Question

Compute the hazard rate function of a gamma random variable with parameters $(\alpha ,\lambda )$and show it is increasing when $\alpha$$\ge 1$and decreasing when $\alpha \le 1$

Short answer

That is, $t_1\le t_2$implies ${\lambda }_X\left(t_1\right)\le {\lambda }_X\left(t_2\right)$when $\alpha -1\ge 0$.

${\lambda }_X\left(t\right)$is increasing when $\alpha \ge 1$similarly, ${\lambda }_X\left(t\right)$is decreasing when $\alpha \le 1$

Step-by-step solution

Determine the Function of the hazard.

Let $X$be a gamma random variable with parameters $(\alpha ,\lambda )$

Then$f\left(x\right)=$$\left\{\frac{\lambda e^{-\lambda x}(\lambda x{)}^{\alpha -1}}{\Gamma \left(\alpha \right)}\right\}$$x\ge 0,x<0$

Simplify the value.

We can simplify it, then we have

${\lambda }_X\left(t\right)=\frac{1}{{\int }_t^{\infty }{e}^{-\lambda (x-t)}{\left(\frac{x}{t}\right)}^{\alpha -1}\mathrm{d}x}$

Let $y=x-t$,

${\lambda }_X\left(t\right)=\frac{1}{{\int }_0^{\infty }{e}^{-\lambda y}{\left(1+\frac{y}{t}\right)}^{\alpha -1}\mathrm{d}y}$

If $\alpha -1\ge 0,$for $t_1\le t_2,y>0$

We have,

Hence,

${\lambda }_X\left(t_1\right)=\frac{1}{{\int }_0^{\infty }{e}^{-\lambda y}{\left(1+\frac{y}{t_1}\right)}^{\alpha -1}\mathrm{d}y}$
$$\begin{gathered} \le \frac{1}{{\int }_0^{\infty }{e}^{-\lambda y}{\left(1+\frac{y}{t_2}\right)}^{\alpha -1}\mathrm{d}y} \\ ={\lambda }_X\left(t_2\right) \end{gathered}$$${\lambda }_X\left(t_1\right)=\frac{1}{{\int }_0^{\infty }{e}^{-\lambda y}{\left(1+\frac{y}{t_1}\right)}^{\alpha -1}\mathrm{d}y}$