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Unformatted text preview: Homework 2 solution 1. a) E b) E
4 5 X1 +X3 = + = , so it's unbiased. 2 2 ++ X1 +X3 +Xn = 3 = , so it's unbiased. 3 X1 +X3 +3X2 X4 = ++3 = 4 , so it's biased 5 5 5  1 . 5 c) E with bias equal to = d) E (3X1  2X3 ) = 3  2 = , so it's unbiased. e) E X1  X2 + Xn/2  Xn1 + Xn =  +  + = , so it's unbiased.
2 2 2 X1 +X3 = + = 2 . Since it's unbiased, M SE X1 +X3 = 2 4 2 2 V X1 +X3 = 2 . 2 2 2 2 2 b) V X1 +X3 +Xn = +9 + = 3 . Since it's unbiased, M SE X1 +X3 +Xn = 3 3 X1 +X3 +Xn 2 + 2 + 2 2 V = = 3. 3 9 2 2 2 2 2 X1 +X3 +3X2 X4 c) V = + +9 + = 12 . So the mean square error is 5 25 25 2 2 2 M SE( X1 +X3 +3X2 X4 ) = 12 + = 13 . 5 25 25 25 d) V (3X1  2X3 ) = 9 2 +4 2 = 13 2 . Since it's unbiased, M SE (3X1  2X3 ) = V (3X1  2X3 ) = 13 2 . e) V X1  X2 + Xn/2  Xn1 + Xn = 5 2 . So M SE X1  X2 + Xn/2  Xn1 2 2. a) V + Xn = 5 as well. We see that the estimator from 1(b) has the smallest MSE. 3. The likelihood function is L() = So ln L() = n +
i=1 en n i=1 n n i=1 xi xi ! .
n xi ln 
i=1 ln xi !. Taking a derivative of the latter and equating it to zero, we obtain n +
n i=1 xi = 0, 1 ^ which solves to = n n xi , which is simply the sample mean. The latter i=1 is not surprising since, for Poisson random variable X, E(X) = . 1 4. The corresponding probability density function is if x < a 1 f (x) = ba , if a x b . 0, if x > b Thus the likelihood function takes the form L(a, b) = 0,
1 , (ba)n 0, if a mini xi and b maxi xi . otherwise So, in order to maximize L(a, b), we have to make a as large as possible and b as small as possible, without letting L(a, b) become zero. This means that a = mini xi and ^ = maxi xi . ^ b 2 ...
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 Spring '08
 Perevalov

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