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### STAT410HW8sol

Course: STAT 410, Spring 2012
School: UIllinois
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Word Count: 440

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410 STAT Homework 8 Solutions 6.1.2 The likelihood function is L() = and the log-likelihood function is n 1 (3)3 n n (X1 X2 ...Xn )2 exp - i=1 Xi / , l() = -n ln((3)) - 3n ln() + 2 i=1 ln(Xi ) - n i=1 Xi . The derivative of the log-likelihood function is l () = - 3n + n i=1 2 Xi , which equals zero if and only if = X/3. The second derivative of the log-likelihood function is l () = 2 3n - 2...

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410 STAT Homework 8 Solutions 6.1.2 The likelihood function is L() = and the log-likelihood function is n 1 (3)3 n n (X1 X2 ...Xn )2 exp - i=1 Xi / , l() = -n ln((3)) - 3n ln() + 2 i=1 ln(Xi ) - n i=1 Xi . The derivative of the log-likelihood function is l () = - 3n + n i=1 2 Xi , which equals zero if and only if = X/3. The second derivative of the log-likelihood function is l () = 2 3n - 2 n i=1 3 Xi = 3n - 2 3 n i=1 Xi , which is negative at = X/3. It follows that the MLE is ^ = X/3. 6.1.3(b) The likelihood function is L() = n (X1 X2 ...Xn )-1 , and the log-likelihood function is l() = n ln() + ( - 1) ln(X1 X2 ...Xn ). The derivative of the log-likelihood function is l () = which equals zero if and only if = -n = ln(X1 X2 ...Xn ) n i=1 n + ln(X1 X2 ...Xn ), -n . ln(Xi ) The second derivative of the log-likelihood function is l () = - for all > 0, so the MLE is ^ = 6.1.5(a) The likelihood function is L() = n < 0 2 -n . ln(Xi ) -n = ln(X1 X2 ...Xn ) n i=1 2n (X1 X2 ...Xn ) 2n 1 for X(n) = max(X1 , X2 , ..., Xn ), and zero elsewhere. This function is positive and decreasing for [X(n) , ) (note that the Xi 's are positive), and zero elsewhere, so the MLE is ^ = X(n) . This can also be seen by graphing L(). 6.1.6 We are to trying estimate 2 P (X 2) = 0 1 -x/ e dx = 1 - e-2/ . We know from Example 6.1.2 that the MLE of is X, so by Theorem 6.1.2, the MLE of P (X 2) is 1 - exp(-2/X). 6.1.9 From a result that was proved in class, the MLE of the unknown Poisson mean is X = 116/55 2.109. Because e- -2 P (X = 2) = , 2! it follows from Theorem 6.1.2 that the MLE of P (X = 2) is e-X X 2 e-2.109 (2.109)2 . 2! 2 6.1.11 (a) The likelihood function is L() = and the log-likelihood function is n l() = - ln(2 2 ) - 2 The derivative of the log-likelihood function is l () = 2 n i=1 (Xi 2 2 n i=1 (Xi - 2 2 1 2 n exp - n i=1 (Xi 2 2 - )2 , )2 . - ) = n i=1 (Xi 2 - ) = n i=1 Xi - n , 2 which equals zero if and only if = X. The second derivative of the log-likelihood function is l () = - n < 0 2 for all . It follows that = X is the unique point of maximum, so the MLE is ^ = X. (b) If X 0, then from part (a) the likelihood is maximized over [0, ) at = X. We know from the expression for the derivative of the log-likelihood function in part (a) that it is negative for > X. It follows that if X < 0 then l() is decreasing on [0, ), so the log-likelihood, and hence the likelihood, is maximized over [0, ) at = 0. The MLE is therefore ^ = max(X, 0). 2
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UIllinois - STAT - 410
STAT 410 Homework 7 Solutions 4.3.2 For z &gt; 0, the cdf of Zn evaluated at z is FZn (z) = P (Zn z) = = = = 1 - P (Zn &gt; z) = 1 - P (n(Y1 - ) &gt; z) 1 - P (Y1 &gt; + z/n) 1 - P (X1 &gt; + z/n, X2 &gt; + z/n, ., Xn &gt; + z/n) 1 - P (X1 &gt; + z/n) P (X2 &gt; + z/n) . P (Xn &gt; +
UIllinois - STAT - 410
STAT 410 Homework 6 Solutions 4.1.9 By Theorem 4.1.1, E(Z) = E(3X - 2Y ) = 3E(X) - 2E(Y ) = 3 - 8 = -5. By Theorem 4.1.2, V ar(X) = 9V ar(X) + 4V ar(Y ) - 12Cov(X, Y ) = (9)(4) + (4)(6) - (12)(1/2)(2)( 6) = 36 + 24 - 12 6 30.6 4.1.13 We have 15 = V ar(X +
UIllinois - STAT - 410
STAT 410 Homework 5 Solutions 3.3.6 Following the hint,we haveFY (y) = P (Y y) = 1 - P (Y &gt; y) = 1 -ye-u du3= 1 - (e-y )3 = 1 - e-3y ,so the probability density function of Y is fY (y) = FY (y) = 3e-3y for y &gt; 0. 3.3.18 Let X have a beta distributio
UIllinois - STAT - 410
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UIllinois - STAT - 410
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UIllinois - STAT - 410
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