# Optimal solution to exponential waiting problem

I have the following experiment and I'm struggling to find what is the optimal solution given some assumptions:

An agent has to complete a certain number of trials and her objective is to maximize the average reward rate. In each trial, a reward $R=1$ (its value doesn't really matter) is available after some time $T$, measured from the beginning of the trial, that follows an exponential distribution with rate $\lambda$. The agent can wait to collect the reward, or forfeit that trial and move on to the next. Between each trial there is always a fixed inter-trial-interval $I$, regardless of what the agent did.

What I want to obtain is the optimal decision rule as a function of $I$, assuming that the agent knows $\lambda$. If I'm not wrong, for a given trial, the expected reward rate as a function of waited time $w$ is:

$$\frac{P(T<=w)}{w+I} = \frac{1-\exp(-\lambda w)}{w+I}$$

which has a maximum at $w^{*} = -W^{-1}(e^{-(I+1)})-(I+1)$, where $W^{-1}$ is the product logarithm function. So one could say that the optimal strategy for one trial is to wait until $w^{*}$.

However, I'm unsure on how to extend this to consider an arbitrarily large number of trials $N$. Would the same strategy apply for every trial?

As $T$ is exponential (memoryless), the distribution of the remainder given the time already passed during the current trial is distributed the same as a new trial. Therefore you always have to wait $T$ more time. But if you restart then you will wait $T+I$. So never restarting is an optional policy for any $I>0$.
• Ah, I see! I think I described the solution for the problem where you don't know if the reward has been delivered or not, and you only have one chance to check. Then you should check at time $w^{*}$. – J. R. C. Aug 28 '17 at 14:16