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.  
 A: 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.
