0
$\begingroup$

If $X$ is a random variable, is there an approach that I can use to upper bound $\mathbb{E}[X^2]$ by a function of $\mathbb{E}[X]$?

If I use Jensen's inequality, I can find a lower bound. But, I need an upper bound. Any idea?

$\endgroup$
  • 3
    $\begingroup$ No: $\mathbb{E}[X^2]$ doesn't even need to be finite. $\endgroup$ – carmichael561 Aug 8 '17 at 1:49
1
$\begingroup$

If $X$ is a real random variable with density $f(x) = 2 x^{-3} \mathbb{1}_{[1, \infty)}(x)$, then $$\mathbb E(X)=\int_{1}^{\infty}2x\cdot x^{-3}\ \mathrm{d}x = 2\int_{1}^{\infty} \dfrac{1}{x^2}\ \mathrm{d}x=2$$ but $$\mathbb E(X^2)=\int_1^{\infty} 2x^2\cdot x^{-3}\ \mathrm{d}x = 2\int_{1}^{\infty} \dfrac{1}{x}\ \mathrm{d}x = \infty$$

Therefore you cannot obtain an upper bound on $\mathbb E(X^2)$ from $\mathbb E(X)$. Indeed you know that $L^2(\mathbb R, \mathcal B(\mathbb R), \mathbb{P}) \subset L^1(\mathbb R, \mathcal B(\mathbb R), \mathbb{P})$ (by Cauchy-Schwarz for example), but the converse is not true.

$\endgroup$
  • 1
    $\begingroup$ I assume you meant "converse"? $\endgroup$ – Clement C. Aug 8 '17 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.