Game Theory Homework

Question

Suppose that a store is high quality with probability 1 3 and low quality with probability 2 3 . A high quality store yields high quality goods with probability 1. A low quality store yields high quality goods with probability 1 3 . Suppose that Ann buys goods from a store once, and observes that they are high quality. What is the probability that Ann bought from a high quality store given that she purchased goods that were high quality? Note: Bayes’ rule states that, for two events X and A, P r (X|A) = P r(A|X)P r(X) P r(A) .

Answer 1

Hey there Rob!

The easiest way to do a Bayesian updating problem is to write down all the possible ways you could have observed what was actually observed. In this case, you observed that you bought a high quality shirt, which could happen two ways (1) you went to a high quality store and bought a high quality shirt (scenario 1), or (2) you went to a low quality store and bought a high quality shirt (scenario 2).

Prob(you went to a high quality store | you bought a high quality shirt )

= Prob(scenario 1) / [Prob(scenario 1) + Prob(scenario 2)]

= (1/3)*1 / [(1/3)*1 + (2/3)*(1/3) = 3/5

I like how it makes intuitive sense, because you are seeing how likely scenario 1 is compared to scenario 1 plus scenario 2.

Hope that helps!

Hanna

Answer 2

Hi Robert M! Hope you are doing well.

Let's start off with a few definitions:

Let Hs denote the High Quality Store event. Then P(Hs)=1/3

Let Ls denote the Low Quality Store event. Then P(Ls)=2/3

Let Hp denote the High Quality Product event. We will need to figure out P(Hp), but we will do that later.

Also the probability of getting a high quality product Hp given that it was purchased at the Hp store is P(Hp|Hs)=1.

The probability of getting a hiqh quality product, Hp,given that it was purchased at the Lp store is P(Hp|Ls)=1/3.

We want to find P(Hs|Hp)--namely the probability that Ann brought from a high quality store, Hs, given that she received a high quality product, Hp.

You are correct, that Bayes Theorem will come in handy here.

P(Hs|Hp)=P(Hp|Hs)*P(Hs)/P(Hp). We know that P(Hp|Hs)=1 and P(Hs) = 1/3. We just need to figure out what P(Hp) is.

P(Hp) = P(Hp,Hs)+P(Hp,Ls)

=P(Hp|Hs)P(Hs) + P(Hp|Ls)P(Ls)

Substituting we have P(Hp) = 1/3 + 2/9 = 5/9.

So, P(Hs|Hp) = (1/3) / (5/9) = 3/5.

Hope that makes sense. If you need any further help, please feel free to contact me.

Thanks,

Bob