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Let T be a gamma random variable with parameters $s>0$ and $\lambda>0$. The PDF of T is $f(t) = \frac{\lambda^st^{s-1}}{e^{\lambda t}\Gamma(s)}$. I need to show that the failure rate of T, $\lambda(t)=\frac{f(t)}{1-F(t)}$ is increasing when $s\geq1$, and decreasing when $s\leq1$.

Since $1-F(t)$ decreases as $t$ increase for any $s>0$, therefore if $f(t)$ is increasing/decreasing then $\lambda(t)$ is increasing/decreasing.

We have $f(t) = \frac{\lambda^st^{s-1}}{e^{\lambda t}\Gamma(s)} = \frac{\lambda^s}{\Gamma(s)} \frac{t^{s-1}}{e^{\lambda t}}$, with $\frac{\lambda^s}{\Gamma(s)}$ constant and positive.

No matter what $s$ is, $\frac{t^{s-1}}{e^{\lambda t}}$ will eventually be decreasing and tend towards $0$. What am I doing wrong?

Heng Wei
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    $f$ decreases to zero as $t \to \infty$, so your approach of looking at the numerator and denominator separately will not work. – angryavian Sep 11 '23 at 01:10

1 Answers1

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The claim

If a function $g(x)$ is monotone decreasing, then $f(x)/g(x)$ increases whenever $f$ increases, and decreases when $f$ decreases.

is not true. Let $f(x) = x$ and $g(x) = 1/x^2$. The function $g$ is decreasing for all $x > 0$, and $f$ is increasing on the same interval, but $f(x)/g(x) = 1/x$ is decreasing.

To understand the behavior of the hazard function $\lambda(t)$, you need to consider its derivative. We know it is continuous for all $t > 0$. So

$$\begin{align} \lambda'(t) &= \frac{d}{dt}\left[\frac{f(t)}{1 - F(t)}\right] = \frac{(1-F(t))f'(t) + f(t)^2}{(1-F(t))^2}. \end{align}$$

Since the denominator is always positive, the derivative is positive (hence $\lambda(t)$ is increasing) if its numerator is positive. I leave the rest as an exercise.

heropup
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