Solution. (i) Find a minimal sufficient statistic for (θ, β) and give your justification. X i=1. By the factorization theorem, ( n

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1 Solution 1. Let (X 1,..., X n ) be a simple random sample from a distribution with probability density function given by f(x;, β) = 1 ( ) 1 β x β, 0 x, > 0, β < 1. β (i) Find a minimal sufficient statistic for (, β) and give your justification. The joint pdf of X = (X 1,..., X n ) is given by f n (X; β, ) = ( 1 n ) 1 β β X n/β β n i I [0,] (X (n) ). By the factorization theorem, ( n X i, X (n) ) is sufficient. Further, if X = (X 1,..., X n ) and Y = (Y 1,..., Y n ) such that f n (X; β, ) = f n (Y ; β, ) for all β and, we then have ( n ) 1 β ( β n ) 1 β β X i I [0,] (X (n) ) = Y i I [0,] (Y (n) ) for all β and, which implies ( n X i, X (n) ) = ( n Y i, Y (n) ). Hence ( n X i, X (n) ) is minimal sufficient. (ii) Suppose that β is known. Show that X (n) is sufficient and complete for, where X (n) is the largest order statistic, i.e., X (n) = max 1 i n X i. By the factorization theorem, X (n) is sufficient. The pdf of X (n) is given by f X(n) (x) = n ( ) x n/β 1 I[0,](x). β Let h be any Borel function such that Eh(X (n) ) = 0. Then we have h(x)x n/β 1 dx = 0, 0

2 for all. Differentiating both sides w.r.t. yields h() n/β 1 = 0 for all > 0. This imples h 0. Therefore X (n) is complete. (iii) Suppose that is known. Find a sufficient and complete statistic for β and give your justification. The joint pdf of X can be expressed as exp{ 1 β β ln X i ξ(β)}h(x) for some function ξ and h. Hence, it is an exponential family with natural sufficient statistic ln X i. Hence n ln X i is sufficient and complete for η = (1 β)/β. Since η and β are one-to-one, n ln X i is also sufficient and complete for β. (iv) Suppose that is known. Drive the uniformly mimimum variance unbiased estimator (UMVUE) for β. We can compute E ln X i = 1 ln(x)x 1/β 1 dx = ln β. 1/β β 0 Take ˆβ = ln 1 n n ln X i. Since ˆβ is a function of the S&C statistic and is unbiased for β, it is the UMVUE of β. 2. Let (X 1,..., X n ) be a random sample from the exponential distribution E(a, ) with parameter < a <, > 0. The probability density function of the distribution is given by f(x; a, ) = 1 (x a) e I (a, ) (x). Let X = 1 n n X i and X (1) = min 1 i n {X i }.

3 (i) Show that n(x (1) a) 0 in probability. X (1) is distributed as E(a, /n) with mean a + /n and variance (/n) 2. By Chebychev s inequality, P ( n X (1) EX (1) > ɛ) nvar(x (1)) ɛ 2 = 2 nɛ 2 0. Hence n(x (1) EX (1) ) p 0. Note that n(ex (1) ) a) = n 0. Then, by continuous mapping theorem, n(x(1) a) p 0. (ii) Show that X X (1) in probability. By the LLN, X p a +. From (i) X (1) p a. By continuous mapping theorem, X X(1). (iii) Show that n( X a ) Z in distribution, where Z is a standard normal random variable. (iv) Show that It follows directly from CLT since E X = a + and Var( X) = 2 /n. n( X X(1) ) X X (1) Z in distribution as well. From (i), n(x (1) a)/ p 0. By (iii) and Slutsky s theorem n( X X(1) ) n( X a ) n(x(1) a) = d Z. From (ii), X X(1) p. By Slutsky s theorem again, n( X X(1) ) X X (1) Z. 3. Let (X 1,..., X n ) be a random sample from the Gamma distribution Γ(α, γ) where α is known. The probability density function of the distribution is given by f(x; a, ) = 1 Γ(α)γ α xα 1 e x/γ I (0, ) (x).

4 (i) Derive the UMVUE of γ. When α is known, the family is an exponential family and T = n X i is sufficient and complete for γ. Since E(X i ) = αγ, E(T ) = nαγ. Hence ˆγ = 1 T is the nα UMVUE of γ (ii) Let F (t; α, γ) be the cumulative distribution funtion of Γ(α, γ). Derive the UMVUE of F (c; α, γ) where c > 0 is fixed. The UMVUE of F (c; α, γ) is given by P (X 1 c T ) = P ( X 1 T c t T ) = P (X 1 T c t ), since X 1 /T is ancilary and hence independent from T, the sufficient and complete statistic. Note that, by the property of the Gamma distribution, X 1 /T Beta(α, (n 1)α). Hence, the UMVUE is given by ˆF (c; α, γ) = (iii) Derive the UMVUE of at a fixed t > 0. Note that Γ(nα) c/t x α 1 (1 x) (n 1)α 1 dx. Γ(α)Γ((n 1)α) 0 df (t;α,γ), the probability density function, dt d dc P (X 1 c) = d dc EP (X 1 c T ) = E d dc P (X 1 c T ). The UMVUE is given by d dc P (X 1 c T ) = d Γ(nα) dc Γ(α)Γ((n 1)α) = { 1 T c/t 0 x α 1 (1 x) (n 1)α 1 dx Γ(nα) Γ(α)Γ((n 1)α) ( c T )α 1 (1 c T )(n 1)α 1, c < T, 0, c T. 4. Let X = (X 1,..., X n ) be a random sample from a double exponential distribution DE(0, ) with probability density function given by f(x; ) = 1 x e I(, ) (x). 2

5 (i) Derive the variance of the double exponential distribution without using neither integration nor moment generating function. You can use the fact that E X i =. The pdf of the distribution can be expressed in the canonical exponential family form: f(x; η) = e x η +ln(η) 1 2 I (, )(x). where η = 1/. Hence Var( X i ) = Var( X i ) = d2 dη 2 ( ln η) = 1/η 2 = 2. The distribution is symmetric about 0 and hence EX i = 0 and Var(X i ) = EX 2 i = E X i 2 = Var( X i )+ (E X i ) 2 = 2 2. (ii) Find the UMVUE of. Show that the UMVUE of achieves the Cremér-Rao lower bound. Since T = n X i is sufficient and complete for and ET = n, the UMVUE of is given by ˆ = T/n. From the solution of (i), Var(T/n) = 2 /n. On the other hand, the Fisher information number about is I() = Var( d d ln f n(x, ) = Var( T 2 1 ) = n2 4 = n 2. Hence Var(ˆ) = I 1 (). (iii) Let ξ = P (X 1 > t) where t > 0 is fixed. Derive the Fisher information number I(ξ). Find the Cremér-Rao lower bound for unbiased estimators of ξ. Note that ξ = ξ() = 1 2 e t/. The Fisher information I(ξ) about ξ satisfies Hence I() = [ξ ()] 2 t I(ξ) = [ 2 2 e t/ ] 2 I(ξ). t I(ξ) = [ 2 2 e t/ ] 2 n = 4n2 e 2t/ = 2 t 2 n [ξ ln(2ξ)] 2 The C-R lower bound for the unbiased estimators of ξ is give by I 1 (ξ) = [ξ ()] 2 I 1 ().

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