characterization of the existence of higher order expectation.

Question

Consider a non-neg r.v $X$ and fix a positive integer $n$, I am trying to show that a necessary and sufficient condition for the $n$th order expectation to exist is that $\sum_{i\in\mathbb{N}}i^{n-1}P(X\ge i)$ converges.

Attempt: We have a well established relation, $\sum_i P(X\ge i)\le EX < 1+\sum_iP(X \ge i)$, I am convinced this is used but i am unsure how to to incorporate $i^{n-1}$ . If we consider a partition of the real line given by $A_i=\{i-1\le X < i\}$ and then let choose $X_i=(i-1)I_{A_i}$ then the series is a discrete r.v. with $E(\sum_i X_i)=\sum iP(X\ge i)$ which is similar to the form we need it in. But im unsure where to go from here.

Answer 1

The quickest way is using Tonelli's theorem to show $$ \mathbb{E}X^n=n\int_0^\infty x^{n-1}\mathbb{P}(X\geq x)\,\mathrm{d}x\tag1 $$ and then sandwich using $\lfloor x\rfloor^{n-1}\leq x^{n-1}\leq 1+2^{n-1}\lfloor x\rfloor^{n-1}$.

Edit (without much measure theory)

If you know the formula $$ \mathbb{E}Y=\int_0^\infty\mathbb{P}(Y\geq y)\,\mathrm{d}y=\int_0^\infty\mathbb{P}(Y>y)\,\mathrm{d}y $$ for a nonnegative random variable $Y$ (some authors refer to this as Cavalieri's principle), then you can deduce (1) by $$ \mathbb{E}X^n=\int_0^\infty\mathbb{P}(X^n\geq y)\,\mathrm{d}y $$ and substituting $y=x^n$, $\mathrm{d}y=nx^{n-1}\,\mathrm{d}x$, noting $X^n\geq x^n\iff X\geq x$.

Answer 2

First of all, note that for any fixed $n \in \mathbb{N}$ we have

$$\sharp \{i \in \mathbb{N}; k^n \leq i < (k+1)^n\} \sim k^{n-1}$$

i.e. there exist constants $c_1,c_2>0$ (depending on $n$) such that

$$c_1 k^{n-1} \leq \sharp \{i \in \mathbb{N}; k^n \leq i < (k+1)^n\} \leq c_2 k^{n-1} \tag{1}$$

for all $k \geq 1$. Now if $X$ is a non-negative random variable, then we can write

$$\sum_{i \geq 1} \mathbb{P}(X^n \geq i) = \sum_{k =1}^{\infty} \sum_{k^n \leq i < (k+1)^n} \mathbb{P}(X^n \geq i). \tag{2}$$

As $\mathbb{P}(X^n \geq i)$ is decreasing in $i$, we have

$$\mathbb{P}(X^n \geq i) \leq \mathbb{P}(X^n \geq k^n) \qquad \text{for all $k^n \leq i < (k+1)^n$}$$

and so

$$\begin{align*}\sum_{i \geq 1} \mathbb{P}(X^n \geq i) &\leq \sum_{k \geq 1} \bigg( \mathbb{P}(X^n \geq k^n) \underbrace{\sum_{k^n \leq i < (k+1)^n} 1}_{\stackrel{(1)}{\leq} c_2 k^{n-1}} \bigg) \\ &\leq c_2 \sum_{k \geq 1} k^{n-1} \mathbb{P}(X \geq k). \tag{3} \end{align*}$$

Similarly, we find from

$$\mathbb{P}(X^n \geq i) \geq \mathbb{P}(X^n \geq (k+1)^n) \qquad \text{for all $k^n \leq i < (k+1)^n$}$$

that

$$\sum_{i \geq 1} \mathbb{P}(X^n \geq i) \geq c_1 \sum_{k \geq 1} k^{n-1} \mathbb{P}(X \geq k+1).$$

Consequently, we have shown that $\sum_{i \geq 1} \mathbb{P}(X^n \geq i)$ is comparable with $\sum_{k \geq 0} k^{n-1} \mathbb{P}(X \geq k+1)$. Since you already know that $X^n$ has finite expectation iff the series $\sum_{i \geq 1} \mathbb{P}(X^n \geq i)$ converges, this proves the assertion.