Expectation of nonnegative random variable when passed through nonnegative increasing differentiable function. Part II: Electric Boogaloo

Question

This is a follow up to my previous question:

Expectation of nonnegative random variable when passed through nonnegative increasing differentiable function

I am now wanting to establish a follow up to the above problem. Specifically, if $X$ is a nonnegative random variable and $g:\mathbb{R}\rightarrow\mathbb{R}$ is a nonnegative, strictly increasing, differentiable function, then

$$\mathbb{E}g(X)<\infty \iff \sum_{n=1}^{\infty}g^{\prime}(n)\mathbb{P}(X>n)<\infty$$

I believe I can show the inequality when $g(x)=x^{p}$ for $p\in\mathbb{N}$, but the case of a general $g$ is more mysterious to me.

My attempt for the converse proceeds in the following way: If you assume that the series converges then (by the linked question)

\begin{equation} \mathbb{E}g(X) = g(0)+\int_{0}^{\infty}g^{\prime}(X)\mathbb{P}(X>x)dx \\ = g(0)+\sum_{n=0}^{\infty}\int_{n}^{n+1}g^{\prime}(x)\mathbb{P}(X>x)dx \\ \leq g(0)+\sum_{n=0}^{\infty}(g^{\prime}(n+1)+g^{\prime}(n))\mathbb{P}(X>n) \\ = g(0)+\left(\sum_{n=0}^{\infty}g^{\prime}(n+1)\mathbb{P}(X>n)\right)+\left(\sum_{n=0}^{\infty}g^{\prime}(n)\mathbb{P}(X>n)\right). \end{equation}

However I am unsure how to proceed from here. I don't see how the middle series would converge without more assumptions on $g$.

Any help with the equivalence in general would be appreciated.

Answer 1

This answer elaborates on my comments (to show the claim is false): Define $g:[0,\infty)\rightarrow \mathbb{R}$ by $$g(x) = 1 + x + \frac{1}{2\pi} \cos(2\pi x + \pi/2)$$ Then
$$g’(x) = 1 - \sin(2\pi x + \pi/2) \geq 0 \quad \forall x \geq 0$$ and for $n\geq 0$ we get $$ g'(n) = 0 \quad \mbox{ if and only if $n$ is an integer}$$ It follows that $g$ is nonnegative and strictly increasing over $x \geq 0$.

Furthermore $g(x)\geq 1 + x -1/(2\pi)\geq x$ and so $$ g(x) \geq x \quad \forall x \geq 0$$ Let $X$ be any nonnegative random variable that satisfies $E[X]=\infty$. We get: $$ g(X)\geq X \implies E[g(X)] \geq E[X] = \infty$$ but $$ \sum_{n=1}^{\infty} g’(n) P[X>n] = 0$$ You can easily extend $g$ to have domain over all real numbers while preserving the non negativity and strictly increasing properties.

*Note: This shows that one direction of the "if and only if" claim is false. The pre-kidney answer shows the other direction is also false.

Answer 2

The equivalence is not true under the assumptions you have stated. The problem is that you can make $g(x)$ nearly constant on each of the intervals $[n,n+1-\epsilon)$ and then have it rapidly increase from $g(n)$ to $g(n+1)$ on the tiny interval $[n+1-\epsilon,n+1]$, which will make the derivatives $g'(n)$ contribute disproportionately to the sum.

For a concrete example of this, take $X\sim \textrm{Exp}(1)$ so that $\mathbb P(X>t)=e^{-t}$ for $t\geq 0$, and take $g(x)$ to be a smoothed and strictly increasing version of $\lfloor x\rfloor $. More precisely, let $g(x)$ be any non-negative strictly increasing differentiable function satisfying the following conditions:

1. $g(x)\leq x$ for all $x\geq 0$
2. $g'(n)\geq e^n$ for all $n\in\mathbb N$.

Note that these conditions do not contradict each other, since we can have $g(x)$ be nearly constant on $[n,n+1-e^{-n}]$ and then rapidly increase by nearly $1$ on an interval of size $e^{-n}$, allowing the derivative to be of size $e^n$ (or bigger). (Explicit formulas can be obtained using a quadratic spline, or smooth bump functions if so desired.)

Since $g(x)\leq x$, it follows that $\mathbb Eg(X)\leq \mathbb EX=1<\infty$. However, since $g'(n)\geq e^n$ it follows that $$ \sum_{n\in\mathbb N}g'(n)\mathbb P(X>n)\geq \sum_{n\in\mathbb N}1=\infty. $$