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(d) For any nonnegative random variable X and $\alpha > 0$, Prove that if $E[X^\alpha] < \infty$, then $\lim_{x \to \infty} x^\alpha P(|X| > x ) = 0$

I tried to show this by using $E[X^\alpha] = \int_\Omega \alpha x^{\alpha-1} P(|X| > x ) dx$. So $x^{\alpha-1}P(|X|>x) \to 0$ but can't proceed further.

(f) Suppose that X is random variable s.t. $E[X^2] < \infty$. Prove that, for any $\eta >0$, $\lim_{x \to \infty} x P(|X| > \eta \sqrt x ) = 0$

I did try Markov inequality so that $xP(|X| > \eta \sqrt x ) \le x \frac{E[X^2]}{x\eta^2} = \frac{E[x^2]}{\eta^2} < \infty$. but after then?? I think fundamental approach is wrong.

Any help will be appreciated. Thanks in advance.

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(d) For any nonnegative random variable X and $\alpha > 0$, Prove that if $E[X^\alpha] < \infty$, then $\lim_{x \to \infty} x^\alpha P(|X| > x ) = 0$

Hint:

$$x^{\alpha} \mathbb{P}(|X|>x) = \int_{|X|>x} x^{\alpha} \, d\mathbb{P} \leq \int_{|X|>x} |X|^{\alpha} \, d\mathbb{P}$$

(f) Suppose that X is random variable s.t. $E[X^2] < \infty$. Prove that, for any $\eta >0$, $\lim_{x \to \infty} x P(|X| > \eta \sqrt x ) = 0$

Hint: Use (d) and the fact that

$$\lim_{x \to \infty} f(x) = \lim_{y \to \infty} f(y^2)$$

(whenever one of the two limit exists).

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    $\begingroup$ Wanted to add that the first hint implies: If $X$ is non-negative with $E[X]$ finite, then $\lim _{t \to \infty}tProb[X>t]=0.$ And this can also be directly applied to (f) for the variable $X^2$ $\endgroup$ – Aravind Jun 17 '18 at 15:41
  • $\begingroup$ @Aravind Yeah, you are right; that's an alternative approach to solve (f). $\endgroup$ – saz Jun 17 '18 at 16:10

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