# $\lim_{x\to\infty}{1-\Phi(x)\over \phi(x)/x}=1$

I have to prove the following:

Let $\phi$ and $\Phi$ denote the standard normal density and distribution functions respectively. Prove that $$\lim_{x\to\infty}{1-\Phi(x)\over \phi(x)/x}=1$$

I am not able to start. $\Phi$ is $\int \phi(x)dx$. How can I calculate the limit without L'Hopital's rule?

• Did you try L'Hopital's rule? Commented Oct 12, 2016 at 20:16
• Not sure what you're asking. I'm not sure it will even work but this is a case with a $0/0$ thing going on, which makes me immediately think of L'Hopital's rule. Differentiate both the numerator and denominator with respect to $x$ and see if the limit of that ratio is 1. Commented Oct 12, 2016 at 20:20
• If that's the case, then perhaps you can motivate the question a little bit. This basically feels like a homework question to me, in which case something like takin' it to the Hospital is the natural answer. Commented Oct 12, 2016 at 20:30
• You are computing the limit of $$x\int_x^\infty \frac{\phi(t)}{\phi(x)}\,dt.$$ Approximate $\phi(t)/\phi(x)=\exp((x^2-t^2)/2)$ as $\exp(-x(t-x))$ and do the integral explicitly. (A rigorous proof uses a two-epsilon argument: the approximation is excellent for small positive values of $t-x$ and you can cut off the integration for higher values.) This at least is more insightful than l'Hopital's Rule, even if less automatic. cc @Jonathan
– whuber
Commented Oct 12, 2016 at 22:15
• Yes, that's the idea. Although it's not as elegant as Dilip's solution, it is interesting to graph the original function $t\to\phi(t)/\phi(x)$ and its approximation: for $t\ge x$, they draw so close together by the time $x=8$ or so that they are indistinguishable. In words: relative to the height of $\phi$ at the value $x$, the function $t\to\phi(t)$ behaves exponentially for $t$ slightly larger than $x$ up until the point (around $t=x+4/x$ or so) that it is indistinguishable from zero.
– whuber
Commented Oct 13, 2016 at 21:07

For $x>0$, write $1-\Phi(x)=Q(x)$ as $$\int_x^\infty \phi(t) dt = \int_x^\infty t^{-1}\cdot t \phi(t) dt$$ and then integrate by parts. You will get that $Q(x)$ is smaller than $\frac{\phi(x)}{x}$ by a term which converges to $0$ as $x\to \infty$. The details can be found in this answer of mine on math.SE. In fact, if you read than answer all the way to the bitter end, you will see that I showed (via integration by parts again!) that for $x > 0$, $$\left.\left.\frac{\phi(x)}{x} \right(1-\frac{1}{x^2}\right)< Q(x) < \frac{\phi(x)}{x}.$$
• How does ${\phi(x)\over x}$ come? It doesnot by partial integration... What am I missing? Commented Oct 13, 2016 at 6:55
• Did you notice that the antiderivative of $t\phi(t)$ is $-\phi(t)$? Did you forget to use the limits? Commented Oct 13, 2016 at 10:21
$$\frac{1-\Phi\left(x\right)}{\phi\left(x\right)}=\frac{1}{\lambda\left(-x\right)}$$ where $$\lambda\left(x\right)=\frac{\phi\left(x\right)}{\Phi\left(x\right)}$$ is the inverse-Mill's ratio. It has the properties that $$\lambda\left(-x\right)>x;$$ $$0\le\lambda'\left(x\right)\le-1.$$ So $$\lim_{x\to\infty}\frac{1-\Phi\left(x\right)}{\phi\left(x\right)/x}=\lim_{x\to\infty}\frac{x}{\lambda\left(-x\right)}$$ Applying L'hospitals rule yields $$\lim_{x\to\infty}\frac{1}{\lambda'\left(-x\right)}.$$ It's known that $$\lim_{x\to\infty}\lambda'\left(-x\right)=1.$$