SOLUTION FOR HOMEWORK 4, STAT 6331 Welcome to your fourth homework. Reminder: if you find a mistake/misprint, do not e-mail or call me. Write it down on the first page of your solutions and you may give yourself a partial credit — but keep in mind that the total for your homeworks cannot exceed 20 points. Now let us look at your problems. 1. a) Here f(x|θ)=(2π)1/2e−(x−θ)2/2I(−∞ <x< ∞) and −∞ <θ< ∞. Write n f(x|θ) −(1/2) (x2−y2)+θn(¯x−y¯) L(x, y|θ) := = e l=1 l l . f(y|θ) P We see that L(x, y|θ) ≡ K(x, y) for all θ and all pairs (x, y) iffx ¯ =y ¯. Thus T = X¯ is Minimal SS (MSS). b) Here f(x|θ)= e−(x−θ)I(x > θ) and θ ∈ (−∞, ∞). Write, n − xl f(x|θ) e l=1 I(x(1) > θ) L(x, y|θ) := = n . (1) −P yl f(y|θ) e l=1 I(y(1) > θ) P (i) Suppose that L(x, y|θ ≡ K(x, y) for all θ ∈ (−∞, ∞). This yields that x(1) = y(1). (ii) Suppose that x(1) = y(1). Then (1) implies that L(x, y|θ) does not depend on θ. The properties (i) and (ii) establish, according to Theorem 6.2.13, that T = X(1) is the minimal sufficient statistic. e) Here f(x|θ) = (1/2)e−|x−θ|I(x ∈ (−∞, ∞)) for θ ∈ (−∞, ∞). Write for ordered observations: f(x|θ) L(x, y|θ) := f(y|θ) (θ−x(l))− (x(l)−θ) e l: x(l)≤θ l: x(l)>θ = . P (θ−y(l))−P (y(l)−θ) e l: y(l)≤θ l: y(l)>θ P P Now note that if at least one pair of ordered observations is such that x(l) = y(l) then L(x, y|θ depends on θ. Observing this, Theorem 6.2.13 and some straightfirward steps (similar to the previous case), show that T =(X(1),...,X(n)) is MSS. 2. Exerc. 6.11. (a) All the considered distributions (random variables) are from a so-called location family where D X = Z + θ, Z ∼ fX (x|θ = 0). (2) You can aslo think about (2) in the following way: there is a random variable Z with distribution corresponding to θ = 0, and then any other RV from the same distribution with parameter θ is generated as shown in (2). 1 Let us prove (2) using moment generating function approach (another approach is to tX calculate fX (x|θ) using our techniques in Chapter 2). Recall that MX (t) = E(e ) and X has the same distribution as Y iff MX (t) ≡ MY (t) for all t in some vicinity of t = 0, say t ∈ (−δ, δ), δ > 0. Thus, to check (2), we need to show that E(etX )= E(et(Z+θ)), t ∈ (−δ, δ) (3) for the case of a location family f(x|θ) =: ψ(x − θ). Here fZ (z|θ)= ψ(z) is the pdf of Z. To prove (3) we write ∞ tX tx MX (t)= E(e )= ψ(x − θ)e dx Z−∞ [make change of variable z = x − θ and continue] ∞ = ψ(z)et(z+θ)dz Z−∞ [now we continue for all t that the integral exists] t(Z+θ) = E{e } = MZ+θ(t). What was wished to show. Now, if Yi = X(n) −X(i) then according to (2) we have Yi = Z(n) −Z(i), and this shows that the distribution of Yi does not depend on θ because the distribition of Z does not depend on θ. We conclude that in all examples of a location family of distributions, statistics Yi are ancillary for the location parameter θ. By the way, can you propose several other ancillary statistics? (b) (i) For the case (a) - normal distribution - the MSS is X¯, and it is complete because normal distribution belongs to an exponential family. Indeed, we can write f(x|θ)= C(θ)h(x)eθnx¯ and then use Theorem 6.2.25. Then Basu’s Theorem yields that X¯ and Yi are independent. (ii) For the case (b) - location exponential family - the MSS is X(1). Is it complete? Let us check. Suppose that for a function g Eθ(g(X(1)) ≡ 0, θ ∈ (−∞, ∞). (4) This is equivalent to ∞ fX(1) (t|θ)g(t)dt ≡ 0, θ ∈ (−∞, ∞). (5) Zθ 2 Well, now is the time to remember the density for X(1). While I know that you remember the formula, let us one more time deduce it. We begin with the corresponding cdf (cumulative