How to show that this integral of the normal distribution is finite?

Question

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.

Answer 1

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).

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.

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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 $k\le 1.$

(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.)

Answer 2

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 $$g(u)=\frac{-(2u+1)\exp\left(\frac{-u^2}2-u\right)}{\pi}<1$$

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.

Answer 3

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 $E_X[h(x)]$

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 $x>0$.

Answer 4

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.

Remarks:

• I appreciate @SextusEmpiricus remarks, which led to improvement of this answer.
• $$\phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$$

Answer 5

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.