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