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Solutions08.pdf

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- New-york
- Cornell University
- Engineering
- Engineering 2700
- Richard
- Solutions08.pdf

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Homework 8 1. a. implies that θ2)( 2 =XE θ= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 2 2 X E . Consider n X i 2 ˆ 2 ∑ =θ . Then () ( ) θ θ θ θ ==== ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∑∑∑ n n nn XE n X EE ii 2 2 2 2 22 ˆ 22 , implying that is an unbiased estimator for θ ˆ θ . b. , so 1058.1490 2 = ∑ i x 505.74 20 1058.1490 ˆ ==θ 3. a. We wish to take the derivative of , set it equal to zero and solve for p. () ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −xn x pp x n 1ln ()( )( ) p xn p x pxnpx x n dp d − − −= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −−++ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 1 1lnlnln ; setting this equal to zero and solving for p yields n x p =ˆ . For n = 20 and x = 3, 15. 20 3 ˆ ==p b. () () ()pnp n XE nn X EpE ===⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 11 ˆ ; thus is an unbiased estimator of p. pˆ c. ()4437.15.1 5 =− 4. a. () 2 1 1 2 1 1)( 1 0 + −= + + =+= ∫ θθ θ θ θ dxxxXE , so the moment estimator is the solution to θ ˆ 2 ˆ 1 1 + −= θ X , yielding 2 1 1 ˆ − − = X θ . Since .325 ˆ ,80. =−== θx b. ()()( ) θ θθ n n n xxxxxf ...1;,..., 211 += , so the log likelihood is . Taking () ( ∑ ++ i xn ln1ln θθ ) θd d and equating to 0 yields ∑ −= + )ln( 1 i x n θ , so 1 )ln( ˆ −−= ∑ i X n θ . Taking ( ) i xln for each given yields ultimately . i x 12.3 ˆ =θ 5. a. [] []() [ ] n i nn n x xxx x x x θ θ θ θ θ θ 2/exp ...2/exp...2/exp 2 1 22 1 1 Σ− =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − . The natural log of the likelihood function is ()() θ θ 2 ln...ln 2 i ni x nxx Σ −− . Taking the derivative wrt θ and equating to 0 gives 0 2 2 2 = Σ +− θθ i xn , so 2 2 i x n Σ =θ and n x i 2 2 Σ =θ . The mle is therefore n X i 2 ˆ 2 Σ =θ , which is identical to the unbiased estimator suggested in Exercise 15. b. For x > 0 the cdf of X if ( ) ( )xXPxF ≤=θ; is equal to ⎥ ⎦ ⎤ ⎢ ⎣ ⎡− − θ2 exp1 2 x . Equating this to .5 and solving for x gives the median in terms of θ : ⎥ ⎦ ⎤ ⎢ ⎣ ⎡− = θ2 exp5. 2 x implies that () θ2 5.ln 2 x− = , so θμ 38630.1 ~ ==x . The mle of μ ~ is therefore () 2 1 ˆ 38630.1 θ . 7. a. () ()5.59,1.5718.13.58 25 396.1 3.58 =±=± b. () ()9.58,7.5759.3.58 100 396.1 3.58 =±=± c. () ()1.59,5.5777.3.58 100 358.2 3.58 =±=± d. 82% confidence 09.18.82.1 2 =⇒=⇒=−⇒ α αα , so 34.1 09. 2 == zz α and the interval is () ()7.58,9.57 100 334.1 3.58 =± . e. () 62.239 1 358.22 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =n so n = 240. 8. a. ()() =±=± 9.328439 25 100645.1 8439 (8406.1, 8471.9). b. 04.08.92.1 2 =⇒=⇒=− α αα so 75.1 04. 2 == zz α 10. d.f. = n – 1 = 7, so the critical value for a 95% C.I. is 365.2 7,025. =t . The interval is () ()8.32,6.276.22.30 8 1.3 365.22.30 =±= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ± .