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David Harper CFA FRM

David Harper CFA FRM
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Here is a recent question I wrote to test Bayes Theorem (which has been reintroduced into the FRM)
Question: 302.1. There is a prior (unconditional) probability of 20.0% that the Fed will initiate Quantitative Easing 4 (QE 4). If the Fed announces QE 4, then Macro Hedge Fund will outperform the market with a 70% probability. If the Fed does not announce QE 4, there is only a 40% probability that Macro will outperform (and a 60% that Acme will under-perform; like the Fed's announcement, there are only two outcomes). If we observe that Macro outperforms the market, which is nearest to the posterior probability that the Fed announced QE 4?
1. Tree Approach
I have typically used a (binomial) tree to visualize this sort of problem. You start the tree with the unconditional (aka, marginal) probabilities which are the probabilities that do not depend on anything else; in this case, the unconditional probability of a QE 4 is 20%, the conditional probability of fund outperformance, Prob[P|Q] = 70%, and their joint Prob[QP] = 20%*70%=14%:



The Bayes (posterior but still conditional) probability question asked is, what is the probability of QE 4 (Q) given that we observe Outperformance (P), Prob(Q|P). It's just the "reverse" of the conditional probability that happens to be supplied by the question: we are given that Prob(P|Q) = 70%, but we want Prob(Q|P).

The tree let's us visualize how Bayes works: Prob(Q|P) is the probability of the joint outcome, Prob(QP), in the numerator, divided by all of the events that include outperformance, Prob(P), which here only includes two: Prob(QP) + Prob(Q'P). Visually, the two yellow nodes constitute all possible nodes that include outperformance (P), so that is our denominator. Now, if we observed (Q) has happened, then the conditional probability that (P) also happens is the same as the joint Prob(QP), so that is the numerator.

2. Probability Matrix
Miller (new to the FRM) employs these handy probability matrixes which, it seems to me, give us another way to grasp Bayes (it's not mechanically any different, I'm sure it's been done somewhere, i just haven't seen it ....).

In the probability matrix approach, here are the question's same setup assumption rendered into cells:



All we need to do is produce the "probability matrix" which in Miller simply means a matrix of joint probabilities. In a joint probability matrix, all of the body cells sum to 1.0. For example, Joint Prob[Q(1),P]=20%*70%=14%; Joint Prob[Q(1),P']=20%*30%=6%; Joint Prob[Q(2),P]=80%*40%=32%. Here is the implied joint probability matrix:



Once the problem's assumptions are "translated " into a joint probability matrix, I find the Bayes Theorem question relatively easier to grasp: the Prob[Q | P] is simply the conditional probability retrieved from the matrix.

Our information is that (P) occurred: this implies our outcome must be located in column (P) which has an overall probability of 46.0% (Our prior information is that P happened and P' did not) . As our joint outcome is the upper cell Prob[Q(1),P] = 14%, our desired Bayes' conditional probability, Prob [Q(1)|P], is simply divides the joint probability cell by the column probability, Prob [Q(1)|P] = 14%/46% = 30.4%.
 

ShaktiRathore

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#3
Hi,
I would like to give a little twist to how one can approach the Bayes theorem Questions although a little bit lengthy but nevertheless helpful,

Consider two sets of Q and Q' in universal set such that only these two events exist.
area Q=20% area Q'=80% so that total area =100%
In case of event Q happening: Now Q has two subsets(events possible) which are P and P' such that P occupies 70% of area of Q and P' occupies 30% of total area of Q in context of Q.
In case of event Q' happening: Now Q' also has two subsets(events possible) which are P and P' such that P occupies 40% of area of Q' and P' occupies 60% of total area of Q' in context of Q'.
Now looking in context of the universal set a a whole,
we see P has a total area of 70% of area of Q+40% of area of Q'=70%*20%+40%*80%=14%+32%=46%
also P' has a total area of 30% of area of Q+60% of area of Q'=30%*20%+60%*80%=6%+48%=54%
Now looking from universal set as a whole any of event Q and Q' can occur and any of P and P' under them,
P(Q|P)=probability of P falling in Q given event P has occurred
P(Q|P)=area of P in Q/ total area of P in both Q and Q'=70% of area of Q/70% of area of Q+40% of area of Q'=14%/46%=30.434%

or one can directly apply the formula as
P(Q|P)=P(P&Q)/P(P)=event P and Q happening simultaneously/total probability of P in universal set context
total probability of P in universal set context= P(P/Q)*P(Q)+P(P/Q')*P(Q')=70% of area of Q+40% of area of Q'=70%*20%+40%*80%=14%+32%=46%(remember total probability rule)
event P and Q happening simultaneously=P(P/Q)*P(Q)=70% of area of Q=70%*20%=14%
P(Q|P)=P(P&Q)/P(P)=P(P/Q)*P(Q)/P(P)=14%/46%=30.434% follow for others in a similar way...

thanks
 

David Harper CFA FRM

David Harper CFA FRM
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Hi Shakti, this is very interesting, have you seen it rendered in Venn diagram? When I get a chance, I may try to render your logic in Venn because it seems like the visual manifestation might be very intuitive, thanks for sharing!
 
