Class, 9/24

Today we first went through the problem set.

On problem 1, I stressed that the data is that the "expert" said that it is a Super Growth Stock, not that it is a SGS (that is something we don't know, so it can't be data). Something that you know to be true is data; something that you want to know but don't is a state of nature. Because of this confusion, at least one group got their tree wrong by ignoring the branch where SGS was false. This means that they missed counting in their calculation the probability that the "expert" said that the stock was a SGS, when it was not (false positive). If the false positive rate is very large, then you cannot ignore this part of the problem!

The only other problem was that the statement says that your friend is no better than a monkey throwing darts at the stock pages in picking stocks, so the prior probability that he actually picked a SGS is only 1/1000. Some people entertained the notion that it was 1/2, but that contradicts the statement of the problem.

Problem 2 and Problem 1 are basically the same, the only difference is that the false positive rate and the false negative rate are different in this problem (they were the same in problem 1).

In problem 3, the main difficulty was that it has two parts. First, after picking out four chocolate chip cookies in a row, the posterior probability that you have the all chocolate chip cookie box is now 42/43, not 1/2, as some assumed. This changes the probability that the next cookie chosen from the box is a CC cookie quite significantly: It is very close to 1 after you get 4 CC cookies out and no raisin oatmeal cookies.

In problem 5, one group tried to solve it by displaying a particular example of independence in a table and showing that the three relationships were satisfied. The problem is that this only shows it for that particular table, but not in general. To use this method, you'd have to do it for every single one of the infinite number of tables that could be conjured up. This is obviously impossible. What I was looking for was something like this (for one of the things requested):

If P(A|B)=P(A), show that P(A,B)=P(A)P(B).

Because P(A,B)=P(A|B)P(B) no matter what, by substitution from the assumed condition that P(A|B)=P(A) we find that P(A,B)=P(A)P(B).

The More Independence problem was no problem!

We finished the class by discussing a slightly more complicated example of a decision tree, where you are personally deciding whether to install an airbag into your car after, for example, it had deployed in an accident. We showed how the expected value or loss (loss in this case) would propagate backwards in the decision tree, allowing us to cut off a more costly branch at the square (decision) box to choose the best outcome.