1) Lecture 1: General
A. Positive question Why is some feature the way it is>
B. Normative question What should some feature be?
C. Quantities we often use:
C.i. Households: Consumption, Savings, Hours worked
C.ii. Firm: Output, Vacancies
C.iii. Governmen
214 Exam 1
1) Lecture 1: Introduction
A. Quality attributes describe the codes fitness for further development/use:
reuse code, easy to change, easy to add
B. Objects support simulation, extensibility, and modifiability
C. Method calls are dispatched to m
21257 Test 1 Notes
1.
Chapter 1.4 Solving LP Graphically
a. Make constraints equalities, and graph the corresponding lines
b. Shade in graph to match inequality constraints
c. Find the corners
d. Evaluate the corners for the solution. Solution must be a
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Lab 1 TV Jammer
* Please read and understand the Lab Readiness Prerequisites and
Lab Etiquette and Procedures documents before starting this lab *
1. Introduction
In this lab you will build the core pieces of a TV jammer. By jammer, we mean a nearunivers
Lecture III24 January 2012
36217
Liz Kulka
Readings: 1.1, 1.2, 1.6
1
Review
Given a sample space , a probability law or distribution P is a function dened over subsets of
such that:
1. P(A) 0 A
2. P() = 1
3. If A1 , A2 , . . . is a sequence of disjoint
Lecture IV26 January 2012
36217
Liz Kulka
Readings: 1.6
1
Counting Methods
Discrete Uniform Distribution: If is nite and all points in are equally likely, then
Pr(A) =
A
A

This is a type of Random Sample.
Four counting methods:
1.1
Permutations
Cons
Lecture V31 January 2012
36217
Liz Kulka
Readings: 1.3  1.5
1
Conditional Probability
Recall the scenario where we took some medicine that had possible side eects.
A = cfw_I will experience side eects
B = cfw_I am a smoker
Pr(A) = 0.1 Unconditional Prob
Lecture VI2 February 2012
36217
Liz Kulka
Readings:
1
Recall
Pr(AB) =
Pr(A B)
Pr(B)
(Will be on the exam!)
P a probability on and A1 , A2 , . . . , An a partition of with P (Ai ) > 0) i.
Law of Total Probability:
n
P (B) =
P (BAi )P (Ai )
i=1
Bayes The
Lecture VII7 February 2012
36217
Liz Kulka
Readings:
Midterm 1 in 10 days, no notes and no calculators permitted.
There will be a handout on BB with equations that we do not need to memorize.
The format will be identical to the practice exam and the dicu
Lecture VIII9 February 2012
36217
Liz Kulka
Readings:
1
Random Variables, Continued
1.1
Review
Random Variables X : R
Range(X) is the set of all possible values of X
pmf (probability mass function) of X, pX : pX (x) = Pr(cfw_X = x), x RAN (X)
Bernoul
Lecture IX14 February 2012
36217
Liz Kulka
Readings: Section 2.4
Material through this lecture will be on Exam 1.
1
Discrete Random Variables
X Bernoulli(p)
X Binomial(n, p)
0<p<1
X Geometric(p)
X N egBin(r, p)
X HyperGeom(R, B, n) or (N, R, n) N = B + R
Lecture XII28 February 2012
36217
Liz Kulka
Readings:
1
Conditional Expectation
Given two rvs X and Y , the conditional expectation of X given Y is the rv E[XY ] obtained as a
function f (Y ) of Y such that if Y = y, f (y) = E[XY = y] (where E[XY = y]
Lecture XIV6 March 2012
36217
Liz Kulka
Readings:
1
Continuous Random Variables (Lots of Integrals!)
Note: reviewing integrals would be a good idea.
A random variable X is continuous if its range is an uncountable set.
1.1
Uniform Random Variables
X U ni
Lecture XIII1 March 2012
36217
Liz Kulka
Readings:
1
Independent Random Variables
X1 , . . . , Xn are independent when
n
pXi (xi ) xj , j [1, n]
pX1 ,.,Xn (x1 , . . . , xn ) =
i=1
Implies:
n
n
V
Xi =
V [Xi ]
i=1
i=1
The variance of a sum of random variab
Lecture XV8 March 2012
36217
Liz Kulka
Readings:
1
Continuous Random Variable
X is a continuous RV when Pr(X A) =
A fX (x)dx
A R. fX is the pdf of X
1. fX (x) 0 x
2.
+
fX (x)dx
cdf of X: FX (c) =
=1
c
fX (x)dx
= Pr(X c)
+
The expected value of X: E[X]
Lecture XVI20 March 2012
36217
Liz Kulka
Readings:
Midterm II: March 29 (next Thursday). Will cover everything since Midterm I.
1
Continuous Random Variable
A random variable X is continuous if there exists a function fX (the pdf of X) such that:
fX (x)d
Lecture XVII22 March 2012
1
36217
Liz Kulka
Review of Integration
Let a b +:
b
xc dx =
a
b
1
xc+1
c+1
c
f (x)dx =
a
b
(1)
a
b
f (x)dx +
a
f (x)dx if a < c < b
(2)
c
b
b
cf (x) + dg(x)dx =
a
b
cf (x)dx +
a
dg(x)dx
(3)
a
Integration by Parts:
b
b
f (x)g (x
Lecture XVIII27 March 2012
1
36217
Liz Kulka
Joint, Marginal and Conditional pdf s
X, Y are uniformly distributed over set A R2
4
A
3
2
L
1
1
2
3
4
1
4
fX,Y (x, y) =
0
(x, y) A
o.w.
Pr(X, Y ) S) =
fX,Y (x, y)dxdy
(x,y):(x,y)S
Pr(X = Y ) =
fX,Y (x, y)dxd
Lecture XIX3 April 2012
36217
Liz Kulka
Readings:
Score
[90, 130)
[80, 90)
Midterm 2:
[70, 80)
[60, 70)
[0, 60)
1
Percentage of class
50%
16%
12%
11%
11%
Conditioning
X, Y continuous r.v.s the conditional pdf of X given Y = y is:
fXY (y) : R R
such tha
Lecture XXI10 April 2012
1
36217
Liz Kulka
Markov Chains
A stochastic process in discrete time is a sequence X0 , X1 , . . . of random variables.
1.1
Trivial Cases
X0 , X1 , X2 are iid from a common distribution if X0 Bernoulli(p) (a Bernoulli process).
Lecture XX5 April 2012
1
36217
Liz Kulka
Covariance
Let X and Y be two rvs. The covariance of X and Y is
Cov [X, Y ] = E [(X E [X])(Y E [Y ])]
If Cov [X, Y ] > 0 X and Y tend to move together:
y
x
If Cov [X, Y ] < 0 X and Y tend to move together:
y
x
E
Lecture XXII12 April 2012
1
36217
Liz Kulka
Discrete Time, Finite State Space Order One Homogeneous
Markov Chains
Sequence of rvs: X0 , X1 , X2 , . . .
Pr(Xn = in Xn1 = in1 , . . . , X0 = i0 ) = Pr(Xn = in Xn1 = in=1 ) = pi,j
independent of n. Also ass