Financial Risk Manager Part 1

Financial Risk Manager Part 1

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What is the probability that the manager is a superstar as at present?

TTanishq



Explanation:

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)=(710)3=3431000=34.3%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)=(12)3=18=12.5%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)β‹…P(S)+P(3B∣O)β‹…P(O)P(3B) = P(3B|S) \cdot P(S) + P(3B|O) \cdot P(O) P(3B)=(3431000β‹…425)+(18β‹…2125)P(3B) = \left(\frac{343}{1000} \cdot \frac{4}{25}\right) + \left(\frac{1}{8} \cdot \frac{21}{25}\right) P(3B)=137225000+21200=16%P(3B) = \frac{1372}{25000} + \frac{21}{200} = 16\%

  4. Apply Bayes' Theorem to find P(S|3B): P(S∣3B)=P(S)β‹…P(3B∣S)P(3B)P(S|3B) = P(S) \cdot \frac{P(3B|S)}{P(3B)} P(S∣3B)=16%β‹…34.3%16%=34.3%=0.343P(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).

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