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27 Feb 2018: 71). Find an element of OK with norm 2a 3b 5 for some. ... a,b 0. Hence prove that the class group of K is cyclic and find its order.

12 Nov 2018: Assume that thereexists some decision rule δ0 such that R(δ0,θ) < for all θ Θ. ... Show thatthe sample mean X̄n is inadmissible for quadratic risk, but that it is still minimax.What happens if Θ = [a,b] for some 0 < a < b <? 9. Let X be

13 Nov 2018: Eo = {x X : B(x) E for some > 0}. ... 8. Use the Contraction Mapping Theorem to show that the equation cos x = x has a unique real solution.Find this solution to some reasonable accuracy using a calculator (remember to

23 Nov 2018: Suppose thatsup. [a,b]BR(x0)‖F‖ R(b a)1 and that ‖F(t,x) F(t,y)‖ K‖x y‖ for some. ... 9. (Some useful properties of the operator norm). Recall that the operator norm on the.

8 Oct 2018: 2. Show that the following sets of subsets of R all generate the same σ-algebra:(a) {(a,b) : a < b}, (b) {(a,b] : a < b}, (c) {(,b] : b R}. ... 10. Let (E,E,µ) be a measure space. Call a subset N E null if N B for some B E withµ(B) = 0.

26 Oct 2018: Suppose for some θ0 in the interior of Θ and every ε > 0small enough we have S(θ0 ε) < 0 < S(θ0 ε), and that Sn has exactly one zero θ̂nfor ... 1. 2. a) Assuming that Φ is one to one, show that the maximum likelihood estimator of φin the model

25 Sep 2018: 2. Consider the three equations in six unknowns given by Ax = b, where. ... that x Rn is an extreme point of the set{x Rn : x 0,Ax b} if and only if for some z Rm,.

6 Nov 2018: a) All words w {0, 1} consisting of either the string 01 repeated some number oftimes (possibly none), or the string 010 repeated some number of times (possiblynone). ... b) All words w {a,b,c} consisting of some number of a’s (possibly none), followed

19 Jan 2018: setU, and µ(U) < for some non-empty open set U. ... 58. Let X,Y be Banach spaces and S,T B(X,Y ). Assume that T is invertible andfor some δ < 1 we have ‖SxTx‖ 6 δ‖Tx‖ for all x

16 Nov 2018: r > 019 or of the form {z C | aRe(z) bIm(z) = c}{}for some a,b,c R with (a,b) 6= (0, 0)20. ... Thus by the discussionabove gfg1 extends z 7 azb0za = z b/a for some a,b C{0}.