APPM/MATH 4/5520
Solutions to Problem Set Four
1. Let us first find the distribution of Yi = ln F (Xi ). For notational simplicity, Ill drop the
subscripts.
FY (y) = P (Y y) = P (F (X) y)
= P (F 1 (F (X) F 1 (y) = P (X F 1 (y)
= F (F 1 (y) = y.
Note that,
APPM 4/5520
Problem Set Seven (Due Wednesday, October 19th)
1. (a) On HW 5, Problem 1, we saw that for a random sample X1 , X2 , . . . , Xn from a distribution with continous and invertible cdf F ,
2
n
X
ln F (Xi ) 2 (2n).
i=1
Recall that the proof was ba
APPM/MATH 4/5520
Problem Set Two (Due Wednesday, September 7th)
1. Suppose that U is a continuous random variable that is uniformly distributed on the interval
(1, 1). That is, U unif (1, 1).
Let > 0 and let
Y =
2
1U
1.
Find the distribution of Y . (Name
APPM/MATH 4/5520
Exam I Review Problems
Exam I: Thursday, September 9th from 6:30 to 9pm in a room TBA.
Optional Extra Review Session: Wednesday, September 28th from 6 to 8 pm in a room TBA.
(Note: I reserved these rooms a long time ago but just discovere
APPM/MATH 4/5520
Solutions to Problem Set One
c
1. (a) F (x) = 1 e(x)
on [0, ) and is 0 otherwise So,
c
f (x) = F 0 (x) = 0 [c(x)c1 e(x) ]
c)
= cc xc1 e(x)
and f is zero otherwise.
All together,
c
f (x) = cc xc1 e(x) I(0,) (x).
This is the pdf for the
Wei
APPM 4/5520
Problem Set Ten (Due Wednesday, November 9th)
1. Let X1 , X2 , . . . , Xn be a random sample from the exp(rate = ) distribution. Verify that
P
S = ni=1 Xi is a sufficient statistic from the definition of sufficiency. (ie: do not use the
Factor
APPM/MATH 4/5520
Exam II Review Problems
The exam will be on Thursday, November 10th from 6:30 to 9:00pm in FLMG 155.
Optional Extra Review Session: will be on Wednesday, November 9th from 6 to 8 pm in MUEN
E113.
1. Let X1 , X2 be a random sample from the
APPM/MATH 4/5520
Problem Set Four (Due Wednesday, September 23rd)
1. Suppose that X1 , X2 , . . . , Xn is a random sample from a distribution with pdf f and cdf F .
Show that
n
ln F (Xi ) 2 (2n).
2
i=1
2. Let X1 , X2 , . . . , Xn be a random sample from t
APPM/MATH 4/5520
Problem Set Five (Due Wednesday, September 30th)
1. Suppose that X1 , X2 , . . . , Xn is a random sample from the (, ) distribution where is
xed and known. Find an unbiased estimator of .
2. Let X1 , X2 , . . . , Xn be a random sample fro
APPM/MATH 4/5520, Fall 2015
Problem Set Three (Due Wednesday, September 16th)
1. Derive the moment generating function for the exponential distribution with rate . Be sure
to include an explanation as to why we need t < .
2. Let X geom0 (p).
(a) Compute t
APPM/MATH 4/5520
Problem Set Nine (Due Wednesday, November 4th)
1. Let X1 , X2 , . . . , Xn be a random sample from the (, ) distribution. Suppose that is
xed and known.
(a) Find the MME of .
(b) Find the MLE of .
(c) Which estimator (MME or MLE) has smal
APPM/MATH 4/5520, Fall 2015
Problem Set One (Due Wednesday, September 2nd)
Welcome to Problem Set One! This is the only assignment where you will encounter some problems
and topics that we have not covered in class. The point is to force you to catch up w
APPM/MATH 4/5520, Fall 2015
Problem Set Two (Due Wednesday, September 9th)
1. Compute the mean of the (, ) distribution by integrating without integrating.
2. Suppose that X is uniformly distributed on the interval (0, 1). (We write X unif (0, 1).)
(a) De
APPM/MATH 4/5520
Problem Set Five (Due Wednesday, October 14th)
1. Recall our denition of a tdistribution: Let Z N (0, 1) and W 2 (n) be independent
random variables. Since the dsitribution of W depends on n, I will write W as Wn . Dene
T = Tn =
Z
.
Wn /
APPM/MATH 4/5520
Problem Set Seven (Due Wednesday, October 14th)
1. Suppose that a random sample of size 12, taken from a N (, 3) distribution, results in a
sample mean of 5.7. Give an 88% condence interval for the true mean .