#5
Can u elaborate more on sample problem of Miller chapter 1 on Bayes theoram? This problem is based on Astra fund of Funds. I am unable to post full question here... But i didnt get the logic how he gets
P[B / S bar]= 0.5. How we can infer this statement form given problem. Is this equation comes from "Other managers are just as likely to beat the market as they are to underperform it".

thanks in advance david
 

David Harper CFA FRM

David Harper CFA FRM
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Hi Abhishek, Here is Miller's question:

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?
I put Miller's example into the tree, he gives three assumptions:
  • unconditional P[Star] = 16%, so P[S']=84%
  • conditional P[Beat | Star] = 75%, so that one-year prob just gets converted into three years, either star manager beats all three years with prob = 75%^3 or not with prob = 1 - 75%^3
  • conditional P[Beat | Not Star; i.e., S'] = 50% in one year, so P[3B|S']=50%^3.
Then it is the same: P[S|3B] = joint Prob[S,3B]/Prob[3B]:
 
#7
Thanks David for prompt reply....
My question was how did he infer conditional P[Beat | Not Star; i.e., S'] = 50% ? As per my understanding of the question there is no sentence which tells us that....
 

ShaktiRathore

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#8
Hi,
from Bayes theorem,
P[Beat | Not Star; i.e., S']*P(S')=joint P(Beat & S')
=> P[Beat | Not Star; i.e., S']=joint P(Beat & S' i.e. Beat & S' happening simultaneously)/P(S')= (.84*.50)/.84=.50
P[Beat | Not Star; i.e., S'] It also infers that what is the probability that S' has occurred. Now we know that S' has occurred conditional on this what is the probability that Beat will occur. Now once S' has occurred there is 50% chance that beat will occur which is our required conditional probability P[Beat | Not Star; i.e., S'].
thanks
 
#9
hi Abhishek,
In the problem it is mentioned that "Other managers(non stars) are just as likely to beat the market as they are to underperform it
 

AnZu

New Member
#12
I am not sure where to post this question, but I am struggling with converting the wording into a probability sometimes--for example, I read something and don't know whether something is a conditional or joint probability. If it says A given B, then I know it is a conditional probability, but for example, in in the 2010 Exam (the BT practice one), Q35.3, I started to draw the tree with P(D) and P(not D). Then I took this sentence, "if a patient has the disease, the test turns a positive result 80% of the time", and drew a branch from the P(D) node with the + node 80%, the - node 20%. I took the sentence to mean "the joint probability of having the disease and testing positive is (.01)*(.8)=.008", when I should have interpreted this sentence as "P(+|Disease)=.8" (per solution). I did something similar with 35.1 as well. Is there a trick to mapping the wording in to the proper probability expression?

Second, if I have more questions from the same exam, what is the protocol for posting questions? Your practice questions have links where we can post relevant questions, but for the practice exams, is it better to post all the questions under e.g. "2010 practice exam questions", or to separate it by topic? (Another question I had, for example was whether we are still expected to know how to do the VaR of a lognormal--this isn't in the GARP books. (I realize this is a few years old exam and the aims might have changed slightly. It's a very short question, so not sure if it warrants creating a new subject thread just to ask this.)

Thanks.
 

dtammerz

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#13
Hi Abhishek, Here is Miller's question:



I put Miller's example into the tree, he gives three assumptions:
  • unconditional P[Star] = 16%, so P[S']=84%
  • conditional P[Beat | Star] = 75%, so that one-year prob just gets converted into three years, either star manager beats all three years with prob = 75%^3 or not with prob = 1 - 75%^3
  • conditional P[Beat | Not Star; i.e., S'] = 50% in one year, so P[3B|S']=50%^3.
Then it is the same: P[S|3B] = joint Prob[S,3B]/Prob[3B]:
How do you know that you have to take the exponent of P[Beat|Star]? Is this a known probability/mathematical function related to time, where the exponent = year?
 

David Harper CFA FRM

David Harper CFA FRM
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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,
 
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