SOLUTION FOR HOMEWORK 3, STAT 4352 Welcome to your third ...

Then the posterior pdf is
f Λ|X (λ|x) = fΛ,X (λ, x)
fX (x) = λ(α+x)−1e−λ(1+1/β) Γ(α)βαf X (x)x!
I(λ > 0). (2)
Now I explain you what smart Bayesian statisticians do. They do not calculate f X (x) or
try to simplify (2); instead they look at (1) as a density in λ and try to guess what family it
is from. Here it is plain to realize that the posterior pdf is again Gamma, more exactly it is
Gamma(α + x, β/(1 + β)). Note that the Gamma prior for the Poisson intensity parameter
is the conjugate prior because the posterior is from the same family.
As soon as you realized the posterior distribution, you know what the Bayesian estimator
is: it is the expected value of this Gamma RV, namely
The problem is solved.
ˆΛB = E(Λ|X) = (α + X)[β/(1 + β)] = β(α + X)/(1 + β).
11. Problem 10.94. This is a curious problem on application and analysis of Bayesian
approach. It is given that the observation X is a binomial RV Binom(n = 30, θ) and
someone believes that the probability of success θ is a realization of a Beta random variable
Θ ∼ Beta(α, β). Parameters α and β are not given; instead it is given that EΘ = θ0 = .74
and V ar(Θ) = σ 2 0 = 3 2 = 9. [Do you think that this information is enough to find the
parameters of the underlying beta distribution? If “yes”, then what are they?]
Now we are in a position to answer the questions.
(a). Using only the prior information (that is, no observation is available), the best MSE
estimate is the prior mean
ˆΘprior = EΘ = .74.
(b) Based on the direct information, the MLE and the MME estimators are the same
and they are
ˆΘMLE = ˆ ΘMME = ¯ X = X/n = 18/30.
[Please compare answers in (a) and (b) parts. Are they far enough?]
(c) The Bayesian estimator with Θ ∼ Beta(α, β) is (see p.345)
ˆΘB =
X + α
α + β + n .
Now, we can either find α and β from the mean and variance information, or use results of
our homework problem 10.74 and get
where
w =
ˆΘB = w ¯ X + (1 − w)E(Θ),
n
n + θ0(1−θ0)
σ 2 0
− 1 =
8
30
30 + (.74)(.26)
9
− 1 .