Statistics 24400
Homework 1 Solutions
Date: October 2014
Problem 1 : Exercise 16
Let Sn be the number of successes in the first n trials of a Bernoulli process.
(a) Note that
P (S5 = 3|S2 = 1) = P (S5 S2 = 2|S2 = 1
Since the number of successes in trials
Stat 24400 Homework 3 Solution
Lian Huan Ng
October 27, 2010
Total points: 130
Problem 2.5.2
The sample space
= cfw_HHHH, HHHT, HHT H, HHT T, HT HH, HT HT, HT T H, HT T T,
T HHH, T HHT, T HT H, T HT T, T T HH, T T HT, T T T H, T T T T
has 16 elements, e
Practice Midterm Exam, Statistics 244
Name:
1. Answer all questions. Points for each question are indicated in brackets.
2. This is a closed book examination. You should have a calculator. You may have a page of notes.
3. Please provide the answers in the
Stat 24400 Homework 5 Solution
Lian Huan Ng
November 5, 2010
Total points: 125 + 20
Problem 3.8.42
(a) [10 pts] The cdf of X is
FX (x) = P (X x|W = 1)P (W = 1) + P (X x|W = 1)P (W = 1)
1
1
= P (T x) + P (T x)
2
2
11
= + (P (T x) P (T < x)
22
so the densit
Stat 24400 Midterm Solution
Lian Huan Ng
November 11, 2010
Question 1
(a) The transformation is y = log x so x = ey and
dx
dy
= ey . Using the change of variable formula,
fY (y ) = fX (ey )
dx
= 2x2 e2y
0
dy
Since the formula for fX is valid only for x >
Stat244 HW2 Solutions
Oct 17 2013
Ang Li
1. 1.8.54 [15]
(a)
c
c
P (Ri+1 ) = P (Ri+1 |Ri )P (Ri ) + P (Ri+1 |Ri )P (Ri )
= p + (1 )(1 p)
= p( + 1) + (1 ).
(b)
c
c
P (Ri+2 ) = P (Ri+2 |Ri+1 )P (Ri+1 ) + P (Ri+2 |Ri+1 )P (Ri+1 )
= P (Ri+1 )( + 1) + (1 )
= p(
Stat 24400 Homework 1 Solution
Ang Li
Oct 8, 2013
1. Problem 1.8.6
a [5 pts] Observe that P (A)+P (B)+P (C) counts each of P (ABC), P (BCA), P (CAB)
twice, and counts P (ABC) three times. The negative term P (AB)P (AC)P (BC)
resolves the double countings
Stat 24400 Homework 3 Solution
Ang Li
Oct 23, 2013
1. Problem 2.5.2
The sample space
=cfw_HHHH, HHHT, HHT H, HHT T, HT HH, HT HT, HT T H, HT T T,
T HHH, T HHT, T HT H, T HT T, T T HH, T T HT, T T T H, T T T T
has 16 elements, each of which has a equal p
Stat 244 HW4 solution
Miaoyan Wang
Nov 4th, 2012
10 pts Problem 2.5.67
(a) To get the density f from the cumulative distribution function F, we
dierentiate F:
d
x 1
x
fX (x) =
F (x) =
exp
, x>0
dx
(b) For x > 0,
P (X x) = P
W
x
1
= P (W x )
1
= F (x )
STAT 244 HW5
Ang Li
Nov. 11, 2013
5 pts Problem 4.7.8
P (X k) =
k=1
P (X = i)
k=1 i=k
i
=
P (X = i) =
i=1 k=1
iP (X = i)
i=1
= E(X)
For geometic random variable X, P (X = i) = (1 p)i1 p, so
(1 p)i1 p = (1 p)k1
P (X k) =
i=k
(1 p)k1 =
P (X k) =
EX =
k=1
k=
Stat 24400 2016 Homework 7 Solution
Total points: 150
1. [20 pts]
(a) The probability mass function of geometric distribution is
f (x|p) = (1 p)x p
for x = 0, 1, . Note that by taking the derivative of log-likelihood function and
equating it to zero, we o
STAT 24400 Statistics Theory and Methods I
Homework 5: Due 3:00PM Thurs, November 10, 2016.
1. Rice, 8.16, parts (b) and (c).
2. Rice, 8.18, parts (b) and (c).
3. Suppose that X is the number of successes in a Binomial experiment with n trials and
probabi
STAT 24400 Statistics Theory and Methods I
Homework 4: Due 3:00PM Thurs, November 3, 2015. 120 Kent
1. Let Z N (0, 1), a standard normal distribution and let X N (, 2 ). Let (z) be
the cdf of Z. Then probabilities P (a < X < b) = P (X < b) P (X < a) can b
STAT 244 (Statistical Theory/Method I), Fall 2016
Homework 6 solutions
November 13, 2016
100 points total. Marks are assigned based on progress made.
