- Let T > 1. We observe a Poisson process of rate 1 on the time interval (0, T ). Each time a point occurs, we may decide to stop. Our goal is to stop at the last point which occurs before time T; if so, we win, and otherwise – i.e. if we never stop, or if we stop at some time t but another point occurs in (t, T ) – we lose. Find the best strategy you can for playing this game. What is its probability of winning? Can you show that it’s optimal? Feel free to argue informally (but convincingly!).
I haven't for very far with this at all. I know that there is a exp(-T) probability that nothing happens and so it's impossible to win. I then tried looking at values and figuring out the probability of winning each time but since one obviously doesn't know how many points will be in the process I can't decide what the probabilities of winning actually are. Any help would be much appreciated!
Consider strategies of the form "stop at the first occurrence after time $c$" where $c<T$. Then you win if and only if there is exactly one occurrence in the interval $[c,T]$: if there is more than one occurrence in that interval, then you stopped too soon, while if there are zero occurrences in that interval, then you did not stop before it was too late (or even worse there were no occurrences in $[0,T]$ at all).
The probability of exactly one occurrence in $[c,T]$ is $(T-c)e^{-(T-c)}$. This is a strictly concave function in $c$, so it has a maximum. The derivative with respect to $c$ is $e^{-(T-c)} ((T-c) - 1)$, and setting it equal to zero gives $c=T-1$. This matches our intuition: the expected time of the next arrival after some time $t$ is $t+1/\lambda=t+1$, so once we are in $[T-1,T]$, we are close to $T$ and should stop. So, among all strategies of the form "stop at the first occurrence after time $c$," the best strategy is choosing $c=T-1$.
Now, what remains is to [informally?] justify why the above type of strategy is the only reasonable strategy...