distribution function) FX(1) (t|θ)=P(X(1) ≤ t)=1 − P(X(1) > t) [remember that the last probability is called the survivor function] n =1 − P(X1 > t,X2 >t,...,Xn > t)=1 − [P(X1 > t)] ∞ = {1 − [ e−(x−θ)dx]n}I(t > θ)=[1 − e−n(t−θ)]I(t > θ). Zt Taking the derivative of the cdf we get the pdf −n(t−θ) fX(1) (t|θ)= ne I(t > θ). Now we plug-in this density in (5), ∞ n e−n(t−θ)g(t)dt ≡ 0, θ ∈ (−infty, ∞). Zθ The last relation is equivalent to ∞ q(θ) := e−ntg(t)dty ≡ 0, θ ∈ (−∞, ∞). Zθ From here we get dq(θ)/dθ ≡ 0 as well. In its turn, this yields (take the derivative) e−nθg(θ) ≡ 0, θ ∈ (−∞, ∞), which finally yields g(θ) ≡ 0 for all θ ∈ (−∞, ∞). We established that X(1) is complete! Conclusion: By Basu’s Theorem the CMSS X(1) is independent of the ancillary statistic Y := {X(n) − X(i), i =1, 2,...,n − 1}. REMARK: Do you see that this exponential distribution is not from the exponential family? This is the reason for the direct proof! (iii) For the double-exponential distribution the statistic T = (X(1),X(2),...,X(n)) is MSS. Clearly it is not independent of Y := {X(n) − X(i), i =1, 2,...,n − 1} because if you know T then you know Y . 3. Exers. 6.19. Let us begin with distribution 1. Let ψ(X) be such that Ep(ψ(X)) ≡ 0 for all p ∈ (0, 1/4). This means that pψ(0)+3pψ(1) + (1 − 4p)ψ(2) ≡ 0, or equivalently p[ψ(0)+3ψ(1) − 4ψ(2)] ≡ −ψ(2), p ∈ (0, 1/4). This is possible for ψ(2) = 0 and ψ(0) = −3ψ(1) = 0. We conclude that ψ(k) is not necessarily equal to 0, so the family of distributions of X is not complete. 3 For the second distribution, similarly assume that Ep{ψ(X)} ≡ 0 for p ∈ (0, 1/2). This yields pψ(0) + p2ψ(1) + (1 − p − p2)ψ(2) ≡ 0, p ∈ (0, 1/2). The last relations yields p2[ψ(1) − ψ(2)] + p[ψ(0) − ψ(2)] + ψ(2) ≡ 0, p ∈ (0, 1/2). A polynom is identically equal to zero on an interval iff all its coefficients are zero. This yields that ψ(k) ≡ 0. Conclusion: The distribution is complete. 4. Exerc. 6.20. (b) We have θ f(x|θ)= , x ∈ (0, ∞), θ> 0. (1 + x)1+θ Note that here Ω = (0, ∞). Write, n n n 1+θ n (1+θ) ln(1+xl) f(x|θ)= θ (1 + xl) = θ e l=1 . lY=1 P We see that the distribution is from an exponential family andΩ:=(0, ∞) contains an open n set, say (1, 2). Thus, by Theorem 6.2.25 the statistic T := l=1 ln(1 + Xl) is CSS. (c) Here P f(x|θ) = log(θ)(θ − 1)θx, x ∈ Ω := (1, ∞). Write, n n n x n θ x f(x|θ) = [log(θ)/(θ − 1)] θ l = [log(θ)/(θ − 1)] e l=1 l . lY=1 P This is the distribution from an exponential family and Ω contains an open set, say (2, 5). n ¯ Thus, by Theorem 6.2.25 the statistic T = l=1 Xl (or X) is CSS. (e). Here we have P 2 f(x|θ)= θx(1 − θ)2−x, x =0, 1,..., θ ∈ [0, 1]. x ! Write, n 2 f(x|θ)= θxl (1 − θ)2−xl xl ! lY=1 n n 2 ln(θ/(1−θ)) x 2n = e l=1 l (1 − θ) . xl ! lY=1 P This is again the distribution from an exponential class with Ω containing an open interval. This X¯ is CSS by Theorem 6.2.25. 4 5. Exerc. 6.21. X is distributed according to f(x|θ)=(θ/2)|x|(1 − θ1−|x|I(x ∈ {−1, 0, 1}), θ ∈ [0, 1]. (a) X is sufficient (of course - this is the observation) but not complete. Let us show this. Suppose that Eθψ(X)) ≡ 0, θ ∈ [0, 1]. This is equivalent to (θ/2)ψ(−1+(1 − θ)ψ(0)+(θ/2)ψ(1) ≡ 0, θ ∈ [0, 1], or θ[(1/2)ψ(−1) − ψ(0) + (1/2)ψ(1)] + ψ(0) ≡ 0, θ ∈ [0, 1]. A polynom is zero over an interval iff all its coefficients are zero. This implies that ψ(0) = 0 ( (1/2)[ψ(−1) + ψ(1)] − ψ(0) = 0 Solving this system yields ψ(0) = 0 and ψ(−a)= −ψ(1). As a results, ψ(k) is not necessarily identical zero. For instance, one may choose ψ(−1) = 5, ψ(0) = 0 and ψ(1) = −5. b),c) Let us begin with c). Write f(x|θ)= e|x| ln(θ/2(1−θ))(1 − θ)I(x ∈ {−1, 0, 1}).