P1.T2.20.2. More probabilities and Bayes rule

[navjyotbirdy]

Hi @David Harper CFA FRM Can you give more explanation to the example in the study notes of Probabilities, I have highlighted my doubts in yellow.

I don't understand on 39.13*75(which is conditional beats of first year?)

Thanks
Navjyot

 

[navjyotbirdy]

Hi @David Harper CFA FRM Is the below questions from study note (probabilities chapter) complete?




Thanks,
Navjyot

 

[David Harper CFA FRM]

Hi Navjyot (@navjyotbirdy ) In #2.3, because we don't know whether the manager is a star/not, the probability she beats next year is a weighted probability: 39.13% probability that she is a star, which we induced via Bayes Pr[S|3B], multiplied by 75% probability that a star beats; plus 60.87% probability that she is not a star, which is given by (1-60.87%), multiplied by the 50% probability that a non-star beats (i.e., 50%). Put is this way, if we only used the a priori (unconditional) probabilities of star/not, which are 16%/84%, then what is the probability of next year's beat? It is 16%*75% + 84%*50%. But uses Bayes (because we have observed 3 beats) we updated the probabilities of start/not to 39.13%/(1-39.13%); i.e., after 3 beats in a row, it is more likely this is a star!

Re GARP'2 #4, it's just SHOWING the answer to the question. Although "Neutral" looks like a typo. Should be Pr(Passage | Constant) = 6.0%/27.0% = 22.22%. That should be the to the question, which is a conditional probability. Dividing the blue joint probability (6.0%) by the orange unconditional probability (27.0%) gives us the conditional probability that answer's the question. Thanks,

 

[JGURR5668]

Hi @David Harper CFA FRM Can you give more explanation to the example in the study notes of Probabilities, I have highlighted my doubts in yellow.

I don't understand on 39.13*75(which is conditional beats of first year?)

Thanks
NavjyotView attachment 3194

Hello @David Harper CFA FRM I have a doubt in the "Bayes Sample Example #2". I don´t get why P[3B|S] is equal to P[B|S]^3. Why we can assume that? Could you me explain me a little more why the equivalence?

Regards.

 

[David Harper CFA FRM]

Hi @JGURR5668 I think we had a good discussion about this here at https://forum.bionicturtle.com/threads/bayes-theorem-two-approaches.6784/ and specifically in my last post at https://forum.bionicturtle.com/threads/bayes-theorem-two-approaches.6784/post-82906 where I wrote

Hi @dtammerz There would be no exponent (^3) if If the question were instead of the garden variety sort as follows:

Question: You are an analyst at Astra Fund of Funds. Based on an examination of historical data, you determine that all fund managers fall into one of two groups. Stars are the best managers. The probability that a star will beat the market in any given year is 75%. Other managers are just as likely to beat the market as they are to underperform it [i.e., non-stars have 50/50 odds of beating, P[B|S'] = 50%]. For both types of managers, the probability of beating the market is independent from one year to the next. Stars are rare. Of a given pool of managers, only 16% turn out to be stars. A new manager was added to your portfolio of funds three years ago. Since then, the new manager has beaten the market every year. What was the probability that the manager was a star when the manager was first added to the portfolio? What is the probability that this manager is a star now? A new manager was added to the portfolio last year and, in her first year, she beat the market. Given she beat the market, what is the probability she is a star?

... in this garden variety version, the question is simply asking for P(S|B) and we can apply Bayes such that P(S|B) = P(S∩B)/P(B) = P(B|S)*P(S)/P(B) = (12%/16%)*16% / 54% = 22.22%.

But the actual question (Miller Chapter 6) asks, " ... For both types of managers, the probability of beating the market is independent from one year to the next. Stars are rare. Of a given pool of managers, only 16% turn out to be stars. A new manager was added to your portfolio three years ago. Since then, the new manager has beaten the market every year. What was the probability that the manager was a star when the manager was first added to the portfolio?"

