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n−1Pn i=1Xi is the Bayesian estimator under the loss functionLwithL(t, θ

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Mathematical Statistics, Winter term 2018/19 Problem sheet 4

10) Let X1, . . . , Xn be independent and identically distributed random variables with Pθ(Xi = 1) =θ= 1−Pθ(Xi = 0), whereθ ∈Θ = (0,1).

Show that t(X1, . . . , Xn) = n−1Pn

i=1Xi is the Bayesian estimator under the loss functionLwithL(t, θ) = (t−θ)2/(θ(1−θ)) and prior distributionπ= Uniform([0,1]).

Hint: Use that R1

0 θk(1−θ)n−k−1dθ/R1

0 θk−1(1−θ)n−k−1dθ = k/n if 0 < k < n (See also the lecture on October 6.) and compute t(0, . . . ,0) and t(1, . . . ,1) separately.

11) Let X1, . . . , Xn be i.i.d. with Xi ∼ N(θ, σ2). For the prior distribution π = N(0, τ2), the posterior distribution is given by Pθ|X=x = N(22 2n,σ22τ22). (See also the lecture on October 7.)

Compute the Bayesian risk of the Bayesian estimator under the squared error loss.

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