Outline of Solutions for HW 9

• 8.17: (b) theta-hat = x-bar, var(theta-hat) = 1/n
  Fisher information = - E [ d^2 log f(x|theta) / d(theta)^2 ] = - E [ -1 ] = 1
  1/[ n*I(theta) ] = 1/n = var(theta-hat).
  (d) theta-hat = x-bar/5, var(theta-hat) = var(X-bar)/25 = var(X)/(25n) = theta^2/(5n)
  I(theta) = - E [ 5/(theta^2) - 2X/(theta^3) ] = -5/(theta^2) + 2*(5*theta)/(theta^3) = 5/(theta^2)
  1/[ n*I(theta) ] = theta^2/(5n) = var(theta-hat).
  
• 8.18: (b) x-bar +/- 1.96/sqrt(n) (d) theta-hat is approx. normally dist'ed with mean theta and variance theta^2/5n, so (theta-hat - theta)/(theta-hat/sqrt(5n)) is approx. N(0,1) by applying Slutsky's theorem and the WLLN (theta-hat -> theta in probability). So a 95% CI is x-bar/5 +/- 1.96*x-bar/(5*sqrt(5n)).
• 9.28: LRT: lambda = sup_{null} f / sup_{alternative} f = f(X|theta=theta') / f(X|theta=x-bar) = exp( -1/2 sum (x_i - theta')^2 + 1/2 sum (x_i - x-bar)^2) = exp( -n/2 [ (x-bar)^2 - 2(x-bar)(theta') + (theta')^2 ] ) = exp( -n/2 (x-bar - theta')2
  LRT rejects when lambda < k, i.e., when |x-bar - theta'| > k'.
  This is not a UMP test. Suppose theta' > 0. Then a one-sided test (of the same size) that rejects when x-bar - theta' < k'' will have larger power for theta=0 (this claim should be verified).
  If you were asked to find the value for k' for a specified size alpha, you would use the normal table (in the book, or on a computer), since you know the distribution of x-bar when theta=theta'.
• 9.34: The numerator and denominator of the LRT statistic are maximized by their respective MLEs, max X_i and X-bar. Thus lambda = (X-bar)^n * e^n / (max X_i)^n. Simplifying, we reject when X-bar/(max X_i) > k.

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