Statistical Theory MT 2009 Problems 1: Solution sketches
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1 Statistical Theory MT 009 Problems : Solutio sketches. Which of the followig desities are withi a expoetial family? Explai your reasoig. (a) Let 0 < θ < ad put f(x, θ) = ( θ)θ x ; x = 0,,,... (b) (c) where α > 0, λ > 0; f(x; α, λ) = λα Γ(α) e λx x α, x > 0, f(x, θ) = e (x θ), x θ. Solutio: (a) where 0 < θ <, ca be writte as f(x, θ) = ( θ)θ x ; x = 0,,,..., (b) (c) f(x, θ) = exp{x log θ + log( θ) + 0}; x = 0,,, so k =, c =, h (x) = x, φ (θ) = log θ, c(θ) = log( θ), d(x) = 0, ad X = {0,,,...} does ot deped o θ, so f is i a -parameter expoetial family. f(x; α, λ) = where α > 0, λ > 0, ca be writte as λα Γ(α) e λx x α, x > 0, f(x; α, λ) = exp { λx + (α ) log(x) + log(λ α /Γ(α)) + 0} so with θ = (α, λ), we ca choose k =, c = c =, h (x) = x, h (x) = log(x), φ (θ) = λ, φ (θ) = α, c(θ) = log(λ α /Γ(α)), d(x) = 0, ad X = (0, ) does ot deped o θ, so f is i a -parameter expoetial family. f(x, θ) = e (x θ), x θ, ca be writte i the expoetial family form with k =, c =, h (x) =, φ (θ) = θ, d(x) = x, but X = [θ, ) depeds o θ, so it is o-regular.
2 . Suppose X, X,..., X is a radom sample from the Pareto distributio f(x, λ) = λαλ x λ+ with x > α, λ > 0, ad α > 0 kow. Fid a miimal sufficiet statistic for λ. Solutio: The likelihood is so T = j= X j is sufficiet. For x, y we have L(λ, x) = (λαλ ) λ α λ = j= xλ+ ( j j= x j), λ+ L(λ, x) L(λ, y) = ( j= y j j= x j ) λ+ is costat i λ if ad oly if t(x) = t(y), so T is miimal sufficiet. 3. Suppose X, X,..., X is a radom sample from the log-ormal distributio with desity { f(x, µ, φ) = x πφ exp } (log x µ) φ with φ > 0 (so that log X j N (µ, φ)). Fid a miimal sufficiet statistic for the parameter θ = (µ, φ). Solutio: L(θ, x) = { ( exp x i )(πφ) / φ } (log x i µ) i= ad the expressio i the expoetial ca be writte as (log xi ) + µ log xi µ φ φ φ, so T = ( log X i, (log X i ) ) is sufficiet. By cosiderig the likelihood ratio we see that T is ideed miimal sufficiet. 4. Suppose X,..., X are idepedet ad expoetially distributed, each with desity fuctio Let X = i= X i ad put f(x; θ) = θ e x/θ, x 0. T = X. i= (X i X) Show that T is a acillary statistic. What does this say about t-tests o expoetial data?
