Many thanks Sir ! I was just taken aback by the sudden calculation of variance and the grid confused me for a while about mean calculation. I was able to work it as per your example cited above ! Many thanks for your help !Hi @vaisman
- The expected values (aka, means) are each/all the sum of [joint probability * value]. For example, in the case of variable (G), its mean = (20% * 3.0) + (60% * 9.0) + (20% * 12.0) = 8.4%. In case it helps, below I represent the assumption situation in a probability matrix; this assumption set is "simpler" in the sense that non-diagonals in the matrix are zeros.
- Variance applies, σ^2(X) = E(X^2) - [E(X)]^2. For example σ^2(B) = E(B^2) - [E(B)]^2 = 50.60 - 7.0^2 = 1.60. I hope that clarifies!
Thank God I found this thread! Was also struggling with the same problem and now it's clear. I think it would be helpful to include the computation for the binomial coefficient in the material itself since this is in Chapter 1 and the binomial distribution is not discussed in detail until Chapter 2.Hi @vaisman
Sure, frankly I try to respond quickly especially if I think we have an error (as above). Plus, we consider it a contribution (to help us fix mistakes or otherwise improve). In regard to the follow-up: We are given the assumption that "outperformers beat the market 75% of the time" which statistically can be written as a conditional probability, Pr[ Beat | Out-performer]. The above is retrieving the probability that the outperforming manager beats the market in exactly 6 out of 10 years. If the manager has a 50% probability of beating each year, you could think of this as the probability of flipping a coin and getting exactly 6 heads out of 10 flips (except this manager has a 75% probability). Because each year's outcome is a Bernoulli (i.e., either beat or miss, only two outcomes) and this is a series of 10 Bernouillis, this a binomial distribution. So, the formula is applying the binomial distribution (https://en.wikipedia.org/wiki/Binomial_distribution) to find the probability of exactly 6 successes out of 10 trials given p = 0.75. The (10 6) is the binomial coefficient (https://en.wikipedia.org/wiki/Binomial_coefficient) that is "inside" the binomial. Here it equals 10!/(6!*4!) = (10*9*8*7)/(4*3*2) = 210; it means there are 210 combinations of six items that can be drawn from a set of 10 items. The 0.75^6*0.25^4 = 0.000695229 is the probability of an exactly (single) sequence of 6 beats + 4 misses; e.g., "BMBMBMBMBB". But we want to count any sequence with six beats, and there are 210 different sequences to the 210* 0.000695229 = 14.60% is the probability of a combination (as opposed to a permutation) of 6 beats out of 10 years. I hope that's helpful!
Hi @David Harper CFA FRM , understood and thanks for the reply.Hi @iamannchi in the ideal, I am sure you are right, but realistically it's hard to manage all such antecedents. I write a lot of questions that, in an attempt to give cross-topic exposure just don't prequalify some concepts/formulas. It's a trade off compounded by a "dynamic" syllabus (ie, that has changed too rapidly). Thanks,