Limit when $E(X^2)=\infty$. 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!
 A: 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.
