5
$\begingroup$

I found the following theorem in Allan Gut, Probability theory (without proof): Let $X$ be a nonnegative random variable and $g$ a nonnegative differentiable strictly increasing function. Then $$Eg(X)=g(0)+\int_{(0,\infty)}g'(x) \mathbb{P}(X>x)\,dx,$$ and $$Eg(X)<\infty\Longleftrightarrow \sum_{n=1}^\infty g'(n)\mathbb{P}(X>n)<\infty.$$

The first statement I could prove (let $X\sim F$):

\begin{eqnarray*} Eg(X)=\int_{(0,\infty)} g(x)dF(x) & =\int_{(0,\infty)} \int_{(0,x)}g'(t)\,dt\,dF(x)+g(0)\\[8pt] & =g(0)+\int_{(0,\infty)}g'(t) \mathbb{P}(X>t)\,dt, \end{eqnarray*} by Fubini. But I am struggling with the second statement. My idea is the following: $$Eg(X)=\int_{(0,\infty)} g(x)\,dF(x)=\sum_{k=1}^\infty \int_{(n_k,n_{k+1})} g(x) \, dF(x).$$

for some increasing sequence satisfying $\cup_{k=1}^\infty [n_k,n_{k+1}] = \mathbb{R^+}$. But i cannot find the appropriate sequence. any suggestions? Thank you.

$\endgroup$
  • $\begingroup$ I suspect the second statement should be $\displaystyle Eg(X)<\infty\Longleftrightarrow \sum_{n=1}^\infty g'(n)\mathbb{P}(X>n)<\infty$ replacing the $k$ with $n$ $\endgroup$ – Henry Apr 25 '14 at 8:10
  • $\begingroup$ Yes you are right.I corrected it. $\endgroup$ – Seneca Apr 25 '14 at 8:55
  • $\begingroup$ What if you apply your idea of decomposition directly on the previous expression $\int_{(0,\infty)}g'(t)\mathbb{P}(X > t)dt$, and then using the monotony and positivity of your functions/distributions to compare it to $\sum g'(n) \mathbb{P}(X > n)$ ? $\endgroup$ – yago Apr 25 '14 at 9:02
  • $\begingroup$ The problem is the missing monotonicity of $g'$. $\endgroup$ – Seneca Apr 25 '14 at 12:53
3
$\begingroup$

Without further assumptions on the monotonicity of the derivative the claim is in general not correct.

Example Let

$$h(x) := \begin{cases} 2 \left(n^2-\frac{1}{n^2} \right) \cdot (x-n)+\frac{1}{n^2} & x \in \left[n,n+\frac{1}{2} \right] \\ n^2- 2 \left(n^2-\frac{1}{(n+1)^2} \right) \left(x-n-\frac{1}{2} \right) & x \in \left[n+ \frac{1}{2},n+1 \right] \end{cases}$$

for $n \in \mathbb{N}$.

$\hspace{90pt}$enter image description here

Then $h$ is a (strictly) positive continous function and therefore

$$g(x) := \int_0^x h(y) \, dy$$

defines a strictly increasing differentiable positive function. In particular, $g'(n)= \frac{1}{n^2}$ and $$g' (x) \geq \frac{n^2}{2}, \qquad x \in \left[n+\frac{1}{4},n+\frac{3}{4} \right]. \tag{1}$$

Now if we consider a random variable $X$ such that $\mathbb{P}(X > x) = \frac{1}{x}$ for $x$ sufficiently large, then $(1)$ shows that

$$\int_{(0,\infty)} g'(x) \mathbb{P}(X>x) \, dx = \infty.$$

On the other hand,

$$\sum_{n} g'(n) \mathbb{P}(X >n) \leq \sum_n g'(n) = \sum_n \frac{1}{n^2} < \infty.$$

$\endgroup$
  • $\begingroup$ Thank you very very much saz! Nice work! I just read your answer and it is a nicely constructed counter example. I really normally believe books and it takes some time till I start thinking about the correctness. So this theorem in the book of Allan Gut is indeed wrong. When $g'$ is also monotone, it is not so difficult to prove and all the examples of functions $g$ in this book are monotone. (the proof of the theorem was an exercise) $\endgroup$ – Seneca May 12 '14 at 16:42
  • $\begingroup$ @Seneca You are welcome. By the way, you can accept (and upvote) helpful answers by clicking on the check mark next to the answer. $\endgroup$ – saz May 12 '14 at 21:32
  • $\begingroup$ Ah I did not know. Of course, I just did it :-) $\endgroup$ – Seneca May 13 '14 at 8:39

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.