5
$\begingroup$

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.

$\endgroup$
10
  • 1
    $\begingroup$ Why not $g(n+1)-g(n) = \Delta g(n)$ ? $\endgroup$
    – Jakobian
    Aug 8 '19 at 16:43
  • $\begingroup$ @Jakobian I considered using that $g(n+1)\approx g(n)+g^{\prime}(n)$, but I am unsure of how well $\Delta g(n)$ compares with $g^{\prime}(n)$. $\endgroup$ Aug 8 '19 at 16:46
  • $\begingroup$ Some boogaloo weirdness happened with the inequality in your equation chain, I would expect the sum of derivatives (which seems incorrect since both derivatives could be zero) to be replaced by the expression Jakobian suggests. (Assuming g’ is continuous so you can use fundamental theorem of calc.) $\endgroup$
    – Michael
    Aug 8 '19 at 22:23
  • $\begingroup$ @Michael since the function is strictly increasing, there shouldn't be anywhere where the derivative is 0. $\endgroup$ Aug 8 '19 at 22:34
  • 1
    $\begingroup$ I hope nobody edits the title of this amazing post $\endgroup$ Aug 9 '19 at 23:20
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $$ \sum_{n\in\mathbb N}g'(n)\mathbb P(X>n)\geq \sum_{n\in\mathbb N}1=\infty. $$

$\endgroup$
7
  • $\begingroup$ Is there a reason that you decided to ignore my comments? They were given 6 hours earlier and also concluded that the claimed result is false. $\endgroup$
    – Michael
    Aug 9 '19 at 22:33
  • $\begingroup$ I have now elaborated on those comments if there was any confusion. $\endgroup$
    – Michael
    Aug 9 '19 at 23:18
  • $\begingroup$ It's worth noting that in your comments you gave a counterexample where $g(X)$ had infinite expectation, but the series would be zero. In this answer we have the opposite. $\endgroup$ Aug 9 '19 at 23:23
  • $\begingroup$ @RobertThingum : Actually, I had already given that note in my answer. Thanks for the suggestion though as it seems we think alike. $\endgroup$
    – Michael
    Aug 9 '19 at 23:30
  • $\begingroup$ @Michael I did not realize there was already a solution in the comments when I composed my answer. Now that I see the context of the question (Theorem 12.1 of Gut's textbook) the more interesting question becomes, how did Allan Gut mess this up so badly? For instance, I notice that part (iv) of Theorem 12.1 is obviously wrong, since at the very least there would need to be an extra factor of $r$ in both bounds. Moreover, I am unconvinced that even with this obvious fix, the intended proof of 12.1 (iv) goes through, since the cases $r<1$ and $r>1$ have different convexity properties. $\endgroup$
    – pre-kidney
    Aug 10 '19 at 3:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.