STAT 400 Lecture AL1 Answers for 6.4, Part 1 Spring 2015 Dalpiaz p.m.f. or p.d.f. f ( x ; ), . – parameter space. 1. Suppose = { 1, 2, 3 } and the p.m.f. f ( x ; ) is = 1: f ( 1 ; 1 ) = 0.6, f ( 2 ; 1 ) = 0.1, f ( 3 ; 1 ) = 0.1, f ( 4 ; 1 ) = 0.2. = 2: f ( 1 ; 2 ) = 0.2, f ( 2 ; 2 ) = 0.3, f ( 3 ; 2 ) = 0.3, f ( 4 ; 2 ) = 0.2. = 3: f ( 1 ; 3 ) = 0.3, f ( 2 ; 3 ) = 0.4, f ( 3 ; 3 ) = 0.2, f ( 4 ; 3 ) = 0.1. What is the maximum likelihood estimate of ( based on only one observation of X ) if … a) X = 1; f ( 1 ; 1 ) = 0.6 f ( 1 ; 2 ) = 0.2 θˆ = 1. f ( 1 ; 3 ) = 0.3 b) X = 2; f ( 2 ; 1 ) = 0.1 f ( 2 ; 2 ) = 0.3 θˆ = 3. f ( 2 ; 3 ) = 0.4 c) X = 3; f ( 3 ; 1 ) = 0.1 f ( 3 ; 2 ) = 0.3 θˆ = 2. f ( 3 ; 3 ) = 0.2 d) X = 4. f ( 4 ; 1 ) = 0.2 f ( 4 ; 2 ) = 0.2 θˆ = 1 or 2. f ( 4 ; 3 ) = 0.1 (maximum likelihood estimate may not be unique)Likelihood function: L ( ) = L ( ; x 1 , x 2 , … , x n ) = ni 1f ( x i ; ) = f ( x 1 ; ) … f ( x n ; ) It is often easier to consider ln L ( ) = ni 1ln f ( x i ; ). Maximum Likelihood Estimator: θˆ = arg max L ( ) = arg max ln L ( ). Method of Moments: E ( X ) = g ( ). Set X = g ( θ~ ). Solve for θ~. 0. Consider a single observation X of a Binomial random variable with n trials and probability of “success” p. That is, knkppkCnk 1 XP )( )(, k = 0, 1, … , n. a) Obtain the method of moments estimator of p, p~. Binomial: E ( X ) = n p X = np~ npX~. b) Obtain the maximum likelihood estimator of p, pˆ. XX 1 XL )( )(nppCnp )(lnlnlnln 1 X XXL )( pnpCnp 11X1Xln X XXXL )(pppnppnpnppppppdd )ˆ( L ln ppdd = 0 npXˆ.2. Let X 1 , X 2 , … , X n be a random sample of size n from a Poisson distribution with mean , > 0. That is, P ( X = k ) = !λ λ kek , k = 0, 1, 2, 3, … . a) Obtain the method of moments estimator of , λ~. E ( X ) = Xλ~ b) Obtain the maximum likelihood estimator of , ˆ. niiniieif1X1 Xλ ; XL!λλλ . niiniin1 1 !λλλ XlnlnXLln. nniidd1 X1Lln λλλ = 0. XX1 λˆnnii. Let θˆ be the maximum likelihood estimate (m.l.e.) of . Then the m.l.e. of any function h ( ) is h ( θˆ ). ( The Invariance Principle ) c) Obtain the maximum likelihood estimator of P ( X = 2 ). P ( X = 2 ) = h ( ) = !λ 22λ e Xλˆ h (λˆ) = ! X22 X e.3. Let X 1 , X 2 , … , X n be a random sample of size n from a Geometric distribution with probability of “success” p, 0 < p < 1. That is, P ( X = k ) = ( 1 – p ) k – 1 p, k = 1, 2, 3, … . a) Obtain the method of moments estimator of p, p~. E ( X ) = 1/p . p~ 1X so niinp1 X X 1~. b) Obtain the maximum likelihood estimator of p , pˆ. nnpppnii 1 L X1 )()( pnpnpnii 1 XL ln)(lnln 1 )( 1Xln 1XL 1 1 )(pppnpnppniiniipndd )ˆ( L ln ppdd = 0 X 1X 1 ˆniinp. pˆ = p~ equals the number of successes, n, divided by the number of Bernoulli trials, nii1 X; c)* Is pˆ an unbiased estimator for p? Since g ( x ) = 1/x , x 1, is strictly convex, and X is not a constant random variable, by Jensen’s Inequality, E (pˆ) = E ( g (X) ) > g ( E (X) ) = g ( 1/p ) = p. pˆ is NOT an unbiased estimator for p.4. Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function f ( x ; ) = otherwise010θ1 θ θ1xx 0 < < . a) Obtain the method of moments estimator of , θ~. 10θ θ1X θ1 θ X E ; dxxxdxxfx. = θ11 01 1 θ 11θ1 θ11 θ 110θ 1 xdxx. θ11 X. XX1 1X1 θ~. b) Obtain the maximum likelihood estimator of , θˆ. Likelihood function: L() = θ θ1 11X X θ1 θ X ;ninniiif. ln L() = niiniinn11 Xln 1θ1θ Xln θθ1θ lnln. niidd n12 Xln θˆ 1 θˆ θˆL θ ln = 0. niin1 Xln 1 θˆ.c) Suppose n = 3, and x 1 = 0.2, x 2 = 0.3, x 3 = 0.5. Compute the values of the method of moments estimate and the maximum likelihood estimate for . 3135.03.02.0 X . 31311XX1 θ~ = 2. 5.0ln3.0ln2.0ln31Xln 1 θˆ1 niin 1.16885. 5. Let X 1 , X 2 , … , X n be a random sample of size n from N ( 1 , 2 ), where = { ( 1 , 2 ) : – < 1 < , 0 < 2 < }. That is, here we let 1 = and 2 = 2 . a) Obtain the maximum likelihood estimator of 1 , 1 θˆ, and of 2 , 2 θˆ. L ( 1 , 2 ) = nii12212 θθθπ 2Xexp 21 = 21212 θθθπ 2Xexp 21 nini, ( 1 , 2 ) . ln L ( 1 , 2 ) = 21212 θθθπln 2X 2 2 niin.The partial derivatives with respect …
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