Is this trader skilled, or just lucky?

Question

Suppose 5% of applicants are genuinely skilled and beat the market with probability 0.60 in any month; the other 95% are unskilled and beat it with probability 0.50 (a coin flip). A candidate beat the market in 9 of their last 12 months.

What is the probability the candidate is genuinely skilled?

Accepted Answer

Set up Bayes' rule with odds. Prior odds of skilled versus unskilled: \frac{P(	ext{skilled})}{P(	ext{unskilled})} = \frac{0.05}{0.95} \approx 0.0526. Likelihood ratio for 9 of 12. The binomial coefficient \binom{12}{9} appears in both the skilled and unskilled likelihoods, so it cancels in the ratio, leaving 	ext{LR} = \frac{0.60^{9}\,0.40^{3}}{0.50^{12}}. Compute: 0.60^{9} = 0.0100777 and 0.40^{3} = 0.064, so the numerator is 0.0100777 	imes 0.064 = 0.00064497. The denominator is 0.50^{12} = 0.00024414. Thus 	ext{LR} = \frac{0.00064497}{0.00024414} \approx 2.642. Update. Posterior odds …

Is this trader skilled, or just lucky?

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