Financial Risk Manager Part 1
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You are an analyst at a large mutual fund. After examining historical data, you establish that all fund managers fall into two categories: superstars (S) and ordinaries (O). Superstars are by far the best managers. The probability that a superstar will beat the market in any given year stands at 70%. Ordinaries, on the other hand, are just as likely to beat the market as they are to underperform it. Regardless of the category in which a manager falls, the probability of beating the market is independent of year to year. Superstars are rare diamonds because only a meager 16% of all recruits turn out to be superstars.
During the analysis, you stumble upon the profile of a manager recruited three years ago, who has since gone on to beat the market every year.
What is the conditional probability for a non-superstar manager given that he/she has beaten the market for 3 years?
Explanation:
Explanation
We need to find P(O | 3B) - the probability that the manager is ordinary given they beat the market for 3 consecutive years.
Given Information:
- P(S) = 0.16 (probability of being a superstar)
- P(O) = 1 - P(S) = 0.84 (probability of being ordinary)
- P(B|S) = 0.70 (probability of beating market given superstar)
- P(B|O) = 0.50 (probability of beating market given ordinary)
Using Bayes' Theorem:
We want P(O | 3B) = P(O) × P(3B|O) / P(3B)
First, calculate P(3B|S) and P(3B|O):
- P(3B|S) = (0.70)^3 = 0.343
- P(3B|O) = (0.50)^3 = 0.125
Now calculate P(3B) using law of total probability: P(3B) = P(S) × P(3B|S) + P(O) × P(3B|O) P(3B) = (0.16 × 0.343) + (0.84 × 0.125) P(3B) = 0.05488 + 0.105 = 0.15988
Now apply Bayes' theorem: P(O | 3B) = P(O) × P(3B|O) / P(3B) P(O | 3B) = (0.84 × 0.125) / 0.15988 P(O | 3B) = 0.105 / 0.15988 ≈ 0.6566 ≈ 0.66
Therefore, the conditional probability that a manager is ordinary given they beat the market for 3 years is approximately 0.66.