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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 from 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.

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A

What is the probability that this manager is a superstar?

B

What is the probability that this manager will beat the market next year?

C

What is the probability that this manager is an ordinary manager?

D

What is the probability that this manager will underperform the market next year?

E

What is the probability that this manager will beat the market in exactly two of the next three years?

F

What is the probability that this manager will beat the market in at least two of the next three years?

Explanation:

Explanation

This is a Bayesian probability problem involving conditional probabilities and Bayes' theorem.

Given Information:

P(S) = 0.16 (Probability a manager is a superstar)
P(O) = 0.84 (Probability a manager is ordinary)
P(B|S) = 0.70 (Probability of beating market given superstar)
P(B|O) = 0.50 (Probability of beating market given ordinary)
Manager has beaten market for 3 consecutive years

Part A: Probability manager is a superstar given 3 years of beating market

Using Bayes' Theorem: $P(S|B^3) = \frac{P(B^3|S) \cdot P(S)}{P(B^3|S) \cdot P(S) + P(B^3|O) \cdot P(O)}$

Where:

$P(B^3|S) = (0.70)^3 = 0.343$
$P(B^3|O) = (0.50)^3 = 0.125$

$P(S|B^3) = \frac{0.343 \times 0.16}{0.343 \times 0.16 + 0.125 \times 0.84} = \frac{0.05488}{0.05488 + 0.105} = \frac{0.05488}{0.15988} \approx 0.343$

Answer: Approximately 34.3%

Part B: Probability manager beats market next year

This is the weighted average of the conditional probabilities: $P(B_{next}) = P(S|B^3) \cdot P(B|S) + P(O|B^3) \cdot P(B|O)$

Where:

$P(O|B^3) = 1 - P(S|B^3) = 1 - 0.343 = 0.657$

$P(B_{next}) = 0.343 \times 0.70 + 0.657 \times 0.50 = 0.2401 + 0.3285 = 0.5686$

Answer: Approximately 56.9%

Key Insights:

Bayesian updating: The manager's track record increases our belief they might be a superstar from 16% to 34.3%
Future performance: The probability of beating the market next year is higher than the unconditional probability but not as high as a confirmed superstar
Independence assumption: Each year's performance is independent given the manager type

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Explanation

Given Information:

Part A: Probability manager is a superstar given 3 years of beating market

Part B: Probability manager beats market next year

Key Insights:

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