Help finding the limit of $ x(1-\Phi(x))$

Question

If $\Phi$ is the standard normal cumulative distribution function, what is the value of this expression

$$\lim_{x\rightarrow \infty} {x (1-\Phi(x))}$$ as $x$ approaches infinity?

While $x$ tends to infinity, the term $(1-\Phi(x))$ tends to zero , so it is not clear if it has a limit.

I think it's pretty clear from context that $\Phi(a<x)$ is supposed to mean the probability that the normal r.v. $a$ is less than $x$, which would properly be written as simply $\Phi(x)$. — John Barber, Jan 24 '18 at 04:00
Thanks.. this is correct. Should I edit the question? @Semiclassical, this should address your concern, it's $x(1-\Phi(x))$ — Jorge Guzman, Jan 24 '18 at 04:01
A standard inequality gives you $\left(1-\frac{1}{x^2}\right)\phi(x)\le x(1-\Phi(x))\le\phi(x)$. — StubbornAtom, Jan 24 '18 at 18:55

score 3 · Answer 1 · answered Jan 24 '18 at 04:11

3

Use de l'Hôpital to transform it into

$$\lim_{x\to\infty}x(1-\Phi(x))=\lim_{x\to\infty}\frac{1-\Phi(x)}{1/x}\overset{H}{=}\lim_{x\to\infty}\frac{-\phi(x)}{-1/x^2}=\lim_{x\to\infty}x^2\phi(x)=\frac{1}{\sqrt{2\pi}}\lim_{x\to\infty}x^2\exp(-x^2/2) \; .$$

Can you work this out from this point forward?

answered Jan 24 '18 at 04:11

Raskolnikov

16,333

Yup! Got it from here... super, thank you! – Jorge Guzman Jan 24 '18 at 04:14

Help finding the limit of $ x(1-\Phi(x))$

1 Answers1