This is the baseball player question. The number of errors is Poisson with mean
theta and theta is Gamma (alpha, beta).
We know alpha/beta = 1/10 and alpha/(beta)^2 = 1/400. This gives alpha=2
and beta=20.
With 1 error in 60 games (observations), I took the posterior distribution
of theta to be Gamma with alpha=3 and beta=80.
The mean of the posterior, and the Bayesian estimate (since we have a
Gamma-Poisson pair) is 3/80=.0375.
With 60 more games, the expected number of errors is 2.25, answer B.
So what's wrong with this approach?
Chris