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MATH70078 Fundamentals of Statistical Inference, 2024-25

Coursework.

Deadline for submission 12 noon (midday) Friday 8 November, via Blackboard.

(1) Let Y1, . . . , Yn be independent, identically distributed, with common density f(y; θ) = θe−θy, y > 0.

Suppose the (non-informative, improper) prior for θ of the form π(θ) ∝ 1/θ, θ > 0.

Find the form of the Bayes estimator of µ = E(Yi) ≡ 1/θ with respect to this prior and the loss function

L(µ, a) = µ2/(µ − a) 2.                         [4]

What is the maximum likelihood estimator µb of µ?              [1]

Calculate the risk of the Bayes estimator and compare it with that of µb. Is µb admissible?                 [5]

[A random variable X with the inverse Gamma distribution has density of the form f(x; α, β) = Γ(α)/βα(x/1) α+1e −β/x, x > 0.]

(2) Let Y1, . . . , Yp (p > 2) be independent random variables such that Yi ∼ N(θi , 1). Write Y = (Y1, . . . , Yp) T and θ = (θ1, . . . , θp) T . Let θ ≡ θ(Y) = (θ1(Y), . . . , θp(Y))T be an estimator of θ, and let g(Y) ≡ (g1(Y), . . . , gp(Y))T = θ − Y .

Let Di(Y) = ∂gi(Y)/∂Yi . Show that

is an unbiased estimator of the risk of θ, under squared error loss  [4]

Suppose the estimator θ is of the form θ = bY , for b ∈ R. Find the value b ∗ of b that minimises the unbiased risk estimator Rb(Y). Compare the estimator b ∗Y with the James-Stein estimator.                [3]

The soft threshold estimator is defined by

where λ > 0 is a constant, to be specified. Show that for this estimator

 [2]

[Here, I(A) = 1 if A holds, = 0 otherwise.]

It has been suggested that a suitable choice of the value of λ is that which minimises R(Y). Explain why determining this value only requires examining the value of R(Y) at a finite number of values of λ.              [3]

Find the form of the estimator θP that minimises the penalized sum of squares

with J(θ) = |{θi : θi ≠ 0}|, the number of non-zero elements of θ.

Calculate the risk of θP for the case p = 1. When is this estimator preferable to the unbiased estimator θ ≡ Y ? Quantify your answer for λ = 1, 2, 5.        [8]

[Total 30]