3 Solutio: Note that Y i = θ X i has the expoetial distributio with parameter, so its distributio does ot deped o θ. Next ote that T = Y. i= (Y i Y ) Hece the distributio of T does ot deped o θ either; it is a acillary statistic. Thus a t-test based o expoetial data would ot reveal ay iformatio about the parameter θ. 5. Let X,..., X be i.i.d. uiform U [ θ, θ + ] radom variables. a) Show that ( ) X (), X () is miimal sufficiet for θ. b) Show that (S, A) = ( (X ) () + X () ), X () X () is miimal sufficiet for θ, ad that the distributio of A is idepedet of θ (so A is a acillary statistic). c) Show that ay value i the iterval [ x (), x () + ] is a maximum-likelihood-estimator for θ. Solutio: a) The likelihood is ( L(θ, x) = θ x,..., x θ + ) ( = θ x (), x () θ + ) which is a fuctio of (θ, x (), x () ), so ( X (), X () ) is sufficiet. For miimal sufficiecy, ote that L(θ, x) = L(θ, y) if ad oly if (x (), x () ) = (y (), y () ), provig the first assertio. b) We have that X () = S A X () = S + A so L(θ, x) is a fuctio of (θ, s, a), hece (S, A) is sufficiet, ad (S, A) is a fuctio of a miimal sufficiet statistic, so must be miimal sufficiet itself. To see that the distributio of A is idepedet of θ, write Y i = X i θ, the Y i U( /, /), ad Y (i) = X (i) θ. Hece A = X () X () = Y () Y (), the differece of order statistics from a distributio which does ot ivolve θ. So the distributio of A does ot deped o θ. c) Follows directly from the likelihood - the likelihood ( L(θ, x) = x () θ x () + ) is costat equal to o the iterval [ x (), x () + ], ad 0 otherwise. 6. The radom variables X,..., X are idepedet with geometric distributio P(X i = x) = p( p) x for x =,,.... Let θ = p. 3
4 (i) Fid the maximum likelihood estimator for p. Cosiderig =, is it ubiased? (ii) Show that ˆθ = X is the maximum likelihood estimator for θ. Is it ubiased? (iii) Compute the expected Fisher iformatio for θ. (iv) Does ˆθ attai the Cramer-Rao lower boud? Solutio: a) We have that L(p, x) = p ( p) xi i= ad so ( ) l(p) = log p + x i log( p); i= differetiate: l (p) = p i= x i ; p the oly solutio for l (p) = 0 is p = x. The secod derivative is l (θ) = p i= x i ( p) < 0, so that ˆp = x For =, is the mle. ( ) E X > p, = p( p)x x x= = p + p( p)x x x= so ˆp is ot ubiased. b) We have that L(θ, x) = θ ( θ ) xi i= ad so l(θ) = log θ + (x i ) log( θ ); i= 4
5 differetiate: l (θ) = (x θ); θ(θ ) the oly solutio for l (θ) = 0 is θ = x. The secod derivative is so that l (θ) = { θ(θ ) (x θ)(θ )}; (θ(θ )) l (ˆθ) = if ˆθ = x >, so if ˆθ >, the mle is ˆθ = x. (ˆθ(ˆθ )) ( ˆθ(ˆθ )) < 0 If x =, the L(θ, x) = θ, ad we kow that θ, so the likelihood is maximized for θ =. c) Calculate that E θ X = θ, V ar θ (X) = θ(θ ), so, from l (θ), I(θ) = = θ (θ ) V ar θ(x) θ(θ ), which equals the iverse of the variace of X. So it attais the Cramer-Rao lower boud. Note: As E θ X = θ, the estimator is ubiased, ad we have see that it attais the Cramer-Rao lower boud, hece it has miimum variace amog all ubiased estimators. 7. Let X,..., X be i.i.d. U[0, θ], havig desity f(x; θ) = θ, 0 x θ with θ > 0. (i) Estimate θ usig both the method of momets ad maximum likelihood. (ii) Calculate the meas ad variaces of the two estimators. (iii) Which oe should be preferred ad why? Solutio: a) We have E θ (X) = θ, so the m.o.m. is θ = X. The likelihood is L(θ, x) = θ (0 x,..., x θ) = θ ( x () θ ) 5
6 ad this is maximized for ˆθ = x (). b) We have by costructio E θ ( θ) = θ, ad V ar θ ( θ) = 4θ = θ 3. For ˆθ, calculate (see secod year probability, or the book by Rice) E θ (ˆθ) = + θ, V ar θ(ˆθ) = ( + ) ( + ) θ. c) The m.l.e. is ot ubiased but asymptotically ubiased, ad has much smaller variace tha the m.o.m., so I would prefer it. You may decide otherwise if you put more emphasis o ubiasedess. 8. Suppose X,..., X are a radom sample with mea µ ad fiite variace σ. Use the delta method to show that, i distributio, (X µ ) N (0, 4µ σ ). What would you suggest if µ = 0? Solutio: From the cetral limit theorem we kow that (X µ) N (0, σ ). We use the fuctio g(s) = s, so that g (s) = s; ow the delta method gives that Stadardizig gives the result. X N (µ, 4µ σ /). If µ = 0 the the result just gives covergece to poit mass at 0, which is ot iformative. Istead we could use that X σ N (0, ), so that X σ X σ χ. χ, ad, i distributio, 6
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