Chapter 5
Markov Chains
5.1 Denitions and Examples
The importance of Markov chains comes from two facts: (i) there are a large number of physical, biological, economic, and social phenomena that can be described in this way, and (ii) there is a well-devel
110
Chapter 4
Limit Theorems
4.1 Laws of Large Numbers
If X1 , X2 , . . . are independent and have the same distribution then we say the Xi are independent and identically distributed, or i.i.d. for short. Such sequences arise if we repeat some experiment
Chapter 1
Basic Concepts
1.1 Dice, Coins, and the World Series
The subject of probability can be traced back to the 17th century when it arose out of the study of gambling games. As we will see the range of applications extends beyond games into business
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Chapter 3
Discrete Distributions
3.1
Examples
A random variable is a numerical value determined by the outcome of an
experiment. We have seen a number of examples.
Roll two dice and let X = the sum of the two numbers that appear.
Roll a die until a 4
Measures of Association and
Hypothesis Testing
by Deborah Rosenberg, PhD and Arden Handler, DrPH
Once best estimates are chosen, both from a statistical and epidemiologic perspective, hypotheses about
the estimated association between a single mean, propo
Homework 2 Solutions
8.) (a) If Al is leading 4 games to 3, and the series ends when someone wins 5 games, the following
three outcomes are possible: Al wins the next game, Bobby wins the next game and Al wins
the subsequent game or Bobby wins both of the
56 (a) There are 522 ways to pick two cards from a deck. If one must be an A and the other
must be either a K, Q, J or 10, then we have four choices for the A, or 4 and sixteen choices
1
for the remaining card, or 116 . So the probability is 4 116 / 522 =
HW 1 Solutions
(2) Suppose we pick a number at random from the phone book and look at the last digit.
(a) The set of outcomes is = cfw_0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and each outcome should be
assigned probability 1/10. (b) No; In a particular phone book, i
17) This is a binomial probability with n = 8 and p = 1/6 (where success is dened as rolling a
three). Then the probability of exactly two 3s is just 8 (1/6)2 (5/6)6 = 0.2605.
2
27) This can easily be done by writing the 6 6 grid of outcomes, yielding 4/2
1.) Let Z = |X Y |. Then P (Z = 0) = 6/36, P (Z = 1) = 10/36, P (Z = 2) = 8/36, P (Z = 3) =
6/36, P (Z = 4) = 4/36 and P (Z = 5) = 2/36.
4.) This is a hypergeometric distribution: Hypergeo(5, 4, 3).
10.) This would be a binomial distribution with p = 1/50
1. For a 90% CIs we use z = 1.64 and for 95% CIs we use z = 1.96. The resulting
intervals are,
3.46 (1.64)(5)(23)1/2 (1.75, 5.17), 3.46 (1.96)(5)(23)1/2 (1.42, 5.50)
5.89(1.64)(7)(12)1/2 (9.20, 2.58), 5.89(1.96)(7)(12)1/2 (9.85, 1.93)
23.56(1.64)(12)(5