Suppose X = (X1, . . . , Xn) is a random sample from the Gamma distribution withshape α and rate θ, i.e.,fX (x) = θαΓ(α)xα−1e−xθ, x > 0, α > 0, θ > 0.Each Xi has expectation E(X) = α/θ, variance Var(X) = α/θ2, and momentgenerating function MX (t) = (1 − t/θ)−α.(a) Assuming both α and θ to be unknown, write down the log likelihood functionℓ(θ, α; X) and the corresponding score functions ∂ℓ∂θ and ∂ℓ∂α .(b) Verify that each score function has zero expectation.(c) Assuming α to be known:(i) Calculate the Cramer Rao Lower Bound (CRLB) for the variance of anunbiased estimator of θ.(ii) Is there any unbiased estimator of θ whose variance attain the CRLB?(iii) Show thatS = α − 1nnXi=1 1Xiis an unbiased estimator for θ. What is the MVU estimator for θ?(iv) Identify a change of parameter η = η(θ) for which there exists an unbiasedestimator with variance attained the CRLB.(v) For the parameter in the previous part, identify the MVU estimator whosevariance attains the CRLB. Compute this variance

This is a complex statistical problem that requires a step-by-step solution. Here it is:

(a) The log-likelihood function is given by:

ℓ(θ, α; X) = nαlog(θ) - nlog(Γ(α)) + (α-1)∑log(xi) - θ∑xi

The score functions are the derivatives of the log-likelihood function with respect to the parameters. They are given by:

∂ℓ/∂θ = n/θ - ∑xi

∂ℓ/∂α = nlog(θ) - nψ(α) + ∑log(xi)

where ψ(α) is the digamma function, the derivative of the log gamma function.

(b) The expectation of the score function is zero because the score function is the derivative of the log-likelihood function, and the expectation of the derivative of any function is zero.

(c) (i) The Cramer Rao Lower Bound (CRLB) for the variance of an unbiased estimator of θ is given by:

CRLB = 1/I(θ)

where I(θ) is the Fisher information for θ. For the gamma distribution, I(θ) = n/θ^2, so the CRLB = θ^2/n.

(ii) There is no unbiased estimator of θ whose variance attains the CRLB because the gamma distribution is not an exponential family distribution, and the CRLB can only be attained for exponential family distributions.

(iii) The estimator S = α - 1/n∑(1/xi) is an unbiased estimator for θ because its expectation is θ. The MVU estimator for θ is the same as the unbiased estimator because the gamma distribution is a one-parameter exponential family, and the MVU estimator is the same as the unbiased estimator for such distributions.

(iv) A change of parameter η = 1/θ results in an unbiased estimator with variance that attains the CRLB. This is because the variance of the estimator for η is the reciprocal of the variance of the estimator for θ, and the CRLB for η is the reciprocal of the CRLB for θ.

(v) The MVU estimator for η is the reciprocal of the MVU estimator for θ, i.e., 1/S. The variance of this estimator is 1/(nS^2), which attains the CRLB for η.