Elad Segment 3

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Segment 3 Problems

1) We can simply enumerate all cases and marginalize. We have 6 scenarios - 3 options for doors for us to pick times 2 options for Monty to open a door, subsequently. Let's label the doors as D1,D2,D3 for clarity. Let's label the hypothesis that there's a car behind door as , and having Monty open door as . So let's consider all the cases and see what the posteriors look like (warning, some menial labor ahead).

Case 1 - We guess D1, Monty shows door no. D2

(priors are equal and uniform)

Case 2 - We guess D1, Monty shows door no. D3

(priors are equal and uniform)

Case 3 - We guess D2, Monty shows door no. D1

(priors are equal and uniform)

Case 4 - We guess D2, Monty shows door no. D3

(priors are equal and uniform)

Case 5 - We guess D3, Monty shows door no. D1

(priors are equal and uniform)

Case 6 - We guess D3, Monty shows door no. D2

(priors are equal and uniform)

First of all, the symmetry argument becomes apparent - we're always computing the same thing, only the variables get renamed. The bottom line is always the same - we're always better off "jumping ship" and switching to the other unopened door. Notice that given two empty non-selected doors (i.e. we initially guessed the door with the car), we still need to make some assumption as to what's the likelihood of picking either of the empty door. To make the point, let's assume that instead of tossing a coin randomly whenever we guess the right door, Monty always chooses the door with the lowest ordinal. Let's see what happens in a sample case then:

We guess D1, Monty shows door no. D2, Monty always picks empty door with lowest ordinal if faced with multiple choices

(priors are equal and uniform)

(if is true, Monty will always open door number 2)

Now, in this new setting computing is slightly trickier. Let's consider the influence of the choice of the participant, events respectively. These are equiprobable EME alternatives.

(if the case we chose door 3 and the car is in door 3 Monty will always pick door number 1, in the current setting)

(if the case we chose door 2 and the car is in door 3 Monty will again always pick door number 1, for lack of any other options)

(if the case we chose door 1 and the car is in door 3 Monty will always pick door number 2, for lack of any other options)

So we get that:

Now, are equiprobable! Under the new policy observing increases the likelihood of both doors, and we are not better off switching doors anymore. So our hypothesis about Monty's door selection policy makes a big difference.

Segment 3 "Think About" Problems

1) Primarily one could make an argument along the lines of "well, during the first stage, there's an equal chance of the car being behind any of the three doors. After you pick Monty picks some available door independently of your choice. Why would his action change the probabilities of your initial choice?". A more cunning argument based on my example above (how changing the assumptions on Monty's policy given a choice of empty doors), I could make the following argument: "Suppose Monty only opened the first (lowest ordinal) free door after you made your pick. I can show that the chances of the guess stay equal in this setting. Since it doesn't matter how Monty picks a door, there should never be a benefit in switching". Other sorts of trickery exist, I suppose, but they lie more firmly in the land of rhetoric than in that of probability theory...

2) Clearly that would depend on Monty's policy of when to offer a switch. For instance, if Monty only offers a switch when you've guessed wrong, the benefits of switching increase. If he only offers a switch when your initial guess is correct, you would clearly be worse off for switching. If he offers the switch randomly, then it's the same as the original setting, in expectation.

3) a) The likelihood of the 2nd child being a girl is 50% - the chances of the other child being of a specific gender are independent of the first child (in classical theory, at least. Genetically speaking I believe there's evidence that while the distribution is 50%-50% in the entire population - and there's a game theoretic reason for it - there are biases for individual parents. And even regardless of these studies, as Bayesians, if I am told someone has 9 daughters, I would estimate the chances of the 10th child being a daughter are higher than 50%, even if that was my original prior).

b) Ah, the plot thickens! Now we are not referring to a specific child but are debating over a joint distribution of the two children. Given that we know one of them is a girl our universe has only three options - (boy, girl), (girl, boy) and (girl, girl). So the chances of that child being a girl are .

c) In (a), we fixed on a specific child and asked whether they were a boy or a girl. Our universe comprised of only two options: (boy) vs. (girl). In (b) however, we have a joint distribution of two variables - two children - given that we know one is a girl, so our universe has three equiprobable states - (boy, girl), (girl, boy) and (girl, girl).

27/1/2014 update

Having discussed this in class, it seems Bill intended us to have it the other way around - 1/3 for (a), 1/2 for (b). The source of the ambiguity is semantic - when are we talking about a single child and when are we talking about two children, when are we fixing one child and examining the other whereas the case where there's ambiguity regarding both. I think this is a very misleading question, that relies too heavily on context that's hard to exactly intuit from how the question is phrased (but of course I might just be cranky because apparently I got it wrong ;-) ).