# Distribution to reflect situation where some waiting leads us to expect more waiting

In reading Blake Master's notes on Peter Thiel's lecture on start ups, I came across this metaphor of the technology frontier:

Picture the world as being covered by ponds, lakes, and oceans. You’re in a boat, in a body of water. But it’s extremely foggy, so you don’t know how far it is to the other side. You don’t know whether you’re in a pond, a lake, or an ocean.

If you’re in a pond, you might expect the crossing to take about an hour. So if you’ve been out a whole day, you’re either in a lake or an ocean. If you’ve been out for a year, you’re crossing an ocean. The longer [the] journey, the longer your expected remaining journey. It’s true that you’re getting closer to reaching the other side as time goes on. But here, time passing is also indicative that you still have quite a ways to go.

My question: is there a probability distribution or statistical framework that best models this situation, especially the bolded part?

The exponential distribution has the property of being "memoryless," i.e. (using your analogy) the length of your journey so far has no effect on the length of the remaining journey. If the density of distribution decays faster than that of the exponential distribution, then a longer journey will mean a shorter remaining journey; conversely, a density that decays slower than exponential (see e.g. subexponetial distributions) will have the property you describe.

Since I think the comparison with memorylessness is clearest, my first suggestion would be to look at other distributions for which the exponential distribution is a special case. That will allow you to control fairly intuitively the magnitude of this effect. The Weibull distribution with shape parameter $<1$ would be a good choice.

• Good answer bnaui. I was planning to say something similar. Jun 8, 2012 at 15:31
• Good answer, thanks. I like the connection to memorylessness and deviations from it. This is a much better explanation than ones I was going on, and which I almost didn't ask this question because of, ask.metafilter.com/152125/Waiting-begets-waiting Jun 8, 2012 at 20:33

The answer by bnaul gives the general property you are looking for. Rather than a Weibull distribution though, I'd suggest the Pareto distribution as the best example. The general pdf is $$f(x) = \frac{\alpha x_m}{x^{\alpha - 1}}$$ with support $[x_m, \infty)$ and $\alpha>0$. This has the nice property that conditional on $x>y$, the distribution has the same shape parameter, but with $y$ as the new minimum.

The distribution has $E[x] = \frac{\alpha x_m}{\alpha - 1}$. Suppose $\alpha=2$. Then, conditional on waiting $T$ days, you should expect the event to happen at time $2T$.

• We can draw two connections here. First, @bnaul's example is illustrative because the exponential is a special case of the Weibull, the latter of which has a monotone hazard function. Depending on the shape parameter, it can cover both the case of "the longer you wait, the longer you expect to wait" and also the case of "the longer you wait, the shorter you expect to have to continue to wait". Your example is nice because the Pareto is the exponentiation of an exponential, and from this fact many of its properties are derived, including the one you mention. Jun 8, 2012 at 19:39
• +1 good answer, thanks. This makes the process a bit more intuitive. Jun 8, 2012 at 20:32