Explanation

Let's define the following events:

• S: the manager is a superstar
• O: the manager is ordinary
• B: the manager beats the market in a given year
• U: the manager underperforms the market in a given year

From the information given:

• P(S) = 0.16 (the prior probability of being a superstar)
• P(O) = 0.84 (the prior probability of being ordinary)
• P(B|S) = 0.7 (probability of beating the market given the manager is a superstar)
• P(B|O) = 0.5 (probability of beating the market given the manager is ordinary)

We want to find P(S|B³), the probability that the manager is a superstar today, given that he has beaten the market for three consecutive years.

Using Bayes' theorem:

$P(S|B^3) = \frac{P(B^3|S) \cdot P(S)}{P(B^3)}$

Calculate each term:

1. P(B³|S) = P(B|S)³ = 0.7³ = 0.343

2. P(B³|O) = P(B|O)³ = 0.5³ = 0.125

3. P(B³) = P(B³|S) × P(S) + P(B³|O) × P(O)
   = (0.343 × 0.16) + (0.125 × 0.84)
   = 0.05488 + 0.105 = 0.15988

4. P(S|B³) = $\frac{0.343 \times 0.16}{0.15988} = \frac{0.05488}{0.15988} = 0.3433$

Therefore, the probability that the manager is a superstar given three consecutive years of beating the market is 0.3433.

This demonstrates how Bayesian updating works - even though the prior probability of being a superstar was only 16%, three consecutive years of strong performance significantly increases this probability to about 34.33%.