For a Gamma distribution with shape parameter $\alpha >1$ and scale parameter $\beta > 1$, one can show that its hazard rate function $h$ is increasing and satisfies

\begin{equation} \lim_{x->\infty} h(x) = \frac{1}{\beta} . \end{equation}

Now, let's assume that $T\sim Gamma(\alpha, \beta)$ with $\alpha, \beta >1$ is the waiting time for an event. Then $h(x)dx$ is approximately the probability that $x \leq T \leq x+dx$, given that $T \geq x$. Therefore, for large $x$, this probability will be approximately constant, and equal to $dx /\beta$. So, essentially, the Gamma distribution becomes almost memoryless after a certain point, just like as if it were an exponential distribution.

I have no problem with the mathematical derivation of the results, but I can't figure out what that means, and why it should be so. I'm wondering if someone has a good explanation of this phenomenon.

  • $\begingroup$ I believe that the limit is for $x \to \infty$ rather than $x \to 0$. $\endgroup$
    – Yves
    Commented May 27, 2015 at 5:50

1 Answer 1


Without loss of generality, assume that $\beta =1$.

Let us assume first that $\alpha$ is an integer $n \geq 1$. Then $\text{gamma}(\alpha,\,1)$ is the distribution of $T_n$ where $$ T_k := X_1 + X_2 + \dots +X_k, $$ and where the $X_i$ are iid standard exponential random variables. Thus $T_n$ is the lifetime of a system consisting in a memoryless item with $n-1$ renewals (or replacements) allowed. The lifetime of the $i$-th item is $X_i$ and the renewals occur at the random successive times $T_k$ for $1 \leq k < n$. The survival function of $T_k$ is given by $$ \text{Pr}\{ T_{k} > t \} = \Gamma(k,\,t)/\Gamma(k) $$ where $\Gamma(s,\,t)$ stands for the upper incomplete gamma function. Now for large $t$ $$ \text{Pr}\left[ T_{n-1} > t \vert T_n > t \right] = \frac{\Gamma(n-1,\,t)/\Gamma(n-1)}{\Gamma(n,\,t)/\Gamma(n)}\sim \frac{t^{n-2} e^{-t}/(n-2)!}{t^{n-1} e^{-t}/(n-1)!} = \frac{n-1}{t} = o(1) $$ provided that $n>1$, and obviously the $o(1)$ statement holds for $n=1$ if $T_0:=0$. This means that conditional on the system still being alive at a large time $t$, with high probability $1 - o(1)$ the $n-1$ allowed renewals occurred before $t$. Thus the residual life at $t$ tends to be distributed as is the lifetime $X_n$ of the $n$-th item.

Consider the case with a non-integer $\alpha > 1$. Then $\text{gamma}(\alpha,\,1)$ is the distribution of $$ S_n := Y + \left[ X_1 + X_2 + \dots + X_{n} \right] = Y + T_n, $$ where $n:= [\alpha]$ is the floor of $\alpha$, the $X_i$ are iid exponential random variables as above, and $Y \sim \text{gamma}(\alpha -n, \,1)$ is independent of the $X_i$. Then $S_n$ can be considered as the lifetime of a system with $n$ allowed renewals and a lifetime $Y$ for the initial item. It is easy to show by conditioning on $Y$ that $\text{Pr}\left[ S_{n-1} > t \vert S_n > t \right] = o(1)$, so the argument above still holds.

Interestingly, the same limit for $h(t)$ holds in the decreasing failure rate case $0 < \alpha < 1$, but requires a different intuitive justification.

An even more intuitive formulation is as follows. As is well known, cats have ten lives; assume further that these lives have i.i.d. exponential durations. The previous result reads as: if you know that a cat is very old, then with a probability close to one you can tell that she is in her tenth life.

  • $\begingroup$ Thank you @Yves, this is a great explanation. Where can I find more information about this topic? $\endgroup$
    – user765195
    Commented May 27, 2015 at 18:44
  • 2
    $\begingroup$ Maybe you will be interested in the fact that for $0 < \alpha < 1$ the gamma distribution is a continuous mixture of exponentials, see Leon J. Gleser 1987. This provides a quite clear understanding the limit of $h(t)$ for large $t$ in this case. $\endgroup$
    – Yves
    Commented May 28, 2015 at 6:23

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