# Proof that $x \Phi(x) + \Phi'(x) \geq 0$ $\forall x$, where $\Phi$ is the normal CDF

As title. Can anyone supply a simple proof that

$$x \Phi(x) + \Phi'(x) \geq 0 \quad \forall x\in\mathbb{R}$$

where $\Phi$ is the standard normal CDF, i.e.

$$\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-y^2/2} {\rm d} y$$

I have so far:

Defining $f(x) = x \Phi(x) + \Phi'(x)$ we get

\begin{align} f'(x) & = \Phi(x) + x \Phi'(x) + \Phi''(x) \\ & = \Phi(x) + x\Phi'(x) - x\Phi'(x) \\ & = \Phi(x) \\ & >0 \end{align}

so it seems that if we can show

$$\lim_{x\to-\infty} f(x) = 0$$

then we have our proof - am I correct?

Clearly $f$ is the sum of two terms which tend to zero, so maybe I have all the machinery I require, and I just need to connect the parts in the right way! Assistance will be gratefully received.

In case anyone is interested in where this question comes from:

Bachelier's formula for an option struck at $K$ with time $T$ until maturity, with volatility $\sigma>0$ and current asset price $S$ is given by

$$V(S) = (S - K) \Phi\left( \frac{S-K}{\sigma S \sqrt{T}} \right) + \sigma S \sqrt{T} \Phi' \left( \frac{S-K}{\sigma S \sqrt{T}} \right)$$

Working in time units where $\sigma S\sqrt{T} = 1$ and letting $x=S-K$, we have

$$V(x) = x \Phi(x) + \Phi'(x)$$

and I wanted a simple proof that $V(x)>0$ $\forall x$, i.e. an option always has positive value under Bachelier's model.

• $\Phi^\prime(x)$ is always $> 0$, and $x \Phi(x) \geq 0$ if $x \geq 0$, so your only worry is negative $x$... – J. M. is a poor mathematician Sep 23 '11 at 14:20
• @SrivatsanNarayanan Yes, I just realized that and deleted my comment. Sorry – Sasha Sep 23 '11 at 14:26
• $\Phi''(x) \ne -x\Phi(x)$. Maybe you're misremembering $\varphi'(x) = -x\varphi(x)$, where $\varphi=\Phi'$. – Michael Hardy Sep 23 '11 at 14:30
• A search for "Miller's inequality" lists this page as the first relevant result :) – Chris Taylor Sep 23 '11 at 15:01
• @Chris: that means it is all the more important that we check that what we post is accurate, for the benefit of future searchers. ;) – J. M. is a poor mathematician Sep 23 '11 at 15:22

Writing $\phi(x) = (2\pi)^{-1/2}\exp(-x^2/2)$ for $\Phi^{\prime}(x)$ $$x\Phi(x) + \phi(x) = \int_{-\infty}^x x\phi(t)\mathrm dt + \phi(x) \geq \int_{-\infty}^x t\phi(t)\mathrm dt + \phi(x) = -\phi(t)\biggr\vert_{-\infty}^x + \phi(x) = 0$$

• Wow! ${}{}{}{}{}$ – Srivatsan Sep 23 '11 at 16:50
• Thanks. The OP asked for a simple proof and this seemed to fit the bill. I had posted a different and more long-winded Answer earlier, but deleted it in view of the some of the comments that appeared while I was writing the Answer out. Then I came up with this one and so I revised my Answer and undeleted it. – Dilip Sarwate Sep 23 '11 at 16:55
• Excellent - thanks a lot! This was just what I needed. – Chris Taylor Sep 23 '11 at 19:44
• @Srivatsan: Yeah. Why didn't I read this before answering? – robjohn Sep 23 '11 at 22:23

As was pointed out by J.M in the comment, our worry is to show the inequality for $x<0$. We will begin with the following observations:

$$\Phi''(x)\geq 0,\:\text{ in }(-\infty,0).$$ Indeed $$\Phi''(x)=-xe^{-x^2/2}>0,\quad\forall x< 0.$$

Now, consider the function $$g(x)=-e^{-x^2/2}\Phi(x).$$ We rearrange the factors to obtain $$g(x)=-\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\Phi(x)=-\sqrt{2\pi}\Phi'(x)\Phi(x).$$ Moreover if we pick $z<x<0$, by our previous observations and the fact that both $\Phi'(x)$ and $\Phi(x)$ are greater than $0$ we can conclude $$g(x)-g(z)=-\sqrt{2\pi}(\Phi'(x)\Phi(x)-\Phi'(z)\Phi(z))<0.$$ But then $g$ is strictly decreasing on $(-\infty,0).$ Hence we can write $g'(x)<0.$ This relation translates in $$-(x\Phi(x)+\Phi'(x))e^{-x^2/2}<0.$$ Divide out by $-e^{-x^2/2}$ to obtain, for any $x<0$, $$x\Phi(x)+\Phi'(x)>0.$$ Hope this is correct.

