)^2 = X^2 + 2XY + Y^2$ but then I didn't know where to go after that. Could anyone help?

  • 1
    $\begingroup$ This might be a stupid comment, but I'm a little bit confused by the notation. What does the $E^{1/2}$ mean? I've never seen the expected value function modified like that. I'm just a student as well and probably don't know the answer, just curious as the notation. $\endgroup$
    – Chris C
    Commented Nov 6, 2015 at 1:42
  • $\begingroup$ Do you mean that you want to prove that $$\sqrt{E[(X+Y)^2]} \leq \sqrt{E[X^2]} + \sqrt{E[Y^2]} = \sqrt{\sigma_X^2+\mu_X^2}+\sqrt{\sigma_Y^2+\mu_Y^2}??$$ $\endgroup$ Commented Nov 6, 2015 at 3:53
  • $\begingroup$ "I deduced it is enough to show...." is, alas, a false deduction. The square root of $E[X^2]+E[Y^2]$ is $\sqrt{E[X^2]+E[Y^2]}$ and not $\sqrt{E[X^2]} + \sqrt{E[Y^2]}$ (assuming that the latter is what you want). $\endgroup$ Commented Nov 6, 2015 at 4:01
  • $\begingroup$ @DilipSarwate yes that is what you are trying to show and yes i guess that is a false deduction... $\endgroup$ Commented Nov 6, 2015 at 18:52

1 Answer 1


You automatically get

$||X + Y|| \le ||X|| + ||Y||$

by Minowski's inequality.

The last part is simple arithmetic showing that:

$\sqrt{A + B} \le \sqrt{A} + \sqrt{B}$ when $A, B > 0$.

which is a simple direct proof.

In fact, I think the whole thing is a direct consequence of Minowski's inequality.


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