Bayes Theorem with Probability Matrix

[jairamjana]

I came across a example question from Schweser which I solved it below.

Because the question specified that the performance in one year is independent of performance in next year. The probability that an Excellent manager will outperform 3 years in a row is (70%)^3 = 34.3% . This is why I reframed the question in Step 2. Now in Step 3 while preparing the Probability Matrix the summation of joint probability comes to 27.40% instead of the usual 100%. I can understand its because I took the third power and changed the structure of matrix.

My question is .. Is this a valid probability matrix because the answer of 40.7% is correct even though the resultant probability matrix does not conform to how it is usually presented..

 

[David Harper CFA FRM]

Hi @jairamjana That's interesting. Your matrix works because it happens to be, for this answer, that you only need the first row and the first row is correct. The flaw in the matrix is that you don't consider all scenarios. You only have scenarios for three consecutive outperforms and three consecutive underperforms; e.g, your matrix sums to less than 100% because you don't include any scenarios that mix an outerperform year(s) and an underperform year(s). For this reason, to get a complete matrix, I would define the second row as something like "Fails to outperform all three years."

In other words, if the first row is 3B = three consecutive beats (to follow Miller's example starting at 6.9), then the second row is 3B'; i.e., not 3B. Then your second row values in a joint probability matrix are 13.14% = 0.20*(1 - 0.7^3) and 70.0% = 0.80*(1-0.50^3) because, for example, the probability of not outpeforming in all three years is given by (1 - probability of outperforming in all three years). For an excellent manager, then, the probability of not outperforming in all three years is given by 1 - 0.70^3 = 65.7%. Then your matrix values sum to 100%, if we go rows first: {6.86, 10.0, 13.14, 70.0}.

I go back and forth between matrix and regular notation. In this case, I happen to find Miller's approach more intuitive. That is, we are given:

• P[p = 0.50] = 80%
• P[p = 0.70] = 20%
• P[3B | p = 0.50] = 0.50^3 = 12.5%
  
• P[3B | p = 0.70] = 0.70^3 = 34.3%
• P[3B] = 80%*0.50^3 + 20%*0.70^3 = 16.86%; i.e., sum of your first row, the unconditional probability of three consecutive beats
• Per Bayes, P[p = 0.70 | 3B ] = joint P[p = 0.70, 3B]/ unconditional P[3B], and since joint P[p=0.70, 3B] = P[p=0.70]*P[3B | p = 0.70], Bayes gives us P[p = 0.70 | 3B ] = P[p=0.70]*P[ 3B | 0.70] / P[3B] = 20%*0.343/16.86% = 40.69%. I hope that's helpful!

 

[jairamjana]

I literally took the third power for the scenarios blindly thinking those to be the only likely scenarios. Thanks a lot @David Harper

 

[David Harper CFA FRM]

Right, understood. I was thinking when you might need the full matrix. You'd need it to answer a question like "A new manager failed to outperform the market for three consecutive years. What is the (posterior) probability the manager is average?" Answer 70%/83.1% = 84.2%. Thanks,

 

[jairamjana]

@David Harper ... Is it better to answer a Bayes Theorem question with a decision tree or probability matrix or use notations (formulas) P(A l B) = (P(B l A) * P(A))/ P(B). The numerator is a joint probability so I thought matrix form is better.. I prefer the matrix approach.. I just want to if I understood how the matrix works.. Can I solve all Bayes theorem problem.. Whether one state like this or even multiple states(say suddenly there is a category "Below Average Manager") ..

Thank you for your help. I have a good grasp of the concept so far..

 

[David Harper CFA FRM]

Hi @jairamjana Every Bayes problem that I have encountered so far, can be solved either with matrix or probability approach. I cannot say which is better. I found the matrix approach required a little more practice, but seems a bit more amenable to solving the problem from any perspective. That is, to me, the glaring advantage of the matrix is that, once it is well defined, it is very easy and robust (i.e., error free) to retrieve any joint/conditional/unconditional probability. On the other hand, for a problem like the one above, when I am not sure how to proceed, I like the "algebraic" approach simply because I can start with the end (the solution). That is, in the question above, the advantage is that you can literally start with defining the first step as "okay we are looking for P[p = 0.70 | 3B ]" ... and collect the known assumptions. I hope that is a little helpful