Answer:
(A)
f(μ|x) = (μ^(x1+x2+x3+x4+x5+2) * e^(-26μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))) / ∫_0^∞ μ^(x1+x2+x3+x4+x5+2) * e^(-26μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3)) dμ
(B)
3/1 = 3
Explanation:
a) The likelihood function of μ is the probability of observing the given data, given a particular value of μ. Since the number of defects in a 400-meter roll of magnetic recording tape has a Poisson distribution with parameter μ, the likelihood function can be expressed as follows:
L(μ|x) = P(X1 = x1, X2 = x2, X3 = x3, X4 = x4, X5 = x5 | μ)
= P(X1 = x1 | μ) * P(X2 = x2 | μ) * P(X3 = x3 | μ) * P(X4 = x4 | μ) * P(X5 = x5 | μ)
= e^(-5μ) * (μ^x1 / x1!) * e^(-5μ) * (μ^x2 / x2!) * e^(-5μ) * (μ^x3 / x3!) * e^(-5μ) * (μ^x4 / x4!) * e^(-5μ) * (μ^x5 / x5!)
= e^(-25μ) * (μ^(x1+x2+x3+x4+x5) / (x1! * x2! * x3! * x4! * x5!))
where x1 = 2, x2 = 2, x3 = 6, x4 = 0, and x5 = 3.
The posterior probability density function of μ can be obtained using Bayes' theorem. According to Bayes' theorem, the posterior probability density function of μ given the observed data x is proportional to the product of the likelihood function and the prior probability density function of μ:
f(μ|x) ∝ L(μ|x) * f(μ)
where f(μ) is the prior probability density function of μ, which is given as μ ~ Ga(3,1). Therefore,
f(μ) = μ^(3-1) * e^(-μ/1) / Γ(3) = μ^2 * e^(-μ)
Substituting the values of L(μ|x) and f(μ), we get
f(μ|x) ∝ e^(-25μ) * (μ^(x1+x2+x3+x4+x5) / (x1! * x2! * x3! * x4! * x5!)) * μ^2 * e^(-μ)
= μ^(x1+x2+x3+x4+x5+2) * e^(-26μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))
Thus, the posterior probability density function of μ given the observed data x is:
f(μ|x) = (μ^(x1+x2+x3+x4+x5+2) * e^(-26μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))) / ∫_0^∞ μ^(x1+x2+x3+x4+x5+2) * e^(-26μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3)) dμ
b) To find the posterior probability mass function of X given the data above, we can use the formula:
P(μ|X) = f(X|μ) * f(μ) / f(X)
where f(X|μ) is the Poisson probability mass function with parameter μ, f(μ) is the Gamma probability density function with parameters α = 3 and β = 1, and f(X) is the marginal probability mass function of X, which can be obtained by integrating the joint density function of X and μ over μ:
f(X) = ∫_0^∞ f(X|μ) * f(μ) dμ = ∫_0^∞ e^(-μ) * μ^(X+2) / (X! * Γ(3)) * μ^2 * e^(-μ) dμ
= Γ(X+3) / (X! * Γ(3))
where X = x1 + x2 + x3 + x4 + x5.
Therefore, we have:
P(μ|X) = f(X|μ) * f(μ) / f(X)
= e^(-5μ) * μ^x1 / x1! * e^(-5μ) * μ^x2 / x2! * e^(-5μ) * μ^x3 / x3! * e^(-5μ) * μ^x4 / x4! * e^(-5μ) * μ^x5 / x5! * μ^2 * e^(-μ) / ∫_0^∞ e^(-5μ) * μ^x1 / x1! * e^(-5μ) * μ^x2 / x2! * e^(-5μ) * μ^x3 / x3! * e^(-5μ) * μ^x4 / x4! * e^(-5μ) * μ^x5 / x5! * μ^2 * e^(-μ) dμ
= (μ^(x1+x2+x3+x4+x5+2) * e^(-55 - μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))) / ∫_0^∞ μ^(x1+x2+x3+x4+x5+2) * e^(-55 - μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3)) dμ
Simplifying the expression, we get:
P(μ|X) = (μ^(x1+x2+x3+x4+x5+2) * e^(-30 - μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))) / 616 * (μ^(x1+x2+x3+x4+x5+2) * e^(-30 - μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3))) dx
Therefore, the posterior probability mass function of X given the observed data is:
P(μ\X) = 616r(x + 16) * (μ^(x1+x2+x3+x4+x5+2) * e^(-30 - μ) / (x1! * x2! * x3! * x4! * x5! * Γ(3)))
c) The Bayesian estimate of μ under the absolute error loss function is given by:
μ_B = E[μ|X] = ∫_0^∞ μ * f(μ|X) dμ
To find the value of μ_B, we can use the fact that the Gamma distribution with parameters α and β has a median of m(α, β) = α/β. Therefore, we can choose the value of μ_B that minimizes the absolute difference between the median of the posterior distribution and the observed data:
|α/β - (x1+x2+x3+x4+x5+3)/31| = |3/1 - (2+2+6+0+3+3)/31| = 0.0645
Hence, the Bayesian estimate of μ under the absolute error loss function is 3/1 = 3.