I am almost sure I have already seen the following result in statistics but I can't remember where.

If $X$ is a positive random variable and $\mathbb{E}(X)<\infty$ then $\varepsilon F^{-1}(1-\varepsilon) \to 0$ when $\varepsilon\to 0^+$, where $F$ is the cdf of $X$.

This is easy to see geometrically by using the equality $\mathbb{E}(X)=\int 1-F$ and by considering a horizontal cut at $\varepsilon$ of the area under the curve of the integrand $1-F$.

Do you know a reference for this result and whether it has a name ?

  • 3
    $\begingroup$ The "more generally" is a straightforward application of integration by parts. That scarcely needs a reference! $\endgroup$
    – whuber
    May 15, 2015 at 16:24
  • $\begingroup$ @whuber I'm asking for a reference about the first result too. $\endgroup$ May 15, 2015 at 16:39
  • 2
    $\begingroup$ You might have seen it, or at least something very much like it, at stats.stackexchange.com/questions/18438. That result is due to a substitution in the integral, which again is so basic one would not expect it to have been especially noted in the literature or given some special name. $\endgroup$
    – whuber
    May 15, 2015 at 16:43
  • 1
    $\begingroup$ @whuber I don't see $\epsilon F^{-1}(1-\epsilon) \to 0$ in your link. Moreover the result I mention is true for a discrete $F$ too (by taking $g$ to be a sequence and replacing $\int$ with $\sum$ in the more general statement). The first result is even true for a general $F$, I think. $\endgroup$ May 15, 2015 at 16:51
  • 2
    $\begingroup$ I believe that this could be used without any reference provided it is stated in more classical terms. Roughly speaking, this is: $x\,\bar{F}(x) \to 0$ for $x \to \infty$ with $\bar{F} := 1 - F$, a direct consequence of: $x \,\text{Pr}\{ X > x \} \leq \mathbb{E}[X \,1_{\{X > x\}}]$ and of dominated convergence. A little work is needed to get the statement for the (left continuous) inverse $F^{-1}$ in the general case where $F$ can have steps. $\endgroup$
    – Yves
    Nov 1, 2015 at 7:28

1 Answer 1


To handle the "little work" suggested by Yves in the comments, geometry suggests a rigorous and fully general proof.

If you like, you may replace all references to areas by integrals and references to "arbitrary" by the usual epsilon-delta arguments. The translation is easy.

To set up the picture, let $G$ be the survival function

$$G(x) = 1-F(x) = \Pr(X \gt x).$$


The figure plots a part of $G$. (Notice the jump in the graph: this particular distribution is not continuous.) A large threshold $T$ is shown and a tiny probability $\epsilon \le G(T)$ has been selected (so that $G^{-1}(\epsilon)\ge T$).

We're ready to go: the value we're interested in, $\epsilon F^{-1}(1-\epsilon) = \epsilon G^{-1}(\epsilon)$ (the one we want to show converges to zero), is the area of the white rectangle with height $\epsilon$ and base from $x=0$ to $x= G^{-1}(\epsilon)$. Let's relate this area to the expectation of $F$, because the only assumption available to us is that this expectation exists and is finite.

The positive part $E_{+}$ of the expectation $\mathbb{E}_F(X)$ is the area under the survival curve (from $0$ to $\infty$):

$$\mathbb{E}_F(X) = E_{+}-E_{-} = \int_0^\infty G(x) dx - \int_{-\infty}^0 F(x) dx.$$

Because $E_{+}$ must be finite (for otherwise the expectation itself would not exist and be finite), we may pick $T$ so large that the area under $G$ between $0$ and $T$ accounts for all, or nearly all, of $E_{+}$.

All the pieces are now in place: the graph of $G$, the threshold $T$, the small height $\epsilon$, and the right-hand endpoint $G^{-1}(\epsilon)$ suggest a dissection of $E_{+}$ into areas we can analyze:

  • As $\epsilon$ goes to zero from above, the area of the white rectangle with base $0\le x \lt T$ shrinks to zero, because $T$ remains constant. (This is why $T$ was introduced; it's the key idea to this demonstration.)

  • The blue area can be made as close to $E_{+}$ as you might like, by starting with a suitably large $T$ and then choosing small $\epsilon$.

  • Consequently, the area left over--which clearly is no greater than the white rectangle with base from $x=T$ to $x=G^{-1}(\epsilon)$--can be made arbitrarily small. (In other words, just ignore the red and gold areas.)

We have thereby broken $\epsilon G^{-1}(\epsilon)$ into two pieces whose areas both converge to zero. Thus, $\epsilon G^{-1}(\epsilon)\to 0$, QED.


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