411_hw02_soln

411_hw02_soln - Stat 411 Homework 02 Solutions 1. (a) The...

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Stat 411 – Homework 02 Solutions 1. (a) The CDF of Unif (0 ) is F θ ( x ) = x/θ , if x (0 ) and the obvious mod- ifications otherwise. Because X 1 ,...,X n are iid Unif (0 ), the CDF of the maximum ˆ θ n = X ( n ) is simply F n,θ ( t ) = P θ ( ˆ θ n t ) = ( t/θ ) n , for t (0 ). (b) Following the hint, since each X i is strictly less than θ , we know that the maximum ˆ θ n is too. Take any ε > 0. Then P θ {| ˆ θ n - θ | > ε } = P θ { θ - ˆ θ n > ε } = P θ { ˆ θ n < θ - ε } = ± θ - ε θ ² n . If ε < θ , the right-hand side converges to 0 as n → ∞ , which proves the con- sistency claim. (Note that if ε > θ , then the left-hand side above is identically 0 and, hence, so is the limit.) 2. (a) If the CDF of ˆ θ n is F n,θ ( t ) = ( t/θ ) n , then the corresponding PDF is obviously f n,θ ( t ) = nt n - 1 n for t (0 ), and = 0 otherwise. Then E θ ( ˆ θ n ) = Z θ 0 tf n,θ ( t ) dt = n Z θ 0 ( t/θ ) n dt = Z 1 0 y n
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411_hw02_soln - Stat 411 Homework 02 Solutions 1. (a) The...

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