Your message:

Dec. 2, 2008

I'm not sure if this was covered in class but after thinking about it (and staring at my BuyAgent code after extensive debugging due to failing tests) I realized that I don't really know how to price derivatives. I am very aware of how to price a constraint, based on a relation number. However, I am unaware of how to price a derivative that has multiple relation numbers. For example, a derivative with type instance [22] will have a very simple price of 4/9, but what happens when the type instance expands to include [22 1 2 3], which price do I choose? I thought maybe assuming an equal distribution of constraints with respect to relation numbers but that is a big assumption to make with almost no guarantee of it actually happen. Ditto for a normal distribution and then using stats to figure out which relation number would be most likely to occur. Then I thought maybe I could cache how often relation numbers occur based on player id and use that as a statistical approach to choosing the price, but there is also no guarantee for that. I don't feel like choosing a price at random is a good solution either. Is there a better way of determining the price of a derivative with multiple relation numbers?

My response:
Yes, this was covered in class. The short answer: Use a linear combination.

Expected fraction of satisfied constraints in formula F under bias b:
E(Z(F,b)) = SUMMATION ti * E(Z(Ri,b)),
where ti is the weighted ratio for relation Ri.
To find the break-even price, you solve:
min [t1, t2, ...] max [0<=b<=1]  E(Z(F,b))

There are several ways of doing this:

Symbolic closed form solution: compute maximum bias,
followed by computing the worst ti.
Tools from MathWorks or Mathematica would automate this.

Numerical, if only a small number of relations is involved.
Finds a good approximation, but the running time is 1000^(r+1)
where r is the number of relations. Assumes a fine grid of 1/1000.

Using linear programming by expressing the design of the symmetric
formula as a linear program. This requires finding a game inside the game,
as described here:

http://www.ccs.neu.edu/home/lieber/courses/csg113/f08/lectures/dec3/The%20Game%20Inside%20the%20Game.ppt

The linear programming approach is probably the easiest way to
approximate the break-even price of a derivative fairly well.
It is an approximation because the linear program has as many
columns as the symmetric formula has variables.