I will push the OP's attempt to a complete proof. If $f(x) = x\Phi(x) + \Phi'(x)$, then, as the OP notes, $f'(x) \geq 0$ for all $x$; i.e., $f$ is monotonically increasing. So it suffices to show that $f(x) \to 0$ as $x \to -\infty$, which implies that $f(x) \geq 0$ for $x \in \mathbb R$, since: $$\lim_{u \to -\infty} f(u) \leq f(x).$$

Now, to show $f(x) \to 0$ as $x \to -\infty$, it is enough to show that $x\Phi(x) \to 0$ and $\Phi'(x) \to 0$ as $x \to -\infty$.

1. $\Phi'(x) = \exp\left(-\frac{1}{2} |x|^2 \right)$ clearly approaches $0$ as $x \to -\infty$.
2. For $x \leq -2$, let $y = |x| \geq 2$. $$|x \Phi(x)| = y \int_{-\infty}^x e^{-t^2/2} dt = y \int_{y}^\infty e^{-t^2/2} \leq y \int_{y}^\infty e^{-t} dt = y e^{-y} ,$$ which approaches $0$ as $x \to -\infty$.

So we are done.

Notes on the name. The quantity $$R(x) = e^{x^2/2} \int_x^{\infty} e^{-t^2/2} dt$$ is called Mill's ratio (see, for e.g., http://www.jstor.org/stable/2236360). I have also seen the inequality $$\int_{x}^\infty e^{-t^2/2} dt \leq \frac{e^{-x^2/2}}{x} \ \ \ \ \ (x > 0),$$ sometimes referred to as Mill's inequality (see Theorem 6 in this lecture notes: http://www.stat.cmu.edu/~larry/=stat705/N3.pdf). Notice that by a change of variables $x \to -x$, this inequality can be seen to be equivalent to the one in the question.

• Thanks, especially for the relationship with Mill's inequality. – Chris Taylor Sep 23 '11 at 19:46
• I proved the equivalent inequality for $x<0$ in $(2)$ in my answer. – robjohn Sep 23 '11 at 21:44
• (+1) Here is a simple proof of the inequality you state in your addendum. L'Hopital shows that it is asymptotically tight as $x \to \infty$. This inequality (and much more) was known to Laplace. – cardinal Sep 23 '11 at 22:05

I will concentrate at $\lim_{x \to -\infty} f(x) = 0$, where $f(x) = x \Phi(x) + \Phi'(x)$.

Consider $$g(x) = \sqrt{2 \pi} f(-x) = \mathrm{e}^{-\frac{x^2}{2}} - x \int_{x}^\infty \mathrm{e}^{-\frac{y^2}{2}} \mathrm{d} y \qquad \text{for} \qquad x > 0.$$ Then $$g(x) = \mathrm{e}^{-\frac{x^2}{2}} - x \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \frac{\mathrm{d} t}{\sqrt{2t}} \, > \, \mathrm{e}^{-\frac{x^2}{2}} - \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = 0$$ where $x \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \frac{\mathrm{d} t}{\sqrt{2t}} < \frac{x}{\sqrt{2 \frac{x^2}{2}}} \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = \int_{\frac{x^2}{2}}^\infty \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^{-\frac{x^2}{2}}$ for $x>0$.

On the other hand $g(x) < \mathrm{e}^{-\frac{x^2}{2}}$ by definition, so $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} \sqrt{2 \pi} f(-x)$ vanishes being sandwiched between 0 and $\mathrm{e}^{-\frac{x^2}{2}}$ which also tends to zero for large $x$.

Taking the derivative of $\Phi(x)$, $$\Phi'(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\tag{1}$$ For $x\ge0$, everything is non-negative, so $x\Phi(x)+\Phi'(x)\ge0$.

For $x<0$, \begin{align} \Phi(x) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\;\mathrm{d}t\\ &\le\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\frac{t}{x}\;\mathrm{d}t\\ &=-\frac{1}{x}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\\ &=-\frac{1}{x}\Phi'(x)\tag{2} \end{align} Therefore, we also get $x\Phi(x)+\Phi'(x)\ge0$.

• I just reviewed my answer and $(2)$ proves the whole thing for $x<0$. Doh! I could remove the rest, but then I think it would be the same as Srivatsan's answer. – robjohn Sep 23 '11 at 21:56
• I removed the rest anyway. – robjohn Sep 23 '11 at 22:05