Numerically, I have noticed that

$$\int_{-\infty}^{\infty} \dfrac{\phi(x)^2}{\Phi(x)}dx < \infty$$ where $\phi$ and $\Phi$ are the standard normal pdf and cdf. However, I do not see how to prove it. I would appreciate any hints.

  • 3
    $\begingroup$ Empirically about $0.903197$ $\endgroup$
    – Henry
    Jan 10, 2022 at 20:51

5 Answers 5


Intuitively, the result is obvious because (a) $\phi$ is a rapidly decreasing function (its magnitude decreases at a quadratic exponential rate) and (b) $\Phi$ is bounded above and, for negative $x,$ is also rapidly decreasing at essentially the same rate as $\phi.$ Thus the fraction $\phi^2/\Phi$ decreases rapidly for both positive and negative $x,$ while remaining bounded in between, whence its integral is very nicely behaved and finite.


The problem, then, is to make this intuition rigorous. The rigor merely parallels the foregoing argument by making suitable quantitative comparisons.

When $x\gt 0,$ $\Phi(x)\ge 1/2$ (by a familiar symmetry argument). Whence (in this case) the integrand is bounded above in magnitude by

$$\bigg|\frac{\phi(x)^2}{\Phi(x)}\bigg| \le 2\phi(x)^2 \lt 2\exp(-x^2)/\sqrt{2\pi}$$

which (because it is proportional to the density of another Normal distribution) has a finite integral.

The harder part is the integral over negative $x.$ But here, the Mills Ratio is

$$R(-x) = \frac{\Phi(x)}{\phi(x)}$$

which, as the linked post explains, is bounded below by $-x/(x^2+1).$ Thus, for large negative $x$ (say, $x \le -1$),

$$\bigg|\frac{\phi(x)^2}{\Phi(x)}\bigg| = \bigg|\phi(x)\left(\frac{1}{R(-x)}\right)\bigg| \le \phi(x) \frac{x^2+1}{|x|} \le 2|x|\phi(x)$$

whose integral also converges (it can integrated exactly using elementary techniques). Figure 2

Since $\phi(x)^2/\Phi(x)$ is bounded on the remaining interval $[-1,0],$ we have established that its integral over the entire real line is bounded in magnitude by the sum of three convergent quantities, whence it is finite, QED.

This argument continues to apply, essentially without change, to any integrand of the form $\phi(x)^k/\Phi(x)$ for $k\gt 1.$ It also shows (look more closely at the upper and lower bounds for Mills' Ratio) that the integral diverges when $k\le 1.$

Figure 3

(For a plot with $k=1$ -- which is just the inverse Mills' Ratio -- see the last figure of https://stats.stackexchange.com/a/166277/919.)


Here is a self-contained elementary argument, by a comparison with the Laplace distribution. We show $$\int_{-\infty}^{\infty}\frac{\phi(x)^2}{\Phi(x)}dx<\frac{1}{2\sqrt{\pi}}+\sqrt{\frac\pi2}\simeq 1.535<\infty$$

The positive side of the integral has the bound $$\int_{0}^{\infty}\frac{\phi(x)^2}{\Phi(x)}dx< \int_{0}^{\infty}\frac{\phi(x)^2}{1/2}dx=\frac{1}{2\sqrt{\pi}}$$

The negative side of the integral has the bound $$\int_{-\infty}^{0}\frac{\phi(x)^2}{\Phi(x)}dx< \int_{-\infty}^{0}\sqrt{2\pi}\frac{f_L(x)^2}{F_L(x)}dx=\sqrt{\frac\pi2}\phantom{1<\infty}$$

Since the Laplace distribution has $f_L(x)=F_L(x)=\frac12 e^x$ on $x<0$, the integral is easy.

To prove that $\phi^2/\Phi<\sqrt{2\pi} f_L^2/F_L$, we start by proving $$g(u)=\frac{-(2u+1)\exp\left(\frac{-u^2}2-u\right)}{\pi}<1$$

graph of g

Using the first-derivative test, the maximum occurs at exactly $u_\max=-(3+\sqrt{17})/4$, and by numerical evaluation $g(u_\max)<1$. Thus for $x<0$, \begin{align} \phi(x)^2\frac{F_L(x)}{f_L^2(x)} &=\frac{\exp(-x^2)}{2\pi}\frac{\frac12 \exp(x)}{\frac14 \exp(2x)}\\ &=\frac{\exp(-x^2-x)}{\pi}\\ &=\int_{u=-\infty}^{x}\frac{-(2u+1)\exp(-u^2-u)}{\pi}du\\ &<\int_{u=-\infty}^{x}\exp(-u^2/2)du\\ &=\sqrt{2\pi}\Phi(x) \end{align} which is what we need.


The integral relates to the expectation value of the normal hazard function

$$E_X[h(x)] = E_X\left[\dfrac{\phi(x)}{1-\Phi(x)}\right] = \int_{-\infty}^{\infty} \left(\dfrac{\phi(x)}{1-\Phi(x)}\right) \phi(x) dx = \int_{-\infty}^{\infty} \left(\dfrac{\phi(x)}{\Phi(x)}\right) \phi(x) dx$$

The last step is due to the symmetry.

