ST 530 February 28, 2005
Name: Midterm #1
All problem parts have equal weight. In budgeting your time expect that some problems will take longer than others.
1. Let X1 , . . . , Xn be an i.i.d. sa
Homework #12 solutions
Stat 530, Spring 2010
1. #8.15 We can use the Neymann-Pearson lemma because H0 and H1 are simple. Recall
from a previous problem (and the second exam) that T = n=1 Xi2 is sucien
Homework #11 solutions
Stat 530, Spring 2010
1. #8.2 Suppose the true rate this year is 15. Then the probability of seeing 10 or fewer
accidents is
10
15k
e15
= .1185,
k!
k=0
which means that it is no
Homework #10 solutions
Stat 530, Spring 2010
1. #7.37 We determined earlier that a sucient statistic for is max(|Xi |). Using
that the cdf for the density for the Xi is FX (x) = ( + x)/(2), we can get
Homework #9 solutions
Stat 530, Spring 2010
1. #4.26 (a)
P (Z z, W = 0) = P (Y z, Y X )
1 z x/ y/
=
e
e
dxdy
0 y
11
=
1 exp z
+
+
and similarly,
P (Z z, W = 1) = P (Y z, Y X ) =
11
1 exp z
+
+
(b) To
Homework #8 solutions
Stat 530, Spring 2010
1. #7.2 (a) We have
n
L(, |x) =
i=1
1 1 1 xi /
11
xe
=
i
()
()
n
1
n
n
expcfw_
xi
i=1
xi /
i=1
and
n
(, |x) = n log() n log( ) + ( 1) log
n
xi
i=1
xi /
Homework #7 solutions
Stat 530, Spring 2010
1. To show that a statistic T (X ) is not complete for p, we need to nd a function g (t)
where Ep [g (T (X )] = 0 for all p (in this case, for both p). We h
Homework #6 solutions
Stat 530, Spring 2010
1. Show that each of the following families of distributions is an exponential family:
(a) Without loss of generality, let = 1. The density is usually writt
Homework #5 solutions
Stat 530, Spring 2010
1. (a) Let X be uniform on (a, b). For a = 4 and b = 5, the true mean and standard
deviation of Y = 1/X are .2231 and .01439, while the estimated values are
Homework #4: solutions
Stat 530, Spring 2010
1. (a) This is easy if you write it down the right way. Write the probabilities in a big
grid as follows:
P (Z n) = P (Z 1)
n=1
P (Z 2)
P (Z 3)
P (Z 4)
P (
Homework #3: solutions
Stat 530, Spring 2010
1. The order statistic X(j ) has cdf
n
P (X(j ) x) =
k=j
n
n
k
nk
k
[FX (x)] [1 FX (x)]
=
k=j
n
k
xk [1 x]nk
for the uniform density, if x (0, 1). The medi
Homework #2: solutions
Spring 2010
1. #5.6 For fun, lets use our three dierent methods. (Each can be done in any
of the three ways we talked about in class.) For (a) Z = X Y , lets use the
conditionin
Homework #13 solutions
Stat 530, Spring 2010
1. #8.55 For part (b), the risk associated with a Type II Error is larger than for part (a).
Note that the risk for > 0 is the same for both parts, but the
Homework #14 solutions
Stat 530, Spring 2010
1. #10.1 Lets try the Method of Moments estimator, as its the easiest to nd:
E (X ) =
1
2
1
x(1 + x)dx = ,
3
1
so = 3X . Also we can show that E (X 2 ) = 1
ST 530 April 11, 2004
Name: Midterm #2
All problem parts have equal weight. In budgeting your time expect that some problems will take longer than others. You only need to answer 10 out of the 11 p
ST 530 April 11, 2004
Name: Midterm #2
All problem parts have equal weight. In budgeting your time expect that some problems will take longer than others. You only need to answer 10 out of the 11 p
ST 530 February 28, 2005
Name: Midterm #1
All problem parts have equal weight. In budgeting your time expect that some problems will take longer than others.
1. Let X1 , . . . , Xn be an i.i.d. sa
ST 530 May 12, 2005
Name: Final Exam
All problem parts have equal weight. In budgeting your time expect that some problems will take longer than others. Remember, answers without proper justificati
Homework #1 solutions
Stat 530 Spring 2010
1. (a) The random variable Z can take values between 1/b and 1/a. Using the CDF
method, we obtain
FZ (z ) = P (Z z ) = P (U > 1/z )
b 1/z
=
ba
for z (1/b, 1/