I have a question regarding this proposition.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an a.e. differentiable function so that $\int \frac{\left|f^{\prime}(x)\right|}{(1+|x|)^s} d x<\infty$ ($s$ is an integer such that $s \geq 1$) and $X$ be a random variable with cdf $F$. Then, we have

$$\mathbb{E}\left[f\left(X\right)\right]=-\mathbb{E}\left[\int_{X}^{\infty} f^{\prime}(x) d x\right]=-\int_{-\infty}^{\infty} f^{\prime}(x) \mathbb{P}\left(X \leq x\right) d x=-\int_{-\infty}^{\infty} f^{\prime}(x) F(x) d x$$

I'm a bit confused by the first equality. Because

$$-\int_{X}^{\infty} f^{\prime}(x) d x = - f(x) \bigl\lvert_0^\infty = \lim _{x \rightarrow \infty} f(x) + f(X)$$.

However, we do not assume that $\lim _{x \rightarrow \infty} f(x) = 0$ or am I missing something?

Source: Hafouta, Y. (2022) Non-uniform Berry-Esseen theorem and Edgeworth expansions with applications to transport distances for weakly dependent random variables. arXiv:2210.07204.

  • $\begingroup$ I was very much confused by the use of $f(x)$ and thought it was the pdf of the CDf $F(x)$. $\endgroup$ Commented Jun 10, 2023 at 18:50

1 Answer 1


enter image description here

The above result from Hafouta (2022) is false without further assumptions like $$ \lim _{x \rightarrow \infty} f(x)= \lim _{x \rightarrow -\infty} f(x)F(x)=0$$ Take $$f(x)=x\qquad F(x)=\Phi(x)=\int_{-\infty}^x \dfrac{\exp\{-\xi^2/2\}}{\sqrt{2\pi}}\text d\xi$$ Then $$\int_{-\infty}^{+\infty} \frac{\left|f^{\prime}(x)\right|}{(1+|x|)^2} \text d x<\infty$$ and $$\mathbb E[X]=0 \qquad \int_{-\infty}^{+\infty} \Phi(x)\,\text dx=+\infty$$

  • 1
    $\begingroup$ Another counter example would be to compare the results from the functions $f(x)$ and $f(x)+1$. They have the same derivative but different expectation. $\endgroup$ Commented Jun 10, 2023 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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