Is the Finite Expectation of Powers Satisfied by Nonnegative Random Variables?

[Euge]

Suppose $X$ is a nonnegative random variable and $p\in (0,\infty)$. Show that $\mathbb{E}[X^p] < \infty$ if and only if $\sum_{n = 1}^\infty n^{p-1}P(X \ge n) < \infty$.

 

[anuttarasammyak]

Spoiler

$$\mathbb{E}[X^p] =\int_0^\infty x^p p(x) dx = [ x^p \sigma(x)]_0^\infty - p \int_0^{\infty} x^{p-1}\sigma(x) dx$$
$$= p \int_0^{\infty} x^{p-1}(1-\sigma(x) )dx$$
where
$$\sigma(x)=\int_0^x p(t) dt,\ \ \sigma(\infty)=1$$
The integral differs from the given series by finite quantity. They both are finite or infinite.

 

[Haborix]

Spoiler

$$\mathbb{E}[X^p] =\int_0^\infty x^p p(x) dx = [ x^p \sigma(x)]_0^\infty - p \int_0^{\infty} x^{p-1}\sigma(x) dx$$
$$= p \int_0^{\infty} x^{p-1}(1-\sigma(x) )dx$$
where
$$\sigma(x)=\int_0^x p(t) dt,\ \ \sigma(\infty)=1$$
The integral differs from the given series by finite quantity. They both are finite or infinite.

I think the tricky part of this problem is showing your second to last sentence. A non-answer spoiler:

Spoiler

Euler-Maclaurin with one of the limits going to infinity has to be treated carefully

 

[Euge]

Hi @anuttarasammyak, in this problem $X$ is not assumed to have a density.

 

[anuttarasammyak]

Hi @anuttarasammyak, in this problem $X$ is not assumed to have a density.

I see. If X is a positive integer I hope the replacement
$$p(x)=\sum_{n=1}^{\infty}\ P_n\ \delta(x-n)$$
where n is positive integer and
$$\sum_{n=1}^{\infty}P_n=1,0 \leq P_n$$
would work in my previous post. The integral and the series coincide.