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 >
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
PEP 604
Summer, 2010
Inclusion and Exclusion Criteria
Inclusion criteria = attributes of subjects that are essential for their
selection to participate.
Inclusion criteria function remove the influence of specific
confounding variables.
eg., fitness, mens
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 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
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 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
Basic Counting Principles
Multiplication Principle
Consider a multistep process in which
Step 1 has n1 possible outcomes,
Step 2 has n2 possible outcomes,
.
Step r has nr possible outcomes.
Then, the entire process has n1 n2 nr possible outcomes.
Inclusio
Page 1
Chapter 8
Poisson approximations
The Bin(n, p) can be thought of as the distribution of a sum of independent indicator
random variables X 1 + . . . + X n , with cfw_X i = 1 denoting a head on the ith toss of a coin.
The normal approximation to the
Counting Methods
Counting principle, permutations, combinations, probabilities
Part 1: The Fundamental Counting Principle
The Fundamental Counting Principle is the idea that if we have a ways of doing something
and b ways of doing another thing, then ther
BINOMIAL vs. POISSON vs. NORMAL DISTRIBUTIONS
Rule of thumb:
Use Poisson to approximate Binomial when n is large and p is small.
Let = np.
Use Normal to approximate Binomial when both np >5 and nq > 5.
Let = np, 2 = npq
Limiting Form of the Binomial is th
Random variables
M. Veeraraghavan
A random variable is a rule that assigns a numerical value to each possible outcome of an experiment. Outcomes of an experiment form the sample space S .
Definition: A random variable X on a sample space S is a function X
Geometric distribution (from http:/www.math.wm.edu/leemis/chart/UDR/UDR.html)
The shorthand X geometric(p) is used to indicate that the random variable X has the geometric
distribution with real parameter p satisfying 0 < p < 1. A geometric random variabl
Inclusion-Exclusion Principle
In the set S = cfw_1, 2, . . . , 100, one out of every six numbers is a multiple of 6, so that the
total of multiples of 6 in S is [100/6] = [16.666 . . .] = 16 (here [x] is the integer part of x).
Similarly, the total number
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
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, HW3 Key
October 14, 2016
Total Score: 120 pts.
1
Problem 1: 10 pts
[10pts] Rice 4.78
Suppose f (x) is a probability density function and symmetric about zero, then it is an even
function (f (x) = f (x). Let k be an odd number then xk f (x) is a
STAT 244 (Statistical Theory/Method I), Fall 2016
Homework 2 solutions
October 9, 2016
120 points total. Marks are assigned based on progress made.
Problem 1 (3 points)
Since the events cfw_X = k and cfw_k T < k + 1 are same by the definition of X, we can
Stat 24400 Homework 5 Solution
November 10, 2016
Total points: 100
1. [20 pts]
(a)
!
n
1
1X
f (x1 , x2 , , xn |) = n n exp
|xi |
2
2 i=1
n
1X
|xi | n ln + constant
l() =
i=1
Pn
|xi |
n
l () =
+ i=12
P
2 ni=1 |xi |
1
00
l () = 2 n
0
0
Set l () = 0, gi
STAT 24400 Statistics Theory and Methods I
Homework 7: Due 10:00AM Monday, November 28, 2016.
Put it in the clearly marked box outside 312 Jones.
The box will be available starting Wednesday, November 23.
1. Suppose X follows a geometric distribution with
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