The highlight of today's lecture was the Secretary Problem. This is the most famous of all problems in the field of optimal stopping. It is credited to Merrill M. Flood in 1949, who called it the fiancée problem. It gained wider publicity when it appeared in Martin Gardner's column of Scientific American in 1960. There is an interesting Wikipedia article about it. One of the interesting things said in this article is that in behavioural studies people tend to stop too soon (i.e. marry too soon, make a purchase too soon). See The devil is in the details: Incorrect intuitions in optimal search.
The story about Kepler's search for a wife is taken from the paper, Who Solved the Secretary Problem, by Thomas S. Ferguson. He also discusses the related game of Googol.
The story about Kepler's search for a wife is taken from the paper, Who Solved the Secretary Problem, by Thomas S. Ferguson. He also discusses the related game of Googol.
A variation of the problem that has never been completely solved is the so-called Robbin's Problem. In this problem we do observe values of candidates, say $X_1,\dotsc, X_h$, and these are assumed to be independent, identically distributed uniform$[0,1]$ random variables. The objective is to maximize the expected rank of the candidate that is selected (best = rank 1, second-best = rank 2, etc). It is known only that, as $h$ goes to infinity, the expected rank that can be achieved under an optimal policy lies between 1.908 and 2.329. This problem is much more difficult that the usual secretary problem because the decision as to whether or not to hire candidate t must depend upon all the values of $X_1,\dotsc, X_t$, not just upon how $X_t$ ranks amongst them.
Following this lecture you can do questions 1–4 and 10 on Example Sheet 1. Question 2 is quite like the secretary problem (and also has a surprising answer). The tricks that have been explained in today's lecture are useful in solving these questions (working in terms of time to go, backwards induction, that a bang-bang control arises when the objective in linear in $u_t$, looking at the cross-over between increasing and decreasing terms within a $\max\{ , \}$, as we did in the secretary problem with $\max\{t/h, F(t)\}$).
A student asked in the lecture in regard to Example 2.3 about the characterization of $a_s$ as the least $x$ such that $F_s(x)-x= -p$.
Certainly $a_s\geq p$, since $F_s(p)-p> -p$. It might be that $a_s=\infty$ if $F_s(x)-x> -p$ for all $x$. This could happen, for example, if $\epsilon_t=1$ for all $t$. In that case the share price is guaranteed to be increasing and so one would never wish to exercise the option before the first day. An optimal policy is described by $(a_0,a_1,a_2,\dotsc)=(p,\infty,\infty,\dotsc)$. I have made a small change on page 7 of the notes so this case is mentioned.
A student asked in the lecture in regard to Example 2.3 about the characterization of $a_s$ as the least $x$ such that $F_s(x)-x= -p$.
Certainly $a_s\geq p$, since $F_s(p)-p> -p$. It might be that $a_s=\infty$ if $F_s(x)-x> -p$ for all $x$. This could happen, for example, if $\epsilon_t=1$ for all $t$. In that case the share price is guaranteed to be increasing and so one would never wish to exercise the option before the first day. An optimal policy is described by $(a_0,a_1,a_2,\dotsc)=(p,\infty,\infty,\dotsc)$. I have made a small change on page 7 of the notes so this case is mentioned.
