STAT 400 Lecture AL1 p m f or p d f 1 Spring 2015 Dalpiaz Answers for 6 4 Part 1 f x Suppose 1 2 3 and the p m f parameter space 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 b c d X 1 f 1 1 0 6 f 1 2 0 2 f 1 3 0 3 1 f 2 1 0 1 f 2 2 0 3 f 2 3 0 4 3 f 3 1 0 1 f 3 2 0 3 f 3 3 0 2 2 f 4 1 0 2 f 4 2 0 2 f 4 3 0 1 1 or 2 X 2 X 3 X 4 maximum likelihood estimate may not be unique Likelihood function L L x1 x2 x n n f x i i 1 ln L It is often easier to consider f x 1 f x n n ln f x i i 1 Maximum Likelihood Estimator arg max L arg max ln L Method of Moments E X g 0 Set X g Solve for Consider a single observation X of a Binomial random variable with n trials and probability of success p That is P X k n C k p k 1 p n k a Obtain the method of moments estimator of p p Binomial E X n p X n p b X p n Obtain the maximum likelihood estimator of p p L p n C X p X 1 p n X ln L p ln n C X X ln p n X ln 1 p X np d X n X X Xp np Xp ln L p dp p 1 p p 1 p p 1 p d ln L p 0 dp X p n k 0 1 n 2 Let X 1 X 2 X n be a random sample of size distribution with mean 0 That is P X k a b k e k k 0 1 2 3 Obtain the method of moments estimator of E X X Obtain the maximum likelihood estimator of n n Xi e L f X i X i i 1 i 1 n from a Poisson n n ln L X i ln n ln X i i 1 i 1 n d ln L d 1 n X i n 0 i 1 Xi i 1 n X Let be the maximum likelihood estimate m l e of Then the m l e of any function h is h c The Invariance Principle Obtain the maximum likelihood estimator of P X 2 2 e P X 2 h 2 X h X 2 e X 2 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 a p Obtain the method of moments estimator of p E X 1 p X 1 so p b k 1 2 3 p 1 X n i 1 X i n Obtain the maximum likelihood estimator of p p n X n L p 1 p i 1 i pn ln L p in 1 X i n ln 1 p n ln p n d n i 1 X i n n p in 1 X i ln L p dp p 1 p p 1 p d ln L p 0 dp p n 1 n X i 1 X i p p equals the number of successes n divided by the number of Bernoulli trials c i 1 X i n Is p 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 p E g X g E X g 1 p p p 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 1 1 x f x 0 0 x 1 otherwise 0 a Obtain the method of moments estimator of 1 1 E X x f X x dx x x 0 1 dx 1 1 1 1 1 1 1 1 1 x x dx 0 1 1 1 0 X b 1 1 1 1 X 1 X X Obtain the maximum likelihood estimator of Likelihood function n L fX X i i 1 1 n 1 X i n i 1 n n 1 ln X i n ln 1 ln X i ln L n ln i 1 i 1 1 d d ln L n 1 n ln X i 0 2 i 1 1 n ln X i n i 1 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 X 0 2 0 3 0 5 3 1 1 1 X 1 3 2 1 X 3 3 n ln X i ln 0 2 ln 0 3 ln 0 5 1 16885 1 n i 1 5 1 3 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 and of 2 2 L 1 2 X 2 i 1 exp 2 2 2 2 n 1 i 1 n exp 2 2 1 ln L 1 2 n 2 i 1 X i 1 2 n 2 2 X i 1 2 i 1 n ln 2 2 2 2 1 2 The partial derivatives with respect to 1 and 2 are n ln L 1 X 1 1 2 i 1 i and n ln L n 1 X i 1 2 2 2 2 2 2 2 i 1 The equation ln L 1 0 has the solution 1 X Setting ln L 2 0 and replacing 1 by X yields 2 n X X n i 1 2 i 1 Therefore the maximum likelihood estimators of 1 and 2 2 are 1 X b and 2 n X X n i 1 2 i 1 Obtain the method of moments estimator of 1 1 and of 2 2 E X 1 E X 2 Var X E X 2 2 2 2 21 Thus Therefore X 1 n X i 1 n i 1 1 X and X 2 n i 1 …
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