Dr. Hackney STA Solutions pg 116

# Dr. Hackney STA Solutions pg 116 - Second Edition 7-19 e....

This preview shows page 1. Sign up to view the full content.

Second Edition 7-19 e. Using the expression for a in part (d), if κ = 3 the second term in the denominator is zero and a = ( n - 1) / ( n + 1), the normal result from part (c). If κ < 3, the second term in the denominator is negative. Because we are dividing by a smaller value, we have a > ( n - 1) / ( n +1). Because Var( S 2 ) = 4 , B > 0, and, hence, a = 1 / ( B +1) < 1. Similarly, if κ > 3, the second term in the denominator is positive. Because we are dividing by a larger value, we have a < ( n - 1) / ( n + 1). 7.46 a. For the uniform( θ, 2 θ ) distribution we have E X = (2 θ + θ ) / 2 = 3 θ/ 2. So we solve 3 θ/ 2 = ¯ X for θ to obtain the method of moments estimator ˜ θ = 2 ¯ X/ 3. b. Let x (1) ,...,x ( n ) denote the observed order statistics. Then, the likelihood function is L ( θ | x ) = 1 θ n I [ x ( n ) / 2 ,x (1) ] ( θ ) . Because 1 n is decreasing, this is maximized at ˆ θ = x ( n ) / 2. So ˆ θ = X ( n ) / 2 is the MLE. Use the pdf of X ( n ) to calculate E
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

Ask a homework question - tutors are online