If the event (beating the market) is independent, and the conditional probability that a star manager beats, P(B|S) = 75.0%, then the probability of a start beating three years in a row is P(B|S)^3 = 75.0%^3 = 42.2%. In the same way that, given a six-sided die, the probability of rolling three sixes is (1/6)^3, or for that matter, the probability of coin-flipping "heads" three times in a row is 0.50^3. This variation still "fits" into the the same Bayes framework but instead of two conditional outcomes (i.e., a star beats with 75% probability or does not with 25% probability), I think we could say the simple outcomes are replaced by two events: a start beats three years in a row with 75%^3 probability and does not beat three years in a row with (1 - 75%^3) probability. I hope that's helpful,

P[3B|S] is equal to P[B|S]^3, I view that simply as the combination of a conditional probability, P{B|S], and the probability of three consecutive independent events? Replace P[B|S] with P[C] such that P[B|S] = P[C] = 75%; e.g., the probability of some event is 75%. If events are independent, the probability of three consecutive events equals 75%^3 = P[C]^3. Maybe it's the notations that confusing? Because we're just saying the Prob[3B] events except they are conditional probabilities, P[B|C] ... so P[3B|C] just means "the probability of three consecutive beats given the probability of each beat is 75% conditional on a star manager, which is 12.0%/16.0%." I hope that helps!

 

[JGURR5668]

Hi @JGURR5668 I think we had a good discussion about this here at https://forum.bionicturtle.com/threads/bayes-theorem-two-approaches.6784/ and specifically in my last post at https://forum.bionicturtle.com/threads/bayes-theorem-two-approaches.6784/post-82906 where I wrote


P[3B|S] is equal to P[B|S]^3, I view that simply as the combination of a conditional probability, P{B|S], and the probability of three consecutive independent events? Replace P[B|S] with P[C] such that P[B|S] = P[C] = 75%; e.g., the probability of some event is 75%. If events are independent, the probability of three consecutive events equals 75%^3 = P[C]^3. Maybe it's the notations that confusing? Because we're just saying the Prob[3B] events except they are conditional probabilities, P[B|C] ... so P[3B|C] just means "the probability of three consecutive beats given the probability of each beat is 75% conditional on a star manager, which is 12.0%/16.0%." I hope that helps!

Thanks David, now it is clear for me. You are right, I got confused for the notation, I was looking for a probability property that helped me to understand how you can use exponential with conditional probabilities, but it is easier if you see this only like and event C that has a 75% chance of occurring and it’s independent over the time.

 

[CanvasEcho]

Hi All, @David Harper CFA FRM, could you please explain the difference in how the same variable is calculated in two problems:

• in the Astra Funds question the probability of beating the market P(B) was calculated as simply adding the joint probabilities of both cases where the managers beat the market (0.12 + 0.42 = 0.54) this is then cubed directly to 0.1575 to calculate P(3B)
• In the Sample Problem #5 (many state problems) the probability of beating the market is calculated by squaring the conditional probabilities first, and then multiplying the result with their respective prior (MO/MI/MU) like a "weighted average".

Why are two different methodologies used?

I read this comment here: https://forum.bionicturtle.com/threads/p1-t2-20-2-more-probabilities-and-bayes-rule.23259/post-85442
but it does not explain then why we do not calculate P(2B) in the Astra Funds question as (0.75^2)*0.16 + (0.5^2)*0.84

Thanks

 

[David Harper CFA FRM]

Hi @CanvasEcho

In Astras #1, the unconditional P(B) = 12% + 42% = 54.0%; retrieved easily by summing the two joint probabilities in the top row within the 2*2 probability matrix. If we asked the analogous question in Astra #2, the answer is the unconditional P(B) = 15% 30% + 5% = 50%; ie., same summation of top row. Notice this is identical to P(1B) = (15%/20%)^1*20% + (30%/60%)^1*60% + (5%/20%)^1*20% = 15% + 30% + 5% = 50%. That's because P(MO|B)*P(B) = joint P(MO ∩ B). There's really no difference.

But Astra's #2 needs to retrieve an unconditional P(2B) rather than the easier unconditional P(1B). In this 2-years case, we could imagine expanding the probability matrix to 3*3 where the three rows are: Beats twice, Beats once, and Beats zero. The top row (beats twice) would be (15%/20%)^2*20% = 11.25% etc ... Then the sum of the top row would be 11.25% + 15.00% + 1.25% = 27.50%. So the only difference, really, here is that the second problem is harder because it's a two-year (two-state) problem. Both are summing joint probabilities but the second problem's joint probabilities, rather than read directly from the 2*2 matrix, necessarily get them via conditional * unconditional. I hope that's helpful,