Solutions to Selected Exercises from Chapter 10 Bain, Ejercicios de Probabilidad

¡Descarga Solutions to Selected Exercises from Chapter 10 Bain y más Ejercicios en PDF de Probabilidad solo en Docsity! Solutions to Selected Exercises from Chapter 10 Bain & Engelhardt - Second Edition Andreas Alfons and Hanno Reuvers Erasmus School of Economics, Erasmus Universiteit Rotterdam Exercise 1 The joint pdf of X1, . . . , Xn is f(x1, . . . , xn;µ) = ∏n i=1 ( e−µµxi xi! ) = e−nµµ ∑n i=1 xi∏n i=1 xi! . Moreover, since X1, . . . , Xn i.i.d.∼ POI(µ) we have S = ∑n i=1Xi ∼ POI(nµ). The pdf of S is thus f(s;µ) = e−nµnsµs s! . The conditional pdf fX|s =  f(x1,...,xn;µ) f(s;µ) = e−nµµ ∑n i=1 xi∏n i=1 xi! e−nµnsµs s! = s! ns ∏n i=1 xi! if ∑n i=1 xi = s 0 otherwise, does not depend on µ, hence S = ∑n i=1Xi is sufficient for µ. Exercise 6 The joint pdf of X1, . . . , Xn is f(x1, . . . , xn; p) = n∏ i=1 ( mi xi ) pxi(1− p)mi−xi = [ n∏ i=1 ( mi xi )] p ∑n i=1 xi (1− p) ∑n i=1mi (1− p) ∑n i=1 xi = ps (1− p) ∑n i=1mi (1− p)s︸ ︷︷ ︸ =g(s;p) [ n∏ i=1 ( mi xi )] ︸ ︷︷ ︸ =h(x1,...,xn) , where s = ∑n i=1 xi. By the factorization criterion, S = ∑n i=1Xi is sufficient for p. Exercise 11 We will use the factorization criterion to answer both subquestion. Note that the joint pdf of X1, . . . , Xn is equal to f(x1, . . . , xn; θ1, θ2) = n∏ i=1 ( 1 θ2 − θ1 I(θ1,θ2)(xi) ) = 1 (θ2 − θ1)n I(θ1,∞)(x1:n)I(−∞,θ2)(xn:n). (a) If θ2 is known, then θ1 is the only parameter to consider. We write f(x1, . . . , xn; θ1) = 1 (θ2 − θ1)n I(θ1,∞)(s)︸ ︷︷ ︸ =g(s;θ1) I(−∞,θ2)(xn:n)︸ ︷︷ ︸ =h(x1,...,xn) with s = x1:n. By the factorization criterion, S = X1:n is sufficient for θ1. 1 (b) We now treat both θ1 and θ2 as unknown parameters. The required factorization is now f(x1, . . . , xn; θ1, θ2) = 1 (θ2 − θ1)n I(θ1,∞)(s1)I(−∞,θ2)(s2)︸ ︷︷ ︸ =g(s1,s2;θ1,θ2) × 1︸︷︷︸ =h(x1,...,xn) . S1 = X1:n and S2 = Xn:n are jointly sufficient for θ1 and θ2. Exercise 13 The joint pdf of X1, . . . , Xn can be written as f(x1, . . . , xn; θ1, θ2) = n∏ i=1 ( Γ(θ1 + θ2) Γ(θ1)Γ(θ2) xθ1−1 i (1− xi)θ2−1 ) = ( Γ(θ1 + θ2) Γ(θ1)Γ(θ2) )n( n∏ i=1 xi )θ1−1( n∏ i=1 (1− xi) )θ2−1 = ( Γ(θ1 + θ2) Γ(θ1)Γ(θ2) )n sθ1−1 1 sθ2−1 2︸ ︷︷ ︸ =g(s1,s2;θ1,θ2) × 1︸︷︷︸ =h(x1,...,xn) , where we defined s1 = ∏n i=1 xi and s2 = ∏n i=1(1 − xi). According to the factorization crition, S1 = ∏n i=1Xi and S2 = ∏n i=1(1−Xi) are jointly sufficient for θ1 and θ2. Exercise 19 The pdf depends