hw11_stat210a_solutions

# hw11_stat210a_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 11 Fall 2006 Issued: Thursday, November 9, 2006 Due: Thursday, November 16, 2006 Graded exercises Problem 11.1 a) The functional h ( F ) = F ( a ) is continuous. For any a R , | F n ( a ) - F ( a ) | < sup x | F n ( x ) - F ( x ) | , hence 0 lim n →∞ | F n ( a ) - F ( a ) | < lim n →∞ sup x | F n ( x ) - F ( x ) | = 0, which establishes continuity. b) The functional h ( F ) = R | F ( t ) - F 0 ( t ) | 2 f 0 ( t ) dt is continuous. First, notice that g ( x ) = | x - y | 2 is a continuos function on x and y . Hence, for any ε > 0 there exists δ such that sup x | F n ( x ) - F ( x ) | ≤ δ ± ± | F n ( x ) - F 0 ( x ) | 2 - | F n ( x ) - F 0 ( x ) | 2 ± ± < ε , for all x . Given that lim n →∞ sup x | F n ( x ) - F ( x ) | = 0, for any ε > 0 there exists N such that n > N sup x ± ± | F n ( x ) - F 0 ( x ) | 2 - | F n ( x ) - F 0 ( x ) | 2 ± ± < ε . As a result, for n > N : 0 ≤ | h ( F n - h ( F ) | ≤ R sup x ± ± | F n ( x ) - F 0 ( x ) | 2 - | F n ( x ) - F 0 ( x ) | 2 ± ± f 0 ( x ) dx ε R f 0 ( x ) dx = ε c) The mean functional is discontinuous. Consider the sequence of random variables given by: X n = ² 0 , with probability 1 - 1 n n 2 , with probability 1 n Let F n denote the cdf of X n and F ( x ) = I ( x 0). We have sup x R | F n ( x ) - F ( x ) | = 1 n achieved for any x [0 , n 2 ]. However, lim n →∞ | h ( F ) - h ( F n ) | = . d) We have that, for all δ > 0 there exists N such that for all n > N , | G n ( x ) - F ( x ) | < δ for all x R . Now, ﬁx F ( x α ) = α so G n ( x ) ( α - δ, α + δ ). Since F - 1 is continuous at α , G n ( x ) ( α - δ, α + δ ) implies F - 1 ( G n ( x )) ( x α - ε, x α + ε ) so the h α functional is continuous. Problem 11.2 a) We have for the mean functional: S h,F ( x ) = h ( ˆ F n,x ) - h ( ˆ F n - 1 ) = 1 n n - 1 X i =1 X i + x n - 1 n - 1 n - 1 X i =1 X i = n - 1 - n n ( n - 1) n - 1 X i =1 X i + x n = x n - 1 n ( n - 1) n - 1 X i =1 X i 1

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b) Let X (1) < X (2) < ··· < X ( k - 1) < X ( k ) < ··· < X (2 k - 1) < X (2 k ) denote the original sample containing n - 1 = 2 k samples. Consider what can happen when a new point x is added to the sample. Three cases can happen: i) x < X ( k - 1) : In that case, the new data set contains 2 k points to the left of
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hw11_stat210a_solutions - UC Berkeley Department of...

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