This hazard function will be approximately asymptotic to $x$ at positive infinity and will be smaller than $x^2$ (except in some region where it is finite).

Since $E_X[x^2]$ is finite so must be $E_X[h(x)]$

hazard function

The big difference between $x^2$, which is already resulting in a finite integral, and the hazard function $h(x)$ indicates that this is not a hard problem and there are probably many approaches to show that the integral does not diverge (as seen in many answers to the question).

For instance, we can more precisely see this asymptotic behaviour of the hazard function in the expression

$$log(1-F(t)) = log(S(t)) = -\int_{-\infty}^x h(t) dt$$

This can be related to the asymptotic behavior of the error function (as $\text{erfc}(x/\sqrt{2})/2 = 1-F(x)$), which has exponential bounds with factor $e^{-x^2}$ for $x>0$.


Intuitive approach
$\Phi(x)$ is changing in the interval $[0,1]$, whereas $\phi(x)^2$ is itself a normal distribution which is integrable. Thus, the only reason why the integral could not be finite, is because it diverges in the limit $x\rightarrow -\infty$. By L'Hôpital's rule: $$ \frac{\phi^2(x)}{\Phi(x)}\sim \frac{2\phi(x)\phi'(x)}{\phi(x)}\sim \phi'(x)= -\frac{2x}{\sigma^2}\phi(x)\\(\text{as }x\rightarrow -\infty) $$ In other words, the asymptotic form of the integrand for $x\rightarrow-\infty$ is $$ \frac{\phi^2(x)}{\Phi(x)}\sim -\frac{2x}{\sigma^2}\phi(x), $$ the integral of which converges in the desired limit.

Rigorous approach
Perhaps, a more rigorous way could be to show that in the limit $x\rightarrow -\infty$ the integrand decays faster than $1/x$,a nd hence convergent. We this apply the L'Hôpital's rule to the following: By L'Hôpital's rule: $$ \lim_{x\rightarrow -\infty}\frac{x\phi^2(x)}{\Phi(x)}= \lim_{x\rightarrow -\infty}\frac{\phi^2(x)+x\phi(x)\phi'(x)}{\phi(x)}= \lim_{x\rightarrow -\infty}\left[\phi(x)+x\phi'(x)\right]= \lim_{x\rightarrow -\infty}\left[1-\frac{x}{\sigma^2}\right]\phi(x)=0. $$Thus, the integral converges in the limit in question.


  • I appreciate @SextusEmpiricus remarks, which led to improvement of this answer.
  • $$\phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$$
  • $\begingroup$ This shows that the term is decaying to $0$ but not that it isn't decaying slower than $1/x$. I guess that you can still use the same technique though, by showing $$\lim_{x\rightarrow -\infty} \frac{x\phi^2(x)}{\Phi(x)}=0$$ $\endgroup$ Jan 12, 2022 at 14:44
  • $\begingroup$ @SextusEmpiricus or one could show that $xe^{-\frac{x^2}{2\sigma^2}}$ decays faster than $1/x$ - either by the second application of the L'Hôpital's rule or by noting that the integral of this quantity is known to converge - which is what I mean by my last remark. Perhaps, I should add this comment to my answer. $\endgroup$
    – Roger V.
    Jan 12, 2022 at 14:54
  • $\begingroup$ I find it a nice short proof, but can we really say that if $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = 0$ and if the integral of g(x) converges, then also the integral of f(x) converges? $\endgroup$ Jan 12, 2022 at 15:00
  • $\begingroup$ Counterexample $\lim_{x \to \infty} 1/x = \lim_{x \to \infty} 1/x^2 = 0$ $\endgroup$ Jan 12, 2022 at 15:01
  • $\begingroup$ @SextusEmpiricus I am talking here about the asymptotic form of the integrand. The limit notation might be somewhat misleading here - I am not equating any unrelated expressions (as in your last comment), I was simply too lazy to write explicitly the coefficient. The approach suggested in your first comment works as well, but it would look less intuitive to me. $\endgroup$
    – Roger V.
    Jan 12, 2022 at 15:50

The integral on any interval $[a, \infty)$ is clearly finite. Now $\varphi(x)$ is convex on $(-\infty, -1]$ and so the tangent line to $\varphi$ at any $x < -1$ lies below $\varphi$. This tangent intersects the $x$-axis in $x+x^{-1}$. This gives the lower bound $$\Phi(x) > -\frac1{2x} \varphi(x)$$ for all $x < -1$ and so $$\frac{\varphi(x)^2}{\Phi(x)} < -2x \varphi(x)$$ for all $x<-1$. This shows that the integral over $(-\infty, -1]$ is also finite.

  • 1
    $\begingroup$ This is a fun proof. An image would be nice, to illustrate this tangent line. $\endgroup$ Jan 12, 2022 at 19:13
  • $\begingroup$ +1 This is one of the most elementary possible answers. $\endgroup$
    – whuber
    Jan 12, 2022 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.