Financial Risk Manager Part 1

Explanation

This question requires applying Bayes' theorem to calculate the conditional probability that a manager is a superstar given they have beaten the market for three consecutive years.

Given Information:

• P(S) = 16% = 4/25 (prior probability of being a superstar)
• P(O) = 21/25 (probability of being an ordinary manager)
• P(B|S) = 7/10 (probability of beating market given superstar)
• P(B|O) = 1/2 (probability of beating market given ordinary manager)

Calculation Steps:

1. Calculate P(3B|S) - Probability of 3 consecutive market beats given superstar: P(3B|S) = \left(\frac{7}{10}\right)^3 = \frac{343}{1000} = 34.3\%$`$2`. **Calculate P(3B|O)** - Probability of 3 consecutive market beats given ordinary manager: P(3B|O) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 12.5%$`$3. **Calculate P(3B)** - Unconditional probability of 3 consecutive market beats: $$P(3B) = P(3B|S) \cdot P(S) + P(3B|O) \cdot P(O)$$ $$P(3B) = \left(\frac{343}{1000} \cdot \frac{4}{25}\right) + \left(\frac{1}{8} \cdot \frac{21}{25}\right)$$ $$P(3B) = \frac{1372}{25000} + \frac{21}{200} = 16\%$$4`. Apply Bayes' Theorem to find P(S|3B): $P(S|3B) = P(S) \cdot \frac{P(3B|S)}{P(3B)}$ $P(S|3B) = 16\% \cdot \frac{34.3\%}{16\%} = 34.3\% = 0.343$

Final Answer:

The probability that the manager is a superstar given three consecutive market-beating years is 0.34 (rounded to two decimal places).