Limit of a probability distribution function times $x$

Question

Let $p(x)$ be a probability density function (i.e. non-negative, integrating to 1).

Assume further that $\displaystyle\lim_{x\to\pm\infty}p(x)=0$.

Is it always true that

$$ \lim_{x\to\infty}\left[xp(x)\right] - \lim_{x\to-\infty}\left[xp(x)\right]=0? $$

If so, why? If not, what is minimal set assumptions on $p(x)$ that will render the statement true?

Answer 1

Some counterexamples to the assertion that every PDF $p$ converges to zero at $\pm\infty$ are explained in the comments, a smooth variant is a "sum-of-bumps" PDF such as$$p(x)=\sum_ng\left(\frac{x-x_n}{\sigma_n}\right),$$ where $g$ is the standard gaussian density, $\sigma_n\gt0$ for every $n$, $\sum\limits_n\sigma_n=1$, and $x_n\to\infty$.

Then $p$ is indeed a PDF (can you check this?), and, for every $n$, $p(x_n)\geqslant g(0)=1/\sqrt{2\pi},$ hence $p$ does not converge to zero at $\infty$.

When $p$ does converge to $0$ at $\infty$, the fact that $xp(x)\to0$ is not guaranteed either, as similar counterexamples show. However:

Let $p$ denote a PDF such that, for $x$ large enough, $x\mapsto p(x)$ is non increasing. Then $xp(x)\to0$.

To show this, note that, for $x$ large enough, $$\int_{x/2}^{\infty}p(u)\,\mathrm du\geqslant\int_{x/2}^{x}p(u)\,\mathrm du\geqslant\int_{x/2}^{x}p(x)\,\mathrm du=\frac12xp(x).$$ The LHS goes to zero hence $xp(x)\to0$, QED. A similar result holds for the asymptotics when $x\to-\infty$.

Answer 2

Note that any pdf that approaches $0$ at a rate of $1/x$ or slower will fail this condition generically. Trivially, $p(-x)=-p(x)$ satisfies this condition. Otherwise, you need $p(x)$ to converge to 0 at a rate faster than $1/x$, for example, at rate $1/x^2$. Then $\lim_{x \rightarrow \infty} x p(x)=0$, as does $\lim_{x \rightarrow -\infty} x p(x)$.