[Math] A property of the hazard function of the normal distribution

normal distributionprobabilityprobability distributionsstatistics

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!

Best Answer

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

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[Math] Distribution of the normal cdf

Let $X$ and $Y$ be the standard normal random variables. Then $$ \mathbb{E}(\Phi(a X + b)) = \mathbb{E}( \mathbb{P}( Y \le a x + b \vert X = x ) ) = \mathbb{P}(Y- a X \le b ) $$ But the combination $Z = Y-a X$ also follows normal distribution (being a linear combination of normals), with zero mean and variance $\mathbb{E}((Y-a X)^2) = 1 + a^2$. Hence $$ \mathbb{E}(\Phi(a X + b)) = \Phi\left(\frac{b}{\sqrt{1+a^2}}\right) $$

Here is numerical checks:

In[14]:= With[{a = 3., 
  b = 1/2}, {NExpectation[CDF[NormalDistribution[], a x + b], 
   x \[Distributed] NormalDistribution[]],
  CDF[NormalDistribution[], b/Sqrt[1 + a^2]]}]

Out[14]= {0.562816, 0.562816}

[Math] Hazard function of $\min(X_1, X_2)$

Answer:

I have given the answer for n random variables which can be simplified to two.

$$F_{min}(x) = P(X_{min} \leq x)$$ $$= 1- P(X_{min} > x)$$$$= 1-P(X_1 > x; X_2 > x;\cdots X_n > x)$$ $$= 1 - [1 - F_1(x)][1 - F_2(x)]\cdots[1 - F_n(x)]$$ For x > 0,$$ F_{i}(x) = 1 - e^{-\lambda x}$$ $$ 1- F_{i}(x) = e^{-\lambda x}$$

$$F_{max}(x) = F_{1}x.F_{2}x.F_{3}x.\cdots F_{n}x$$

Hazard function $= \frac{P(x)}{1-F(x)}$.(probability function/survival fuction) For this to be minimum, we need to have $F_{min}(x)$. Do the calculation appropriately.

Thanks

Satish

Best Answer

Related Solutions

[Math] Distribution of the normal cdf

[Math] Hazard function of $\min(X_1, X_2)$

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