View Full Version : Bayesian Analysis in Mahler Notes

Krieger

August 8th 2005, 05:44 PM

Hello,

I'm going through the Bayesian Analysis portion of Mahler's Study Notes, and I'm stalled on this example.

"Assume there are a total of 100 multi-sided dice of which 60 are 4-sided, 30 are 6-sided and 10 are 8-sided."

The n-sided die is numbered from 1 to n. Your friend picks a die randomly and rolls a 3. You are to calculate the expected roll of the next die, given the first roll is 3, i.e. calculate E(second roll | first roll = 3).

Mahler computes * = P(picking n-sided die | first roll = 3) for n=4,6,8, using Bayes Theorem. He also computes ** =E(second roll | picking n-sided die), easily enough by the definition of expectation.

He multiplied * by ** for each of the n values and sums them to get the desired quantity of E[second roll | first roll = 3).

Can somebody explain to me how this is an application of Bayes Theorem for Total Probability. I don't understand how multiplying * and ** and summing the results would get what we want.

Thanks

wat

August 8th 2005, 06:42 PM

Hello,

I'm going through the Bayesian Analysis portion of Mahler's Study Notes, and I'm stalled on this example.

"Assume there are a total of 100 multi-sided dice of which 60 are 4-sided, 30 are 6-sided and 10 are 8-sided."

The n-sided die is numbered from 1 to n. Your friend picks a die randomly and rolls a 3. You are to calculate the expected roll of the next die, given the first roll is 3, i.e. calculate E(second roll | first roll = 3).

Mahler computes * = P(picking n-sided die | first roll = 3) for n=4,6,8, using Bayes Theorem. He also computes ** =E(second roll | picking n-sided die), easily enough by the definition of expectation.

He multiplied * by ** for each of the n values and sums them to get the desired quantity of E[second roll | first roll = 3).

Can somebody explain to me how this is an application of Bayes Theorem for Total Probability. I don't understand how multiplying * and ** and summing the results would get what we want.

Thanks

Well, there's two parts you need to consider, then.

First, think about the * times the ** part. What is that calculating? Essentially it's a conditional expectation - what's the expected value of the second roll, given that you've picked a certain die? The reason why it's a Bayesian application is because Bayes Theorem applies to using results to calculate the expected value. You're given that the first roll was a 3. Use that information to calculate the expected value. Maybe the number is different than if you hadn't rolled the first dice at all and just wanted to calculate the a priori expected value. Consider this example, though: suppose the first roll was a 5. Then, the numbers change. Why? Well, you know you can't have the 4-sided die, so the probability of rolling a 5 given a 4-sided die = 0. So, you don't need to factor in the expected value of the 4-sided die in your final results.

The second part is the easy one - the summing of the expected values. Essentially, you're just finding the expected values if you have a 4-sided, 6-sided or 8-sided die, and summing those 3 values together for a total expected value.

Krieger

August 8th 2005, 07:27 PM

Well, there's two parts you need to consider, then.

First, think about the * times the ** part. What is that calculating? Essentially it's a conditional expectation - what's the expected value of the second roll, given that you've picked a certain die? The reason why it's a Bayesian application is because Bayes Theorem applies to using results to calculate the expected value. You're given that the first roll was a 3. Use that information to calculate the expected value. Maybe the number is different than if you hadn't rolled the first dice at all and just wanted to calculate the a priori expected value. Consider this example, though: suppose the first roll was a 5. Then, the numbers change. Why? Well, you know you can't have the 4-sided die, so the probability of rolling a 5 given a 4-sided die = 0. So, you don't need to factor in the expected value of the 4-sided die in your final results.

The second part is the easy one - the summing of the expected values. Essentially, you're just finding the expected values if you have a 4-sided, 6-sided or 8-sided die, and summing those 3 values together for a total expected value.

If we want E(second roll | first roll is 3) we would need to compute

sum {possible values of second roll * P(possible value | first roll is 3)}. I don't see how this is equivalent to sum{ E(possible value of second roll | chosen die) * P(chosen die | first roll is 3)}. I tried expanding the conditional probabilities to work from one equation to the other to prove equivalence, but I hit a dead end. :-(

wat

August 8th 2005, 08:01 PM

If we want E(second roll | first roll is 3) we would need to compute

sum {possible values of second roll * P(possible value | first roll is 3)}. I don't see how this is equivalent to sum{ E(possible value of second roll | chosen die) * P(chosen die | first roll is 3)}. I tried expanding the conditional probabilities to work from one equation to the other to prove equivalence, but I hit a dead end. :-(

Right, but given that your first roll is a 3, how do you know which die you rolled? The whole idea is to figure out which die you're using, given that you've rolled a 3.

I guess another way to express what the problem solution writes is:

SUM { possible values of second roll * P(possible values of second roll | n-sided die) * P(n-sided die | first roll = 3) }

= SUM { E(possible values of second roll | n-sided die) * P(n-sided die | first roll = 3) }

Krieger

August 8th 2005, 10:00 PM

Interestingly enough, Loss Models clarified this for me. But is it true that for events A,B,C, that P(A|C) P(C|B) = P(A|B) ?

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