on k = 1 unknown parameter, namely µ. However, if we expand the square in the exponential, that is f(x;µ) = 1√ 2π|µ| e − (x−µ)2 2µ2 = 1√ 2π|µ| e − x 2−2µx+µ2 2µ2 = 1√ 2π|µ| e − ( x2 2µ2 − xµ+ 1 2 ) = e− 1 2 √ 2π|µ| e − x2 2µ2 + x µ , then we see that the exponential contains two summands of the form qj(µ)tj(x). The N(µ, µ2) is thus not a member of the REC. Exercise 20 (a) The pdf can be written as f(x; p) = px(1− p)1−x = (1− p) ( p 1− p )x = (1− p)ex ln( p 1−p ), such that it is a member of the REC with c(p) = 1 − p, h(x) = 1, q1(p) = ln ( p 1−p ) , and t1(x) = x. Hence S = ∑n i=1Xi is a complete sufficient statistic for p. (b) The pdf can be written as f(x;µ) = e−µµx x! = 1 x! e−µex lnµ, such that it is a member of the REC with c(µ) = e−µ, h(x) = 1 x! , q1(µ) = lnµ and t1(x) = x. Hence S = ∑n i=1Xi is a complete sufficient statistic for µ. 2 (b) Having found S n = − 1 n ∑n i=1 ln(Xi) as an UMVUE for 1 θ , we might try n S as an UMVUE for θ. This estimator is still a function of S only but it is no longer clear that it is unbiased. We have to compute E ( n S ) . If X has the pdf f(x; θ), then Y = − ln(X) has the pdf fY (y; θ) = fX(e−y; θ) | − e−y| = θ(e−y)(θ−1)e−y = θe−θy, y > 0. From Table B.2 we can see that Y ∼ GAM ( 1 θ , 1 ) . Using the properties of MGFs we also find S = ∑n i=1 Yi = − ∑n i=1 ln(Xi) ∼ GAM ( 1 θ , n ) . The pdf of S is thus fS(s) = 1( 1 θ )n Γ(n) sn−1e−θs, s > 0, and we can compute E (n S ) = ∫ ∞ 0 n s 1( 1 θ )n Γ(n) sn−1e−θsds = n ∫ ∞ 0 ( Γ(n− 1) Γ(n− 1) ) 1( 1 θ )n Γ(n) s(n−1)−1e−θsds = nθ Γ(n− 1) Γ(n) ∫ ∞ 0 1( 1 θ )n−1 Γ(n− 1) s(n−1)−1e−θs︸ ︷︷ ︸ pdf of GAM( 1 θ ,n−1) ds = nθ Γ(n− 1) Γ(n) = n n− 1 θ, where we have used the fact that the pdf of the GAM( 1 θ , n− 1) integrated over its support should be equal to 1. This calculations suggest that θ̂ = n− 1 n n S = n− 1 S = − n− 1∑n i=1 ln(Xi) , is an estimator for θ which is (1) unbiased, and (2) a function of S only. According to Lehmann-Scheffé, Theorem 10.4.1 on page 346 of B&E, this is also an UMVUE. Note 1 : It also possible to solve the integral directly. Note 2 : One could have seen immediately from Jensen’s inequality that n S will give a biased estimator for θ. It was thus clear from the start that a correction was necessary. Exercise 31 (a) The likelihood and log-likelihood are L(θ) = ∏n i=1 f(xi; θ) = θn ( ∏n i=1(1 + xi)) −(1+θ) and lnL(θ) = n ln(θ)− (1 + θ) ∑n i=1 ln(1 + xi), respectively. The first and second derivative of the log-likelihood are: d dθ lnL(θ) = n θ − n∑ i=1 ln(1 + xi) d2 dθ2 lnL(θ) = − n θ2 < 0, for all θ. Because the second derivative is always negative, we find the ML estimator as follows: n θ̂ − n∑ i=1 ln(1 +Xi) = 0 ⇒ θ̂ = n∑n i=1 ln(1 +Xi) . 