
Financial Risk Manager Part 1
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Determine the probability that the manager is a superstar today, given that he has beaten the market for three consecutive years.
Explanation:
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:
-
P(B³|S) = P(B|S)³ = 0.7³ = 0.343
-
P(B³|O) = P(B|O)³ = 0.5³ = 0.125
-
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 -
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%.