Problem 1 (15 points)
(b) The log likelihood is
`() = log
n
Y
!
( +
1)xi
= n log( + 1) +
of . Since
d`
d
STAT 244, HW1 Key
October 7, 2016
Total Score: 85 pts.
1
Problem 1: Rice, Ch1 Prob 20 - 10 pts
52 cards can be arranged/shuffled among themselves in 52! ways. There are 48 non aces and 4 ace
cards. Treat 4 ace cards as 1 card. We then have 49 cards which
Stat 24400 Homework 6 Solution
Miaoyan Wang
Dec 5, 2012
Total Points:130
Problem 8.10.7
(b) [10 pt] The log likelihood is
n
p(1 p)xi 1 = n log p + n(n 1) log(1 p)
x
l(p) = log
i=1
n
1
where xn = n i=1 xi is the sample mean. To maximize it, we need to cons
Expectation of geometric distribution
What is the probability that X is nite? fX (k) = (1 p)k1p k=1 k=1 = p (1 p)j j=0 1 = p 1(1p) =1 Can now compute E(X): E(X) = k (1 p)k1p k=1 = p (1 p)k1 + (1 p)k1 + k=1 k=2 (1 p)k1 + k=3 = p[(1/p) + (1 p)/p + (1 p)2/p
The geometric distribution
So far, we have seen only examples of random variables that have a nite
number of possible values. However, our rules of probability allow us to
also study random variables that have a countable [but possibly innite]
number of p
Monte Carlo Simulation: IEOR E4703 c 2004 by Martin Haugh
Fall 2004
Generating Random Variables and Stochastic Processes
1 Generating U(0,1) Random Variables
The ability to generate U (0, 1) random variables is of fundamental importance since they are usu
Lecture 13
Agenda
1. Geometric Random Variable
2. Negative Binomial Distribution
Geometric Random Variable
Consider an experiment which consists of repeating independent Bernoulli
trials until a success is obtained. Assume that probability of success in e
Math/Stat 394
F.W. Scholz
Poisson-Binomial Approximation
Theorem 1: Let X1 and X2 be independent Poisson random variables with respective parameters
1 > 0 and 2 > 0. Then S = X1 + X2 is a Poisson random variable with parameter 1 + 2 .
Proof:
P (X1 + X2 =
123
CHAPTER 4 Probability
SECTION 5 Principles of Counting
The key to solving many problems in mathematics and specifically probability lies in
counting the number of ways an event can occur or, equivalently, counting the number of
possible ways there are
Chapter 3: DISCRETE
RANDOM VARIABLES AND
PROBABILITY DISTRIBUTIONS
Part 4: Geometric Distribution
Negative Binomial Distribution
Hypergeometric Distribution
Sections 3-7, 3-8
The remaining discrete random variables well
discuss:
Geometric
Negative Binom
Chapter 8 The Normal Distribution
8 THE NORMAL
DISTRIBUTION
Objectives
After studying this chapter you should
appreciate the wide variety of circumstances in which the normal
distribution can be used;
be able to use tables of the normal distribution to so
Chapter 7-The Normal Probability Distribution
7.1 Properties of the Normal Distribution
Definition: Probability Density Function A probability density function is an equation used to compute probabilities of continuous random variables that must satisfy t
Lecture 5: Continuous Probability Distributions
(Text Sections 2.32.4)
Normal Distribution
A random variable X has a Normal(, 2 ) distribution if it has pdf
f (x) =
1
2
2
e(x) /2 ,
2 2
< x, < , > 0.
The normal cdf does not have a closed form.
Example: M
STT 430/630/ES 760 Lecture Notes:
Chapter 4: Continuous Distributions
1
February 23, 2009
Chapter 4: Continuous Distributions
In Chapter 3, the notion of a continuous sample space was introduced. Recall that continuous probability
models are useful for mo
Stat 244 Solution to HW 7
Miaoyan Wang
Dec. 2, 2011
10 pts Problem 9.11.2
a Simple hypothesis
b Simple hypothesis
c Composite hypothesis
d Composite hypothesis
Grading Scheme. 10 marks, 2.5 each.
20 pts Problem 9.11.3
(a)
= P (|X 50| > 10; p = .5)
X 100p