1
HW1 Solution
Prob 1
)i+j /(1
a) P (W > i + j |W > i) = P (W > i + j )/P (W > i) = (1
P (W > j )
)i = (1
)j =
b) Considering the memoryless property, we have for any integer n > 1
P (W > n) = P (W > 1) P (W > n
1).
Then inductively, we have P (W > n) =
Lecture 5 Stat 426
Moments and Expectations
Sec 4.1-4.3, 4.5
1. Review
() = () ()
1.1 If X is a discrete RV with pmf f and range R , and Y=g(X), then
provided the sum converges absolutely.
E(Y) = ()()
1.2 If X has density f, and Y=g(X), then
provided that
Stat 426 Winter 2016, Homework 2
Selected practice problems (no solutions provided, not for submission or grading):
Chap 2, Q4, 7, 8, 11, 31; Chap 4, Q 2, 3; Review Lecture 2
Turn in the solutions to the following questions:
Problem 1. Chap 2. Q 14
Two bo
Some Inequalities and the Weak Law of Large Numbers
Moulinath Banerjee
University of Michigan
August 30, 2012
We rst introduce some very useful probability inequalities.
Markovs inequality: Let X be a non-negative random variable and let g be a
increasing
2009 Fall Stat 426 : Homework 1
Moulinath Banerjee University of Michigan Announcement: The homework carries 50 points and contributes 5 points to the total grade. Your score on the homework is scaled down to out of 5 and recorded. 1. A geometric random v
Stat 426 W2014
Homework 1 Solution
Problem 1
A lie detector test has a 90% detection rate if a person is lying, and a 5% detection rate if a
person is not lying. Suppose that 1% of all individuals in a certain population tell lies. Two lie
detector tests
Stat 426 : Homework 3.
Moulinath Banerjee
October 28, 2012
Announcement: The homework carries a total of 41 points. The maximum possible score is
40 points.
1. (a) Consider the standard estimator of 2 based on X1 , X2 , . . . , Xn i.i.d. N (, 2 ). This i
Fall 2014, Stat 426 : Homework 1
Moulinath Banerjee
University of Michigan
Announcement: The homework carries 80 points and is due on Sept
29th in class. It contains two supplementary problems that you dont
need to turn in.
1. A geometric random variable
1
HW2 Solution
Prob 1
(i) If we denote the number of heads based on the later N tosses as N1 , the total number of
heads will be N + N1 . And,
1
1
3
E (N + N1 ) = E (N ) + E (N1 ) = n + E (E (N1 |N ) = n + E (0.5N ) = n
2
2
4
And by Theorem B on Pg151 of
Midterm 1, Winter 06
Moulinath Banerjee
University of Michigan February 22, 2006
Announcement: The exam carries 30 points but the maximum you can score is 25. (1) If X and Y are two uncorrelated random variables, are they necessarily independent? On the o
Stat 426
Solutions for Homework 3
1. Let Zi = Xi Yi . Thus, E[Z] = 1
central limit theorem,
(X
2 and V ar(Z) =
Y ) (1 2 )
Z (1
p
=p 2
2
2
( 1 + 2 )/n
( 1+
2 )
2
2 )/n
2
2
1+ 2
n
. By the
d
! N (0, 1)
2. Since X1 , . . . , X5 0 N (6, 0.2) i.i.d., 4(Xi )2 =
Stat 426 Winter 2016
Homework 8 Solution
Chapter 8, Q 5, 19, 23, 31, 37 Review lectures 9 and 10
Practice problems: Chapter 8, Questions, 1, 3, 19, 23, 31 (Not for submission)
Problem 1. The Beta distribution is useful for modeling random variables that a
1
HW1 Solution
Prob 1
a) P (W > i + j|W > i) = P (W > i + j, W > i)/P (W > j) = P (W > i + j)/P (W > i) =
(1 )i+j /(1 )i = (1 )j = P (W > j).In this derivation, we have used the fact that
P (W > k) = (1)k for any integer k 0, from the properties of the ge
Stat 426 Lecture 1:
Review of Probability
1
Chapter 1
Introduction
A sound background in probability is essential to understanding the theory
of statistics
2
The Essentials
is the sample space, the set of all possible outcomes of a random experiment. The
Homework 1 (Due 5-13)
Instructions: Turn in at the beginning of class. Late homework not accepted.
