In my problem set there was an exercise involving optimal stopping theory. Here is the problem: There is an exam, a list of $n$ questions and $n$ students. Student $A$ knows answers to $k$ of them. He sees all the people who took the exam before him and knows whether or not they got one of those $k$ questions he knows. He can choose the moment to take his exam (basing on the knowledge about people going in before him). What strategy should he take to maximise his chances to pass the exam?
I think I know how to solve it, I'm just a bit surprised with the result. I show my solution below.
Let $Z_t=\begin{cases}1 & A \mbox{ knows the answer to } t \mbox{-th question} \\ 0 & \mbox{otherwise} \end{cases}, t=1,\dots,n$.
We notice that $\mathbb{E}Z_t=\mathbb{P}(Z_t=1)$, so we want to find such a stopping time $\sigma$ that $\mathbb{E}Z_{\sigma}=\sup_{\tau}\mathbb{E}Z_{\tau}$, where $\tau$ is any stopping time.
Now let $X_t=\begin{cases}1 & (t-1)\mbox{-th question was one of } k \mbox{ known by } A \\ 0 & \mbox{otherwise} \end{cases}, t=2,\dots, n, X_1=0$.
And $F_t=\sigma(X_1,\dots,X_t)$ - natural filtration.
I want to construct Snell envelope, but $Z$ is not adapted to $\mathbb{F}$, so I introduce $Y_t=\mathbb{E}(Z_t|F_t)$ and finding an optimal stopping time for $Y$ is equivalent to finding an optimal stopping time for $Z$ and $Y$ is of course adapted to $\mathbb{F}$.
Now we construct Snell envelope:
$U_n=Y_n$ $U_t=\max(Y_t,\mathbb{E}(Y_{t+1}|F_t))$ and the optimal stoping time $\sigma$ is given by: $\sigma=\min \{ t \ge 1 : U_t=Y_t \}$.
And now after some calculations I get:
$$Y_t=\frac{k-\sum_{i=1}^t X_i}{n-t+1}$$.
Now the interesting part:
$$\mathbb{E}(Y_{t+1}|F_t)=\mathbb{E}(\frac{k-\sum_{i=1}^t X_i}{n-t}|F_t)-\mathbb{E}(\frac{X_{t+1}}{n-t}|F_t)=\frac{n-t+1}{n-t}Y_t-\frac{1}{n-t}\frac{k-\sum_{i=1}^t X_i}{n-t+1}=\frac{n-t+1}{n-t}Y_t-\frac{1}{n-t}Y_t=Y_t$$, so we get that $Y$ is a martingale and
$U_t=\max(Y_t,\mathbb{E}(Y_{t+1}|F_t))=\max(Y_t,Y_t)=Y_t$
and
$\sigma=\min \{ t \ge 1 : U_t=Y_t \}=\min \{ t \ge 1 : Y_t=Y_t \}=1$. Is this solution correct and it doesn't matter when he takes his exam? His odds stay the same?