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08Ma2bHw5Sol

# 08Ma2bHw5Sol - Math 2b Winter 2009 Solutions to Homework 5...

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Math 2b, Winter 2009: Solutions to Homework 5 1. The Raleigh distribution is given by f ( r | θ ) = r θ 2 exp - r 2 2 θ 2 . Find the mle for a sample of size n from this distribution. Solution. Let X 1 , . . . , X n be independent random variables from the Raleigh distribution for some parameter θ > 0. Since these variables are independent, their joint density function is the product of the individual density functions. Therefore we have f ( x 1 , . . . , x n , θ ) = n i =1 x i θ 2 exp - x 2 i 2 θ 2 . A maximum likelihood estimate of θ is a value of θ which maximizes f ( x 1 , . . . , x n , θ ). First note that if any x i is 0, this expression will be 0 and hence any θ > 0 will maximize. We now assume that x i > 0 for all i . To show that f ( x 1 , . . . , x n , θ ) has a maximum, we must verify that the limits as θ goes to 0 and exist and are not maxima. We first have that that lim θ →∞ f ( x 1 , . . . , x n , θ ) = 0 since the first part of the product goes to 0 and the exponential part goes to 1. Also, lim θ 0 x i θ - 2 exp( x 2 i / 2 θ 2 ) = lim θ 0 - 2 x i θ - 3 ( - x 2 i / θ 3 ) exp( x 2 i / 2 θ 2 ) = lim θ 0 2 x i exp - x 2 i 2 θ 2 = 0 for all i by l’Hˆopital’s Rule, so therefore lim θ 0 f ( x 1 , . . . , x n , θ ) = n i =1 lim θ 0 x i θ - 2 exp( x 2 i / 2 θ 2 ) = n i =1 0 = 0. Since f ( x 1 , . . . , x n , θ ) takes positive values in general, 0 is not a maximum. Therefore there must be a maximum at some θ (0 , ), and we will now find it. Since log is a monotonically increasing function, f ( x 1 , . . . , x n , θ ) will have maxima as the same points as log f ( x 1 , . . . , x n , θ ) = n i =1 log( x i ) - log( θ 2 ) - x 2 i 2 θ 2 = n i =1 log( x i ) - n log( θ 2 ) - 1 2 θ 2 n i =1 x 2 i . Di ff erentiating log f ( x 1 , . . . , x n , θ ) with respect to θ , we obtain - 2 n θ + 1 θ 3 n i =1 x 2 i . Since this is defined for all θ > 0, maxima must occur at points where this equals 0. Multiplying through by θ 3 , this means - 2 n θ 2 + n i =1 x 2 i = 0. 1

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Thus θ = ± 1 2 n n i =1 x 2 i , and since θ > 0, we must have θ = 1 2 n n i =1 x 2 i . Since we’ve shown that a maximum must exist, it must occur at this point, so therefore the maximum likelihood estimate of θ is ˆ θ = 1 2 n n i =1 x 2 i . 2. The results of 600 independent trials of an experiment that yields an outcome 1, 2, 3, or 4 are as follows Outcome 1 2 3 4 # of trials 280 199 93 28 Assume that for some q between 0 and 1, the probabilities of outcomes 1 to 4 on each trial are (1 - q ) 3 , 3 q (1 - q ) 2 , 3 q 2 (1 - q ), and q 3 . Calculate a maximum likelihood estimator of q . (Hint: This is similar to Example A in Rice’s section 8.5.1.) Solution. Let n denote the total number of trials, and let X 1 , X 2 , X 3 , and X 4 be random variables which count the number of outcomes 1, 2, 3, and 4, respectively. The joint density function of X 1 , . . . , X n is f ( x 1 , . . . , x n , q ) = n ! 4 i =1 x i ! ( (1 - q ) 3 ) x 1 ( 3 q (1 - q ) 2 ) x 2 ( 3 q 2 (1 - q ) ) x 3 ( q 3 ) x 4 .
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