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Let $X$ be a random variable taking values in $\mathbb{R}$ such that $E(X^2)=\infty$. I want to show that $$\lim_{M\to\infty} \frac{\mathbb{E}(X\mathbf{1}_{|X|\leq M})^2}{\mathbb{E}(X^2\mathbf{1}_{|X|\leq M})} = 0$$

The monotone convergence theorem suggests that as $M\to \infty$ the denominator diverges. But this is not enough because the numerator might also diverge. Another idea that came to mind was to write $E(X\mathbf{1}_{|X|\leq M})^2 \leq M^2$ and prove that $$\lim_{M \to \infty} \frac{1}{M^2}\int_{|X|\leq M}X^2dP=\infty$$ which unfortunately doesn't hold. Any help/hint is appreciated!

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Jan 27, 2023 at 12:23

1 Answer 1

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WLOG assume that $X \geq 0$ (say, by replacing $X$ with $|X|$).The main idea to prove the original statement is to note that the small values of X don't matter too much and can be ignored because they won't affect the value of the expectation too much. We make this rigorous as follows. Fix $\epsilon> 0$.

We collect some basic facts: since $X$ is a non-negative real-valued random variable, there exists $C > 0$ such that $P(X\geq C) \leq \sqrt{\epsilon/2}$. This follows from the fact that $X$ is a finite random variable. Our second fact is that there exists $N>0$ such that if $M \geq N$ then $\frac{\mathbb{E}(X \mathbf{1}_{X \leq C})}{\lVert X \mathbf{1}_{X \leq M}\rVert_{L^2}} \leq \epsilon/2$. This follows from the fact that the denominator goes to infinity as $M\to \infty$ (and the denominator is monotonically increasing).

We simply compute now, with any $M \geq N$ fixed: $$\frac{\mathbb{E}(X \mathbf{1}_{X \leq M})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}}= \frac{\mathbb{E}(X \mathbf{1}_{X \leq C})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}} + \frac{\mathbb{E}(X \mathbf{1}_{C < X \leq M})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}}$$ $$\leq \epsilon/2 + \frac{\mathbb{E}(X \mathbf{1}_{C < X \leq M})}{\lVert X\mathbf{1}_{C \leq X \leq M} \rVert_{L^2}}$$ $$\leq \epsilon/2 + \frac{\lVert X \mathbf{1}_{C < X \leq M} \rVert_{L^2}}{\lVert X \mathbf{1}_{C < X \leq M} \rVert_{L^2}} \left (P(C < X \leq M)\right)^{1/2}$$ $$\leq \epsilon/2 + \sqrt{P(X \geq C)} \leq \epsilon/2 + \epsilon/2 = \epsilon$$ where the inequality on the third line results from Cauchy-Schwarz. This holds for all $M \geq N > 0$ for some fixed $N$ so we deduce that $\limsup_{M \to \infty}\frac{\mathbb{E}(X \mathbf{1}_{X \leq M})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}} \leq \epsilon$, and since this holds for all $\epsilon > 0$ we see that $\lim_{M\to\infty} \frac{\mathbb{E}(X \mathbf{1}_{X \leq M})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}} = 0$ as claimed.

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    $\begingroup$ @GiorgosGiapitzakis My equation, $\frac{\mathbb{E}(X \mathbf{1}_{X \leq M})}{\lVert X\mathbf{1}_{X \leq M} \rVert_{L^2}}$, is just the square root of your expression with the numerator squared (note the L2 norm has an implicit square root on the outside). This was done to get the linearity after the first equality sign. If one were to square this, we'd get an annoying cross term which would be unpleasant to deal with, and once we have the result as posted, we can get your result by simplying squaring and noting that $x \mapsto x^2$ is a continuous function. $\endgroup$ Commented Jan 26, 2023 at 3:43
  • $\begingroup$ With this in mind, I think it's best to not redo the computation because it'll be worse, not better. Have I misunderstood and/or do you agree? $\endgroup$ Commented Jan 26, 2023 at 3:44
  • $\begingroup$ Yeah I missed that. Your solution looks great, thanks! $\endgroup$ Commented Jan 26, 2023 at 16:14

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