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Is the hazard rate of the standard normal distribution convex? Can you give a reference?

Mehmet
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    For the sake of self-containment: the question is whether the function $\displaystyle x\mapsto \frac{\varphi(x)}{1-\Phi(x)} = \frac{e^{-x^2/2}}{ \int_x^\infty e^{-u^2/2} , du}\vphantom{\dfrac{\displaystyle\int}{\displaystyle\int}}$ is convex. The function here denoted $\varphi(x)$ is the numerator you see above divided by$ \sqrt{2\pi}$ and $\Phi(x) = \int_{-\infty}^x \varphi(x),dx \vphantom{\dfrac{\displaystyle\int}{\displaystyle\int}}$, and the factor $\sqrt{2\pi}$ has been canceled from the numerator and the denominator. ${}\qquad{}$ – Michael Hardy Jul 04 '15 at 20:22
  • Can you define exactly what you mean by "hazard rate"? That will help people answer your question even if they don't know all the definitions (e.g. I know the normal distribution but I don't know hazard rate). – user2566092 Jul 04 '15 at 20:22
  • @MichaelHardy Thanks, I'll take your word for it – user2566092 Jul 04 '15 at 20:23
  • I would be very surprised if the hazard rate is convex because the normal distribution is log-concave, not concave. Furthermore your hazard function is not symmetric. However I haven't been able to construct these pieces into a definitive proof that your function is not convex. – user2566092 Jul 04 '15 at 20:27

2 Answers2

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$\phi $ is the standard normal density, $\Phi$ is the cdf, $\Phi'=\phi$, $\bar{\Phi}=1-\Phi $ is the survivor function, and $\lambda =\phi /\bar{\Phi}$ is the hazard rate. We know that $% \lambda(s) >s$.

Lemma $\lambda(s) >\frac{3s}{4}+\frac{\sqrt{s^{2}+8}}{4}.$(Baricz / J. Math. Anal. Appl. 340 (2008) 1362--1370, Thm. 2.3)

Proposition The hazard rate $\lambda $ is convex, $\lambda ^{\prime \prime }>0$.

Proof We prove this by differentiation. Note that $\lambda ^{\prime }(s)=\frac{-s\phi(s) }{\bar{\Phi}(s)}+\frac{\phi ^{2}(s)}{\bar{\Phi}^{2}(s)}=\lambda(s) \left( \lambda(s) -s\right) >0$ and then \begin{eqnarray*} \lambda ^{\prime \prime }(s) &=&\lambda(s) \left( \lambda(s) -s\right) \left( \lambda(s) -s\right) +\lambda(s) \left( \lambda(s) \left( \lambda(s) -s\right) -1\right) \\ &=&\lambda(s) \left( s^{2}-3s\lambda(s) +2\lambda ^{2}(s)-1\right) \\ &=&\lambda(s) \left( \left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right) -1\right) . \end{eqnarray*}

This is positive iff $\left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right) -1>0 $. If $s\leq 0$, then $$ \left( 2\lambda(s) -s\right) \left( \lambda(s) -s\right) -1\geq 2\lambda ^{2}\left( 0\right) -1=\frac{4}{\pi }-1>0. $$ For $s>0$, we find the upper root of the polynomial $\lambda \rightarrow \left( 2\lambda -s\right) \left( \lambda -s\right) -1$ , which is $\frac{3s}{4}+\frac{\sqrt{s^{2}+8}}{4}$. We conclude by the lemma that $\lambda(s)$ is greater than the upper root. This implies that $\lambda ^{\prime \prime }(s)>0$ and hence that $\lambda $ is convex.

Mogens Fosgerau
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  • Your proof is incorrect. From lemma it follows that $2 \lambda(s)-s < \sqrt{s^2+4}$ hence for $s>0$ the inequality should be reversed. I.e. $(2\lambda(s)-s)(\lambda(s)-s)-1< \text{smth} > 0$ which does not prove anything. – D F Aug 02 '21 at 09:25
  • I agree, sorry! I have corrected the proof. – Mogens Fosgerau Aug 03 '21 at 11:03
  • @DF Is there something wrong with the existing proofs, or are you just looking for more details? I think the given proofs do give details, so you might have to be more specific. – Sarvesh Ravichandran Iyer Aug 06 '21 at 11:26
  • @TeresaLisbon There was a wrong proof when I started the bounty, but then Morgens Fosgerau came and corrected it. Now everything's fine. – D F Aug 08 '21 at 05:22
  • @DF Good to know, I was just wondering why the bounty when there were a couple of answers. Good to know. – Sarvesh Ravichandran Iyer Aug 08 '21 at 05:22
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Yes, the hazard rate

$$ h(x) = \frac{\varphi(x)}{1-\Phi(x)}$$

of the normal distribution is convex. The second derivative is given by

$$h''(x) = \varphi(x) \frac{\left((1-\Phi(x))\frac{x-\mu}{\sigma}-\frac{3}{2}\sigma\varphi(x)\right)^2 - (1-\Phi(x))^2-\frac{1}{4}\sigma^2\varphi(x)^2}{(1-\Phi(x))^3\sigma^2}$$

This gives $h''(\mu) = \frac{\sqrt{2}}{\pi^{\frac{3}{2}}\sigma^3}(4-\pi) > 0$ and it is also elsewhere positive.

A more interesting question might be whether the log of the survival function $log(1-\Phi)$ is concave or convex since this gives hints about the tail behavior of the distribution. The normal distribution is very light-tailed so we expect a concave log survival function. And indeed, it is concave which is equivalently to an increasing hazard rate. We have

$$ h'(x) = \varphi(x) \frac{\varphi(x)-x(1-\Phi(x))}{(1-\Phi(x))^2} \geq 0 \;.$$

The last inequality holds since it is sufficient to show $$g(x) = \varphi(x)-x(1-\Phi(x)) \geq 0 \;.$$ At $x=0$, $g(x)=\varphi(0)>0$ and at $x=\infty$, $g(x)=0$. In order to be negative the derivative of $g(x)$ would have to have roots which it doesn't have:

$$g'(x) = \Phi(x)-1 < 0 .$$

Soeren Soerensen
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