13 - or one that relies on the sample mean being an...

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Unformatted text preview: or one that relies on the sample mean being an unbiased estimate of θ/ 2: ˆ Θ = 2 n n X i =1 X i . Solution to Problem 9.9. The PDF of X i is f X i ( x i ) = n 1 , if θ ≤ x i ≤ θ + 1, , otherwise. The likelihood function is f X ( x 1 , . . . , x n ; θ ) = f X 1 ( x 1 ; θ ) ··· f X n ( x n ; θ ) = n 1 , if θ ≤ min i =1 ,...,n x i ≤ max i =1 ,...,n x i ≤ θ + 1, , otherwise. Any value in the feasible interval max i =1 ,...,n X i − 1 , min i =1 ,...,n X i maximizes the likelihood function and is therefore a ML estimator. Any choice of estimator within the above interval is consistent. The reason is that min i =1 ,...,n X i converges in probability to θ , while max i =1 ,...,n X i converges in probability to θ + 1 (cf. Example 5.6). Thus, both endpoints of the above interval converge to θ . Let us consider the estimator that chooses the midpoint ˆ Θ n = 1 2 max i =1 ,...,n X i + min i =1 ,...,n X i − 1 of the interval of ML estimates. We claim that it is unbiased. This claim can be verified purely on the basis of symmetry considerations, but nevertheless we provide a detailed calculation. We first find the CDFs of max i =1 ,...,n X i and min i =1 ,...,n X i , then their PDFs (by differentiation), and then E [ ˆ Θ n ]. The details are very similar to the ones for the preceding problem. We have by straightforward calculation, f min i X i ( x ) = n ( θ + 1 − x ) n − 1 , if θ ≤ x ≤ θ + 1, , otherwise, f max i X i ( x ) = n ( x − θ ) n − 1 , if θ ≤ x ≤ θ + 1, , otherwise. Hence E min i =1 ,...,n X i = n Z θ +1 θ x ( θ + 1 − x ) n − 1 dx = − n Z θ +1 θ ( θ + 1 − x ) n dx + ( θ + 1) n Z θ +1 θ ( θ + 1 − x ) n − 1 dx = − n Z 1 x n dx + ( θ + 1) n Z 1 x n − 1 dx = − n n + 1 + θ + 1 = θ + 1 n + 1 . 121 Similarly, E max i =1 ,...,n X i = θ + n n + 1 , and it follows that E [ ˆ Θ n ] = 1 2 E max i =1 ,...,n X i + min i =1 ,...,n X i − 1 = θ. Solution to Problem 9.10. (a) To compute c ( θ ), we write 1 = ∞ X k =0 p K ( k ; θ ) = ∞ X k =0 c ( θ ) e − θk = c ( θ ) 1 − e − θ , which yields c ( θ ) = 1 − e − θ . (b) The PMF of K is a shifted geometric distribution with parameter p = 1 − e − θ (shifted by 1 to the left, so that it starts at k = 0). Therefore, E [ K ] = 1 p − 1 = 1 1 − e − θ − 1 = e − θ 1 − e − θ = 1 e θ − 1 , and the variance is the same as for the geometric with parameter p , var( K ) = 1 − p p 2 = e − θ (1 − e − θ ) 2 . (c) Let K i be the number of photons emitted the i th time that the source is triggered. The joint PMF of K = ( K 1 , . . . , K n ) is p K ( k 1 , . . . , k n ; θ ) = c ( θ ) n n i =1 e − θk i = c ( θ ) n e − θs n , where s n = n X i =1 k i ....
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This note was uploaded on 01/11/2011 for the course MATH 170 taught by Professor Staff during the Spring '08 term at UCLA.

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13 - or one that relies on the sample mean being an...

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