$F$ is the standard normal CDF and $X\sim F$. My question is, can we compute:


directly, or is it the case that we can only bound it from above by


for some suitably small $\epsilon>0$.


2 Answers 2


$F^{-1}(0)$ diverges, which means it doesn't exist. You can't use it in an equation. You can only talk about $$\lim_{\delta\to 0^+}F^{-1}(\delta) = -\infty$$ which has the usual definition for infinite limits.

Therefore, you can show that

$$\lim_{\delta\to 0^+} P(X<(F^{-1}(\delta)+F^{-1}(0.5))/2) = 0$$

  • $\begingroup$ Sorry for being sloppy with limits. I am sure, though, that the r.h.s. of your display should be zero, too? $\endgroup$ Apr 1, 2015 at 8:37
  • $\begingroup$ I prefer this answer (easier for me to understand) $\endgroup$
    – user603
    Apr 1, 2015 at 11:09
  • 1
    $\begingroup$ @ChristophHanck: To be fair, your answer is the more intuitive one. I guess the advantage of the formalism of limits when dealing with infinity is that it encodes the way in which we approach infinity, which is useful when things get more complicated, e.g., $\lim_{x\to \infty} \frac{x}{x}$ versus $\lim_{x\to \infty} \frac{x}{x^2}$. $\endgroup$
    – Neil G
    Apr 2, 2015 at 9:56
  • $\begingroup$ I sure agree about this advantage (+1). As for intuition, yours was intuitive enough for user42397 at least :-). $\endgroup$ Apr 2, 2015 at 10:07

We can compute this as zero:

$F^{-1}(0)$ is the 0-quantile of the $N(0,1)$-distribution, hence $-\infty$ (which value is such that 0% of the probability mass is to the left of it?). Taking the average of this with some finite number like the median $F^{-1}(0.5)=0$ as in $(F^{-1}(0)+F^{-1}(0.5))/2$ still produces $-\infty$, and the probability that a standard normal r.v. (in fact, any r.v.) takes values less than $-\infty$ is 0: $$ P(X<(F^{-1}(0)+F^{-1}(0.5))/2)=P(X<-\infty)=0 $$ R seems to agree with my logic:

[1] 0
  • $\begingroup$ the /2 should be outside of the next to last paranthesis: pnorm((qnorm(0)+qnorm(.5))/2) $\endgroup$
    – user603
    Apr 1, 2015 at 11:07
  • 2
    $\begingroup$ Right, thanks. Did not matter for the final result here, so that unfortunately slipped. Fixed. $\endgroup$ Apr 1, 2015 at 11:14

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