# Intuitive meaning of the limit of the hazard rate of a gamma distribution

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.

• I believe that the limit is for $x \to \infty$ rather than $x \to 0$. – Yves May 27 '15 at 5:50

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.
• 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. – Yves May 28 '15 at 6:23