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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?

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    $\begingroup$ 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
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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.

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    $\begingroup$ 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
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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.

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    $\begingroup$ 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

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