I am wondering about showing that the limit: $$ \lim_{x \to \infty} x\overline{F}(x) =0 $$ where $\overline{F} =1-F$ is the tail distribution function, $\overline{F}(x)=1−F(x)$, where $F$ is the cumulative distribution function

As $x \to \infty$, $\overline{F} \to 0$, so we have indeterminate form, I rewrite as: $$ \lim_{x \to \infty} \frac{\overline{F}(x)}{1/x} $$ and use L'Hôpital's rule: $$ \lim_{x \to \infty} \frac{f(x)}{1/x^2} $$ but this requires knowledge of $f$ as $x \to \infty$ which I don't have.

How do I evaluate this limit?

  • 1
    $\begingroup$ You should clarify your assumptions: the claimed result is not true in general (e.g. for Pareto), but holds when $X$ is positive $\mathbb{E}[X] < \infty$. Hint: use $x \text{Pr}\{ X > x \} \leq \mathbb{E}[X 1_{ \{X > x\} }]$. $\endgroup$ – Yves Nov 10 '15 at 13:24
  • $\begingroup$ @Solitary nitpicking a little bit, but the condition is actually slightly weaker requiring integrability. For example, one can show $x^p \Pr\{|X| > x\} \to 0$ implies $E[|X|^q] < \infty$ for all $q$ strictly less than $p$. But it is not true for $q = p$ in general. Off the top of my head, I think the density proportional to $1 / [x^{p+1} \log x]$ for $x > 2$ gives the counterexample, but I confess that I haven't done the math. $\endgroup$ – guy Nov 10 '15 at 15:30
  • $\begingroup$ This is proven in a paper with a silly name, the darth vader rule on page 2. This paper isn't about your question exactly, but they do answer your question in it. $\endgroup$ – RayVelcoro Nov 10 '15 at 21:25

Assuming that the expectation exists and for convenience that the random variable has a density (equivalently that it is absolutely continuous with respect to the Lebesgue measure), we are going to show that

$$\lim_{x\to\infty} x \left [1-F(x)\right]=0$$

The existence of the expectation implies that the distribution is not very fat-tailed, unlike the Cauchy distribution for instance.

Since the expectation exists, we have that

$$E(X)=\lim_{u\to \infty} \int_{-\infty}^u xf(x) \mathrm{dx} = \int_{-\infty}^{\infty} x f(x) \mathrm{dx} < \infty$$

and this is always well-defined. Now note that for $u \geq 0$,

$$\int_{u}^{\infty} x f(x) \mathrm{dx} \geq u \int_{u}^{\infty} f(x) \mathrm{dx} = u \left[1-F(u) \right]$$

and from these two it follows that

$$\lim_{u \to \infty} \left[ E(X) - \int_{-\infty}^u xf(x) \mathrm{dx} \right] = \lim_{u\to \infty} \int_{u}^{\infty} x f(x) \mathrm{dx}=0$$

as in the limit the term $\int_{-\infty}^u xf(x) \mathrm{dx}$ approaches the expectation. By our inequality and the nonnonegativity of the integrand then, we have our result.

Hope this helps.

| cite | improve this answer | |
  • 4
    $\begingroup$ Thank you (+1). Re relaxing the assumption: when, for instance, $F$ is a Cauchy distribution, then the limiting value of $x(1-F(x))$ is $1/\pi$, not zero. For Student $t$ distributions with parameter less than $1$ ($1$ denotes the Cauchy), this limit is infinite. $\endgroup$ – whuber Nov 10 '15 at 14:23

For any nonnegative random variable $Y$ , we have (see (21.9) of Billingsley's Probability and measure): $$E[Y] = \int Y dP = \int_0^\infty P[Y > t] dt. \tag{*}$$ For $M > 0$, replacing $Y$ by $XI_{[X > M]}$ leads from $(*)$ to $$\int XI_{[X > M]} dP= MP[X > M] + \int_M^\infty P[X > t] dt \geq MP[X > M]. \tag{**}$$

Assume that $X$ is integrable (i.e., $E[|X|]< \infty$), then the left hand side of $(**)$ converges to $0$ as $M \to \infty$, by the dominated convergence theorem. It then follows that $$0 \geq \limsup_{M \to \infty} MP[X > M] \geq \liminf_{M \to \infty} MP[X > M] \geq 0.$$ Hence the result follows.

Remark: This proof uses some measure theory, which I think is worthwhile as the proof assuming the existence of densities doesn't address a majority class of random variables, for example, discrete random variables such as binomial and Poisson.

| cite | improve this answer | |
  • 4
    $\begingroup$ The proof does not really require that $X$ is integrable, but only that $X 1_{ \{X > x_0\}}$ be such for some finite $x_0$, hence $X$ can have an heavy left tail. The identity from Billingsley's book is not really needed else since $X 1_{ \{X > x\}}$ tends to $0$ for $x \to \infty$ with probability one. $\endgroup$ – Yves Nov 10 '15 at 15:58
  • $\begingroup$ @Yves@guy Yes, good point. Integrability is just one sufficient condition but never a necessary one. However, it might be the most succinct and normal condition imposed to derive the relation asked by OP. $\endgroup$ – Zhanxiong Nov 10 '15 at 16:13
  • $\begingroup$ OK. Succinct alternative: $\mathbb{E}(X_+) < \infty$. $\endgroup$ – Yves Nov 10 '15 at 16:27
  • $\begingroup$ @Yves Of course :) $\endgroup$ – Zhanxiong Nov 10 '15 at 16:45

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.