Prove that $\mathbb{E}\vert X \vert^{a}<\infty$ iff $\sum\limits_{n=1}^{\infty}n^{a-1}\mathbb{P}(\vert X \vert \geqslant n)<\infty$

Question

Suppose $X$ is a random variable, $a>0$ is a constant. Prove that $\mathbb{E}\vert X \vert^{a}<\infty$ iff $\sum\limits_{n=1}^{\infty}n^{a-1}\mathbb{P}(\vert X \vert \geqslant n)<\infty$.

This is a exercise from Mathematical Statistics. Jun Shao. Second edition. EX1.54

For convenience, we suppose that $X$ is non-negative.

I know how to prove in the case $a=1$, i.e. $\mathbb{E}X <\infty$ iff $\sum\limits_{n=1}^{\infty}\mathbb{P}(X \geqslant n)<\infty$.

Using the inequality below: $$ \sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\} \leqslant \mathbb{E}X \leqslant 1+\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\} $$

This is because: \begin{align} \sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}&=\sum_{n=1}^{\infty}\sum_{k\geqslant n}\mathbb{P}\{k\leqslant X < k+1\}=\sum_{k=1}^{\infty}k\mathbb{P}\{k\leqslant X <k+1\} \\ &=\sum_{k=0}^{\infty}\mathbb{E}[k I\{k\leqslant X <k+1\}]\leqslant \sum_{k=0}^{\infty}\mathbb{E}[XI\{k\leqslant X <k+1\}] \\ &=\mathbb{E}X \\ & \leqslant \sum_{k=0}^{\infty}\mathbb{E}[(k+1)I\{k\leqslant X <k+1\}] \\ &=\sum_{k=0}^{\infty}(k+1)\mathbb{P}\{k\leqslant X <k+1\} \\ &=\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}+\sum_{k=0}^{\infty}\mathbb{P}\{k\leqslant X <k+1\}=\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}+1 \end{align}

So, $\mathbb{E}X <\infty$ iff $\sum\limits_{n=1}^{\infty}\mathbb{P}(X \geqslant n)<\infty$.

But for $a>0,a\neq 1$, I have no idea how to change the inequality above to help me to prove the proposition.

Answer 1

You can use the more general result that encompasses this specific problem:

Result $1.$ Let $(\Omega, \boldsymbol{\mathfrak A}, \mu)$ be a $\sigma$-finite measure space and let $f$ be a nonnegative real-valued $\boldsymbol{\mathfrak A}$-measurable function on $\Omega.$ Let $\varphi$ be a nonnegative real-valued increasing, absolutely continuous with continuous derivative almost everywhere. Then $$\int_\Omega \varphi(f(x))~\mathrm d\mu=\int_{[0,~\infty)}\mu\{x\in \Omega:f(x)\geq t\}~\varphi'(t)~\mathrm d\lambda.\tag 1\label 1$$ Also, define for all $n\in\mathbb N, $ \begin{align} L_n &= \inf_{t \in [n, ~n+1]}\varphi'(t),\\ H_n &= \sup_{t \in [n,~ n+1]}\varphi'(t).\end{align} If $$\frac{L_n}{H_n}\to 1,\tag{a}\label a$$ then $$\int_\Omega \varphi(f(x))~\mathrm d\mu<\infty\iff \sum_{n=1}^\infty \varphi'(n)\mu\{x\in \Omega:f(x)\geq n\}<\infty.\tag 2\label 2$$

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Remarks:

$\bullet$ For proving $\eqref 1,$ check $\rm [I]$ and for that of $\eqref 2,$ check this Math.SE post - basically the derivation of the latter uses $\eqref 1$ and the assumption $\eqref a. $

$\bullet$ For context, Allen Gut in his book Probability: A Graduate Course stated a similar result (and as he is infamous for his hand-wavy writing style, he gave no proof) in Theorem $12.3$ without any assumption of the form $\eqref a. $ That result is, however, false, for it doesn't rule out instances of $\varphi$ with erratic behavior owing to rapid change in $\varphi'$ over small intervals possibly invalidating $\eqref 2 $ (for example, see these counter-examples). $\eqref a$ tames the behavior of the derivative and as such circumvents possible counter-examples.

$\bullet$ My initial answer was wrong (courtesy to Taylor) for $n^b\Pr[|X|\geq n]\to 0\nRightarrow \mathbb E[|X|]^b<\infty.$ Take, for example, $$\Pr[X=n]=\frac{k}{n^2\ln n}, ~n\in\{2, 3,\ldots\}$$ (cf. $\rm [II]$).

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References:

$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific Publishing, $2014, $ Theorem $23.68, $ pp. $609-610.$

$\rm [II]$ An Introduction to Probability and Statistics, Vijay K. Rohatgi, A. K. Md. Ehsanes Saleh, John Wiley & Sons, $2001, $ Remark $3.2.9, $ pp. $74-75.$

Answer 2

For convenience, we suppose that $X$ is non-negative.

We only need to prove that: \begin{equation} \sum_{n=1}^{\infty}n^{a-1}\mathbb{P}(X\geqslant n) \leqslant \mathbb{E}(X^a) \leqslant \sum_{n=1}^{\infty}n^{a-1}\mathbb{P}(X\geqslant n)+c \end{equation} where $c$ is a constant.

Here is the reason: \begin{align} \sum_{n=1}^{\infty}n^{a-1}\mathbb{P}\{X\geqslant n\}&=\sum_{n=1}^{\infty}n^{a-1}\sum_{k\geqslant n}\mathbb{P}\{k\leqslant X < k+1\} \\ & \leqslant \sum_{n=1}^{\infty}\sum_{k\geqslant n}k^{a-1}\mathbb{P}\{k\leqslant X < k+1\} \qquad (\text{We need the condition that $a>1$}) \\ &=\sum_{k=1}^{\infty}k^a\mathbb{P}\{k\leqslant X <k+1\}=\sum_{k=0}^{\infty}\mathbb{E}[k^a I\{k\leqslant X <k+1\}] \\ &\leqslant \sum_{k=0}^{\infty}\mathbb{E}[X^aI\{k\leqslant X <k+1\}] \\ &=\mathbb{E}(X^a) \end{align}

The other side of the inequality I haven't proved it.

Answer 3

Modifying Inforz's work a little: \begin{align} \sum_{n=1}^{\infty}n^{a-1}\mathbb{P}\{|X|\geqslant n\} &= \sum_{n=1}^{\infty}n^{a-1}\sum_{k\geqslant n}\mathbb{P}\{k\leqslant |X| < k+1\} \\ &= \sum_{n=1}^{\infty}\sum_{k\geqslant n}\mathbb{E}\left[ n^{a-1}I\{k\leqslant |X| < k+1\}\right] \\ &\leqslant \sum_{n=1}^{\infty}\sum_{k\geqslant n}\mathbb{E}\left[ |X|^{a} I\{k\leqslant |X| < k+1\}\right] \tag{$n \le k \le |X|$}\\ &= \sum_{n=1}^{\infty}\mathbb{E}\left[ |X|^{a} I\{n \leqslant |X| \}\right] \\ &= \mathbb{E}\left[ |X|^{a} \right] \end{align}