Different way of writing expectation of function of random variable

Question

Let $X$ be a discrete random variable. We can write its expectation as $$ EX = \sum_{n=0}^\infty P(X > n).$$

Now, let $p\in (0,1)$ and then, $$ E(X^p) = \sum_{n=0}^\infty n^p P(X =n).$$

My question is, whether there exists a similar formula like the first one in this case, i.e.

$$ E (X^p) = \sum_{n=0}^\infty \ ... P(X > n)\ ?$$

Edit: Or at least an inequality $E (X^p) \leq \sum_{n=0}^\infty \ ... P(X> n)$.

Edit 2: Specifically, assume $P(X>n ) \leq \frac{1}{n^{1/2}}$. I want to conclude that $$ EX^{p-\epsilon} \leq \sum_{n=1}^\infty \frac{1}{n^{1/2+\delta}} P(X > n) \leq \sum_{n=1}^\infty \frac{1}{n^{1+\delta}} $$ holds true.

Answer 1

When $X \geq 0$, you can write $$ \begin{align} E[X] &= \sum_{n=0}^{\infty} n^p \Pr(X = n) \\ &= \sum_{n=1}^{\infty} n^p (\Pr(X>n-1)-\Pr(X>n)) \\ &= \sum_{n=1}^{\infty} n^p \Pr(X>n-1) - \sum_{n=1}^{\infty} n^p \Pr(X>n+1) \\ &= \sum_{n=0}^{\infty} (n+1)^p \Pr(X>n) - \sum_{n=0}^{\infty} n^p \Pr(X>n) \\ &= \sum_{n=0}^{\infty} ((n+1)^p-n^p) \Pr(X>n) , \end{align}$$ where repeated adjustment has been done on the summation indices to make everything line up. As a check, putting $p=1$ gives $(n+1)^1-n^1 = 1 $, recovering the original formula.

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Regarding the edit, we have by the MVT $$ (n+1)^p-n^p = p\xi^{p-1} $$ for some $\xi$ between $n$ and $n+1$. In particular, it is smaller than $p(n+1)^{p-1}$, and for $n>0$, $(n+1)/n = 1+1/n \leq 2$, so we could say it is $\leq p2^{p-1}n^p$. Then if $\Pr(X>n) \leq n^q$, we have $$ E[X^p] = \sum_{n=0}^{\infty} ((n+1)^p-n^p) \Pr(X>n) \leq 1-\Pr(X=0)+p2^{p-1} \sum_{n=1}^{\infty} n^{p+q-1} , $$ where we have separated off $n=0$ to avoid potential singularities.

Answer 2

Yes. Suppose $X$ is non-negative. Use the identity $$ X^p=\sum_{k=1}^\infty (k^p-(k-1)^p) I(X\ge k).\tag{0} $$ which is a telescoping sum where $I$ is the indicator function. At this point use the monotone convergence theorem to take expectations of both sides to yield $$ EX^p=\sum_{k=1}^\infty (k^p-(k-1)^p)P(X\geq k).\tag{1} $$ With $p=2$ we obtain the special case $$ EX^2=\sum_{k=1}^\infty(2k-1)P(X\ge k).\tag{2} $$ We can re-index (1) to obtain $$ EX^p=\sum_{k=0}^\infty ((k+1)^p-k^p)P(X\geq k+1)=\sum_{k=0}^\infty ((k+1)^p-k^p)P(X>k).\tag{3} $$ Note that (0) is a discrete analogue of the identity $$ X^p=\int_{0}^\infty px^{p-1} I(X\geq x)\, dx. $$