2. Suppose that a random sam
APPM/MATH 4/5520
Problem Set Eight (Due Wednesday, October 28th)
iid
1. Let X1 , X2 , . . . , Xn N (, 2 ). Find a 100(1 )% condence interval for 2 based on the
sample variance and a 2 critical value(s).
2. Let X1 , X2 , . . . , Xn be a random sample from
APPM/MATH 4/5520
Problem Set Four (Due Wednesday, September 21st)
1. Let X1 , X2 , . . . , Xn be a random sample from the P areto() distribution. (By default, the
Pareto distribution refers to the continuous Pareto distribution.)
(a) Find the distribution
APPM 4/5520
Problem Set Eight (Due Wednesday, October 26th)
1. Let X1 , X2 , . . . , Xn be a random sample from the distribution with pdf
f (x; ) =
(2) 1
x (1 x)1 I(0,1) (x).
[()]2
Find the method of moments estimator (MME) of .
2. Let X1 , X2 , . . . , X
APPM/MATH 4/5520
Solutions to Problem Set Three
1.
M (t) = E[etX ] =
R
e
tx f (x) dx
=
R tx
x dx
0 e e
=
R
0
e(t)x dx
Note that we need t < or else this integral will not converge. In the case that t < , we can
integrate without integrating by making thi
APPM/MATH 4/5520
Solutions to Problem Set Two
1.
E[X] =
R
=
R
=
R
x
0
0
x
f (x) dx
1
1 ex dx
() x
x
x
 ecfw_z
1
()
dx
like the
pdf of
( + 1, )
=
(+1) 1
()
R
0
1
+1 x1 ex dx
( + 1)
cfw_z

this is the pdf
of the ( + 1, )
and so it integrates
to 1
APPM/MATH 4/5520
Solutions to Problem Set One
1. (a) We need the integral of the pdf from to to be 1. This is how we will find c.
R
1 =
f (x) dx
R0
0 dx
=
= c
R1
0
+c
R1
ex dx + c
0
ex dx + c
R
1
R
1
ex+2 dx
ex+2 dx
= c(e 1) + c e = c(2e 1)
Thus, we hav
APPM 4/5520
Solutions to Problem Set Seven
1. Small sample, normality, have 2 the confidence interval is
X z/2 .
n
Here, x = 5.7, = 3, n = 12, = 0.12 and z/2 = z0.06 = 1.555 (1.55 or 1.56 are okay
too). So, the confidence interval is
3
5.7 1.555
12
which
APPM 4/5520
Solutions to Problem Set Six
1. Note that
E
Wn
1
1
= E[Wn ] = n = 1
n
n
n
1
1
2
Wn
= 2 V ar[Wn ] = 2 2n = 0 as n
V ar
n
n
n
n
P
So, by That Theorem, we know that Wn /n 1.
Since continuous functions preserve convergence in probability, we also
APPM/MATH 4/5520
Solutions to Problem Set Nine
b = 1/X.
1. First, you find the MLE for , as we have done before. It is
Note that
() = P (X > 1) = e .
Using the invariance property of MLEs, the MLE for () is
b = e1/X
d
() = ()
.
2. Yes, its weird that th
APPM/MATH 4/5520
Solutions to Problem Set Five
1. We know that
E[X] = E[X1 ] =
.
So, (1/)X is an unbiased estimator of 1/. (Note that is known so its okay that we use
it in our estimator!)
We need to get the in the numerator. Lets try 1/X.
"
1
E
=E
X
wher
APPM/MATH 4/5520
Solutions to Problem Set Nine
1. We need to find the conditional pdf fXY (xy). for this, we need tha marginal pdf of Y .
fY (y) =
R
f (x, y) dx
= 8y I(0,1) (y)
= 8y I(0,1) (y)
So, the conditional pdf for X given Y is
fXY (xy) =
=
R
R
APPM/Math 4/5520
Solutions to Final Exam Review Problems, 119
1.
E[Icfw_X1 >3 ] = 0 P (Icfw_X1 >3 = 0) + 1 P (Icfw_X1 >3 = 1)
= P (Icfw_X1 >3 = 1) = P (X1 > 3) = e3
2. (a) S is sufficient for if, given S, the joint distribution of X1 , X2 , . . . , Xn no
APPM/MATH 4/5520
Final Exam Review Problems
The final exam is on Sunday, December 13th in our normal classroom from 4:30am to 7:00pm.
It is not cumulative.
The exam will have 6 problems and you must choose and complete 5 out of 6 problems. There
is no g
APPM/MATH 4/5520
Exam I Review Problems
Exam I: Thursday, October 1st from 6:30 to 9pm in MATH 100.
Optional Extra Review Session: Wednesday, September 30th from 6 to 8 pm in ECCR 108.
The actual exam will have 6 problems and you will only have to do 5 of