5 (b) The pdf can be written as f(x; θ) = θ(1 + x)−(1+θ) = θe−(1+θ) ln(1+x), such that it is a member of the REC with c(θ) = θ, h(x) = 1, q1(θ) = −(1 + θ), and t1(x) = ln(1 + x). Hence S = ∑n i=1 ln(1 +Xi) is a complete sufficient statistic for θ. (c) For the numerator of the CRLB, τ(θ) = 1 θ yields τ ′(θ) = − 1 θ2 . The following results are helpful to find the denominator: ∂2 ∂θ2 ln f(x; θ) = − 1 θ2 , E ( ∂2 ∂θ2 ln f(X; θ) ) = E ( − 1 θ2 ) = − 1 θ2 . Overall, the CRLB is [τ ′(θ)]2 −nE ( ∂2 ∂θ2 , ln f(X; θ) ) = 1 θ4 n θ2 = 1 nθ2 . (d) It seems intuitive to use 1 θ̂ = 1 n ∑n i=1 ln(1 + Xi) to estimate θ. This estimator is already a function of S but we do not know yet whether it is biased or not. If X has the pdf f(x; θ) = θ(1 + x)−(1+θ), then the pdf of random variable Y = ln(1 +X) is given by fY (y; θ) = fX(ey − 1; θ) |ey| = θ(ey)−(1+θ)ey = θe−θy, y > 0. Hence Y ∼ GAM( 1 θ , 1) and in turn S = ∑n i=1 Yi = ∑n i=1 ln(1 + Xi) ∼ GAM( 1 θ , n) (see Example 6.4.6 in the book). Since E(S) = n θ , the estimator T = S n = 1 n ∑n i=1 ln(1 +Xi) is unbiased for τ(θ) = 1 θ such that it is also an UMVUE. (e) The CRLB for θ is 1 −nE ( ∂2 ∂θ2 ln f(X; θ) ) = 1 n θ2 = θ2 n such that the asymptotic distribution of the MLE θ̂n is θ̂n − θ√ CRLB = θ̂n − θ θ/ √ n d−→ Z ∼ N(0, 1). The CRLB for 1/θ was derived in part (c). The asymptotic distribution of the MLE τ(θ̂n) = 1 θ̂n of τ(θ) = 1 θ is thus 1 θ̂n − 1 θ√ CRLB = 1 θ̂n − 1 θ 1√ nθ d−→ Z ∼ N(0, 1). (f) This exercise is similar to Exercise 25(b). Given that T = S n = 1 n ∑n i=1 ln(1 + Xi) was an UMVUE for 1/θ, it seems reasonable to figure out whether n/S can be an UMVUE for θ. It is clearly a function of the complete sufficient statistic so it only remains to check whether it is unbiased. Because the pdf of S is given by fS(s) = 1( 1 θ )n Γ(n) sn−1e − s 1 θ = 1( 1 θ )n Γ(n) sn−1e−θs s > 0, 6 we find E (n S ) = ∫ ∞ 0 n s 1( 1 θ )n Γ(n) sn−1e−θsds = n ∫ ∞ 0 ( Γ(n− 1) Γ(n− 1) ) 1( 1 θ )n Γ(n) s(n−1)−1e−θsds = nθ Γ(n− 1) Γ(n) ∫ ∞ 0 1( 1 θ )n−1 Γ(n− 1) s(n−1)−1e−θs︸ ︷︷ ︸ pdf of GAM( 1 θ ,n−1) ds = nθ Γ(n− 1) Γ(n) = n n− 1 θ. An UMVUE for θ is thus n−1 S = n−1∑n i=1 ln(1+Xi) . 7