1
$\begingroup$

I have a problem that I can't figure out. Define $$\Gamma\left(x\right):=\frac{\phi(x)}{1-\Phi(x)}$$ where $\phi(x)$, $\Phi(x)$ are the density respectively cumulative distribution function of the standard normal distribution. Hence $\Gamma(x)$ is the hazard function of the normal distribution. Define $$\delta\left(x\right):=\Gamma\left(x\right)\left(\Gamma\left(x\right)-x\right)$$

Is $\delta\left(x\right)$ increasing in $x$, i.e. $\frac{\partial \delta (x)}{\partial x}>0$?

I know $\Gamma(x)$ is monotonously increasing in $x$, but I can't manage to show the same for $\delta(x)$. I have already plottet the function and indeed it is increasing. Does anybody know a reference for this result, or the actual solution to the problem?

Thanks a lot!

$\endgroup$
4
  • $\begingroup$ Have you calculated the derivative of $\delta$ yet? Expressing it in terms of $\phi$ and $\Phi$ sounds like a reasonable thing to try. $\endgroup$ Nov 25, 2014 at 13:50
  • $\begingroup$ Thanks for the comment. Yes, I have tried that. What I get is $ \delta'\left(x\right)=-3x\left(\frac{\phi(x)}{1-\Phi(x)}\right)^{2}+2\left( \frac{\phi(x)}{ 1-\Phi(x)}\right)^{3}-\frac{\phi(x)}{1-\Phi(x)}=-3x \Gamma\left(x\right)^{2}+2 \Gamma \left(x\right)^{3}- \Gamma \left(x \right) $. Not sure, how to continue from here. $\endgroup$ Nov 25, 2014 at 15:06
  • $\begingroup$ Did you look at it numerically; your result seems to hold. $\endgroup$
    – Math-fun
    Nov 26, 2014 at 14:40
  • $\begingroup$ Can you check the derivative? I find the same except that I have $...-Γ(x)(1-x^2)$ $\endgroup$
    – Jimmy R.
    Nov 26, 2014 at 14:41

1 Answer 1

3
$\begingroup$

The function $\delta$ is the derivative of $\Gamma$, so the question amounts to whether the hazard $\Gamma$ is convex. This is in fact the case. I have proved this in an answer to another question here: Standard normal distribution hazard rate

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .