1 M3S3/M4S3 STATISTICAL THEORY II WORKED EXAMPLE: TESTING FOR THE PARETO DISTRIBUTION Suppose that X1;:::;Xn are i.i.d random variables having a Pareto distribution with pdf θcθ f (xjθ) = x > c Xjθ xθ+1 and zero otherwise, for known constant c > 0, and parameter θ > 0. (i) Find the ML estimator, θbn, of θ, and ¯nd the asymptotic distribution of p n(θbn ¡ θT ) where θT is the true value of θ. (ii) Consider testing the hypotheses of H0 : θ = θ0 H1 : θ 6= θ0 for some θ0 > 0. Determine the likelihood ratio, Wald and Rao tests of this hypothesis. SOLUTION (i) The ML estimate θbn is computed in the usual way: Yn Yn θcθ θncnθ Ln(θ) = fXjθ(xijθ) = θ+1 = θ+1 i=1 i=1 xi sn Yn where sn = xi. Then i=1 ln(θ) = n log θ + nθ log c ¡ (θ + 1) log sn n l_ (θ) = + n log c ¡ log s n θ n and solving l_n(θ) = 0 yields the ML estimate · ¸ " #¡1 log s ¡1 1 Xn θb = n ¡ log c = log x ¡ log c : n n n i i=1 The corresponding estimator is therefore " #¡1 1 Xn θb = log X ¡ log c : n n i i=1 Computing the asymptotic distribution directly is di±cult because of the reciprocal. However, consider Á = 1/θ; by invariance, the ML estimator of Á is 1 Xn 1 Xn Áb = log X ¡ log c = (log X ¡ log c) n n i n i i=1 i=1 2 which implies how we should compute the asymptotic distribution of θbn - we use the CLT on the random variables Yi = log Xi ¡ log c = log(Xi=c), and then use the Delta Method. To implement the CLT, we need the expectation and variance of Y = log(X=c): Now ³ c ´θ F (x) = 1 ¡ x > c X x so that FY (y) = P [Y · y] = P [log(X=c) · y] = P [X · c expfyg] = 1 ¡ expf¡θyg y > 0 and hence Y » Exponential(θ). By standard results 1 1 E [Y ] = = Á V ar [Y ] = = Á2; fY θ fY θ2 and therefore, by the CLT, p b L 2 n(Án ¡ ÁT ) ¡! N(0;ÁT ) where ÁT = 1/θT . Finally, let g(t) = 1=t so thatg _(t) = ¡1=t2. Then, by the Delta Method p b L 2 2 n(g(Án) ¡ g(ÁT )) ¡! N(0; fg_(ÁT )g ÁT ) so that, as g(ÁT ) = 1=ÁT = θT p b L 4 2 2 n(θn ¡ θT ) ¡! N(0; f1=ÁT gÁT ) ´ N(0; θT ): (ii) For the likelihood ratio test: b b b bn nθn θn+1 h i Ln(θn) θnc =sn b b b ¸n = 2 log = 2 log θ +1 = 2n log(θn/θ0) + (θn ¡ θ0) log c ¡ (θn ¡ θ0)mn L (θ ) n nθ0 0 n 0 θ0 c =sn where mn = (log sn)=n. For the Wald test: 2 Wn = n(θbn ¡ θ0)I(θbn)(θbn ¡ θ0) = nI(θbn)(θbn ¡ θ0) where I(θ) is the Fisher information for this model. From ¯rst principles l(θ) = log θ + θ log c ¡ (θ + 1) log x 1 l_(θ) = + log c ¡ log x θ 1 Äl(θ) = ¡ θ2 so I(θ) = 1/θ2; in fact, we could have deduced this from the results derived in (i). Thus à !2 θbn ¡ θ0 Wn = n : θbn 3 Note that, although the Fisher Information is available in this case, it may be statistically advantageous in a ¯nite sample case to replace I(θbn) by Ibn(θbn), derived in the usual way from the ¯rst or second derivatives of ln. That is, we might use µ ¶ 1 Xn 1 Xn 1 2 1 Xn Ib (θb ) = S(x ; θ)2 = + log c ¡ log x = (y ¡ y)2 n n n i n b i n i i=1 i=1 θn i=1 where yi = log xi ¡ log c. Alternately, using the second derivative, 1 Xn 1 Xn Ib (θb ) = ¡ Äl (θ) = 1=θb2 = 1=θb2 = I(θb ): n n n i n n n n i=1 i=1 For the Rao test: in the single parameter case T ¡1 2 Rn = Zn[I(θ0)] Zn = Zn=I(θ0) where µ ¶ 1 Xn 1 Xn 1 Z = p S(x ; θ ) = p + log c ¡ log x n n i 0 n θ i i=1 i=1 0 so that if yi = log xi ¡ log c as before ( ) ( ) Xn 2 Xn 2 (yi ¡ 1/θ0) (yi ¡ 1/θ0) i=1 i=1 Rn = = 2 : nI(θ0) n/θ0 SUPPLEMENTARY EXERCISES: Suppose that c is also an unknown parameter. Find ² the ML estimator for c, bc ² the weak-law (ie probability) limit of bc ² an asymptotic (large n) approximation to the distribution of bc. ² the pro¯le likelihood for θ. Hint: Recall, when considering the likelihood for c, that xi > c for all i. Then, think back to M2S1 Chapter 5, and extreme order statistics..