1) Evaluate
x4 e2x dx
0
No need to show any work. Relate to the gamma distribution.
2) Suppose the random variable X has the density given by
fX (x) =
a) Fi
Stat 426 Winter 2016
Homework 7 Solution
Problem 1 Question 1 Chapter 7
Consider a population consisting of five values, which are 1,2,2,4,8. Find the population mean
and variance. Calculate the sampling distribution of the mean of a sample of size 2 by g
Stat 426 Fall 2014
Homework 2
Let () =
, 1 1 , and () = 0 otherwise, where 1 1. Show that
Problem 1. Chap 2, Q34
1+
2
() is a density and find the corresponding cdf. Find the quartiles and median of the
distribution in terms of .
() 0 and () =
1
1
F(x)=
Homework 6
Practice problems: Chapter 8, Q 5, 19, 23, 31, 37
1. The Beta distribution is useful for modeling random variables that are restricted to the interval [0,1].
The Beta distribution with parameters (a,b), and its expected value and variance are g
374
Chapter 11
Table 11.17
Incomplete Block Designs
Partial output from PROC GLM for analysis of an incomplete block designdetergent
experiment
The SAS System
General Linear Models Procedure
Dependent Variable: Y
Sum of
Mean
Source
DF
Squares
Square
Model
11.10
Table 11.18
375
Using SAS Software
Partial output from ESTIMATE and LSMEANS for an incomplete block
designdetergent experiment with detergent 9 as the control treatment
The SAS System
General Linear Models Procedure
Parameter
Det 9-1
Det 9-2
Estimat
372
Chapter 11
Incomplete Block Designs
Using these formulae, we nd that the least squares estimates of the linear and quadratic
trend contrasts for step frequency (adjusted for subjects) are
DL
and DQ
17.125
7.125 .
The linear trend is positive, suggesti
11.9
Example 11.8.2
369
Factorial Experiments
Sample size to achieve specied power
Suppose a test of the null hypothesis H0 : cfw_i all equal is required to detect a difference in
the treatment effects of
1 unit with probability 0.95, using signicance lev
11.9
371
Factorial Experiments
step frequency is increased, but that the linear trend is not the same for the two step heights.
The experimenters wanted to examine the average behavior of the two factors, so despite
this interaction, they decided to exami
11.10
Table 11.20
Using SAS Software
377
SAS program to plot data adjusted for block effectsplasma experiment, day one
only.
* This program requires 3 runs, adding more information in each run;
DATA ONE;
INPUT BLOCK TRTMT Y;
LINES;
1 4 0.459
1 5 0.467
: :
Stat 426 (003)
Solutions for Homework 4
1. The length of the confidence interval is
q(1 2 , N (0, 1) q(1 , N (0, 1) = 1 (1
n
n
Let f (1 ) = 1 (1
+ 1 )
standard normal distribution.
1 (1 ) and be the density of the
1
d
f (1 ) =
1
d1
( (1
Solving
d
f (1 )
Stat 426 (003)
Solutions for Homework 5
1. (a) E[X1 ] = /2, so bM OM = 2X n . E[bM OM ] = and V ar(bM OM ) =
4V ar(X1 )/n = 2 /(3n).
Q
(b) L() = 1n ni=1 1(0 Xi ) = 1n 1(maxi Xi ). L() achieves
the maximum when = maxi Xi and so bM LE = maxi Xi .
n
(c) P (m
Stat 426 : Homework 4.
Moulinath Banerjee
November 8, 2015
Announcement: The homework carries a total of 60 points (10 6).
(1) A level 1 C.I. for the normal mean , when is assumed known, is given by:
X n q(1 2 , N (0, 1) , X n q(1 , N (0, 1) ,
n
n
where
Stat 426 : Homework 5.
Moulinath Banerjee
November 20, 2014
Announcement: The homework carries a total of 80 points (20 4).
(1) Problem 53, Chapter 8.
(2) (i) Let y1 , y2 , . . . , yn be n given numbers. Let gn (a) := ni=1 |yi a|. Show that (i) if
n is