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?

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

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.

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

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