Let's say I'm selling widgets, which people have the right to return if they are defective.
Consider that the probability of a particular widget returned on any particular day after its sale is $ p $ (ie. its equally likely to be returned on any day after its sale). Obviously a given widget can only be returned once. Given p, I know I can calculate the expected time until an item is returned as follows:
With probability p, T=1. (The return happens tomorrow.) If it doesn't happen tomorrow, then we expect to wait the same amount of time starting then. This gives us:
$ E(T) = p + (1-p)(E(T)+1) $
And we can solve for $E(T)$ to get:
$ E(T) = 1/p $
However, what if if I don't accept returns after N days, and people know this so even if they want to return after N days - they won't even try (so I don't know that a return would have occurred).
The problem is that I want to come up with a way to estimate $p_n$ given an actual duration of time until a return is received - $ d_n $.
My eventual goal is to come up with an overall estimate $ P $ by taking the mean of a variety of values $ p_1, p_2, ... $ I get for a number of independent widget returns.
Now I realize it may be preferable to take the mean of $ d_1, d_2, ... $ and calculate $ P=1/D $ from that, but some external constraints prevent me from taking that approach, I have to take the mean of the probabilities, the $ p_N $ values.
With no return window this is easy, we just reverse $ E(T) = 1/p $, to get (where $d$ is the observed duration until a return): $p=1/d$.
Now the question: If I don't get a return within the $ N $ day time window, meaning that I don't know the actual return duration, but I know its greater than $ N $, what $ p $ value do I use to update my mean, to account for this observation?
Could it be $ p=0 $?