limit of $x \left[1-F(x) \right]$ as $x \to \infty$

Question

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?

Answer 1

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.

Answer 2

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 $M > 0$, substituting $Y$ in $(*)$ with $XI_{[X > M]}$ to \begin{align} \int XI_{[X > M]} dP &= \int_0^\infty P[XI_{[X > M]} > t]dt \\ &= \int_0^M P[XI_{[X > M]} > t]dt + \int_M^\infty P[XI_{[X > M]} > t]dt \\ &= MP[X > M] + \int_M^\infty P[X > t] dt \geq MP[X > M]. \tag{$**$} \end{align}

The third equality holds because for every $t \in [0, M]$, $\{XI_{[X > M]} > t\} = \{X > M\}$, and for every $t > M$, $\{XI_{[X > M]} > t\} = \{X > t\}$. To wit, say, if we want to show for every $t \in [0, M]$, $\{XI_{[X > M]} > t\} = \{X > M\}$, just note \begin{align} \{XI_{[X > M]} > t\} &= (\{XI_{[X > M]} > t\} \cap \{X > M\}) \cup (\{XI_{[X > M]} > t\} \cap \{X \leq M\}) \\ &= (\{X > t\} \cap \{X > M\}) \cup (\{0 > t\} \cap \{X \leq M\}) \\ &= \{X > M\} \cup \varnothing = \{X > M\}. \end{align}

Similarly it is easy to show for every $t > M$, $\{XI_{[X > M]} > t\} = \{X > t\}$.

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.