# Difference between revisions of "stat340s13"

## Introduction, Class 1 - Tuesday, May 7

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### Four Fundamental Problems

1. Classification: Given an input object X, we have a function which will take in this input X and identify which 'class (Y)' it belongs to (Discrete Case)
2. Regression: Same as classification but in the continuous case
3. Clustering: Use common features of objects in same class or group to form clusters.(in this case, y is unknown)
4. Dimensionality Reduction

### Applications

Most useful when structure of the task is not well understood but can be characterized by a dataset with strong statistical regularity
Examples:

• Computer Vision, Computer Graphics, Finance (fraud detection), Machine Learning
• Search and recommendation (eg. Google)
• Automatic speech recognition, speaker verification
• Text parsing
• Face identification
• Tracking objects in video
• Financial prediction, fraud detection
• Medical diagnosis

### Course Information

General Information

• No required textbook, recommended: "Simulation" by Sheldon M. Ross
• Computing parts of the course will be done in Matlab, but prior knowledge of Matlab is not essential (will have a tutorial on it)
• Announcements and assignments will be posted on Learn.
• Other course material on: http://wikicoursenote.com/wiki/
• Log on to both Learn and wikicoursenote frequently.
• Email all questions and concerns to UWStat340@gmail.com. Do not use your personal email address!

Primary contributor: Put a summary of the lecture up within 48 hours.

General contributor: Elaborate on concepts, add example, add code, add pictures, reference, corrections etc… withing 2 weeks within 1 week.

Must do both All primary contributions are now considered general contributions you must contribute to 50% of lectures for full marks

• A general contribution can be correctional (fixing mistakes) or technical (expanding content, adding examples, etc) but at least half of your contributions should be technical for full marks

Do not submit copyrighted work without permission, cite original sources. Each time you make a contribution, check mark the table. Marks are calculated on honour system, although there will be random verifications. If you are caught claiming to contribute but didn't, you will lose marks.

### Tentative Marking Scheme

Item Value
Assignments (~6) 30%
WikiCourseNote 10%
Midterm 20%
Final 40%

In reality, it is often complicated to identify the distribution.

## Sampling (Generating random numbers), Class 2 - Thursday, May 9

### Introduction

Some people believe that activities such as rolling a dice and flipping a coin are not truly random but are deterministic – the result can be reliably calculated using things such as physics and math.

A computer cannot generate truly random numbers since computers can only run algorithms, which are deterministic in nature. They can, however, generate Pseudo Random Numbers; numbers that seem random but are actually deterministic. Although the Pseudo Random Numbers are deterministic, these numbers have a sequence of value and all of them have the appearances of being independent uniform random variables.

### Multiplicative Congruential Algorithm

This is an algorithm used to generate uniform, pseudo random numbers. It is also referred to as the Linear or Mixed Congruential Methods. We define the Linear Congruential Method to be xk+1=(a*xk + b) mod m. Given a "seed" x0, we can obtain values for x1, x2, ..., xn recursively. The Multiplicative Congruential Method may also refer to the special case where b=0.

Algorithm 1
xk+1 = xk mod m

Example
Let x0 = 10, m = 3
Step 1: 1 = 10 mod 3
Step 2: 1 = 1 mod 3
Step 3: 1 = 1 mod 3
This method generates a sequence of identical integers, hence we need a better algorithm.

Algorithm 2 (Multiplicative Congruential Algorithm)
xk+1 = (a*xk+b) mod m

Example
Let a = 2, b = 1, m = 3, x0 = 10
Step 1: 0 = (2*10+1) mod 3
Step 2: 1 = (2*0+1) mod 3
Step 3: 0 = (2*1+1) mod 3
This method generates a sequence with a repeating cycle of two integers.

MatLab for Multiplicative Congruential Algorithm:
Before you start:

>>clear all
>>close all

>>a=17
>>b=3
>>m=31
>>x=5
>>mod(a*x+b,m)
ans=26
>>x=mod(a*x+b,m)


(Note: Keep repeating this command over and over again and you will seem to get random numbers – this is how the command rand works in a computer.)

>>a=13
>>b=0
>>m=31
>>x(1)=1
>>for ii=2:1000
x(ii)=mod(a*x(ii-1)+b,m);
end
>>size(x)
ans=1    1000
>>hist(x)


(Note: The semicolon after the x(ii)=mod(a*x(ii-1)+b,m) ensures that Matlab will not show the entire vector of x. It will instead calculate it internally and you will be able to work with it. Adding the semicolon to the end of this line reduces the run time significantly.)

This algorithm involves three integer parameters a, b, and m and an initial value, x0 called seed. A sequence of numbers is defined by x(k+1) = a*x(k) + b mod m. Mod m means take the remainder after division by m.

Note: For some bad a and b, the histogram may not be uniformly distributed.

Example: a=13, b=0, m=31
The first 30 numbers in the sequence are a permutation of integers for 1 to 30 and then the sequence repeats itself. Values are between 0 and m-1. If the values are normalized by dividing by m-1, then the result is numbers uniformly distributed in the interval [0,1]. There are only a finite number of values (30 in this case). In Matlab, you can use function "hist(x)" to see if it is uniformly distributed.

Examples:[From Textbook]
If x0=3 and xn=(5xn-1+7)mod 200, find x1,...,x10.
Solution:
x1= (15+7) mod 200= 22
x2= 117 mod 200= 117
x3= 592 mod 200 = 192
x4= 2967 mod 200= 167
x5= 14842 mod 200= 42
x6= 74217 mod 200 = 17
x7= 371092 mod 200= 92
x8= 1855467 mod 200= 67
x9= 9277342 mod 200 = 142
x10= 46386717 mod 200 = 1117

Typically, it is good to choose m such that m is large, and m is prime. Careful selection of parameters helps generate relatively "random" output values, where it is harder to identify patterns.

When we used m=40, our results were not satisfactory in producing an output resembling a uniform distribution. It is better when m is prime. Care must also be taken when choosing suitable a and b values.

Values are between 0 and m-1. If the values are normalized by dividing by m-1, their result is numbers uniformly distributed on the interval [0,1].

From the example shown above, we can see why to create a good random number sequence, we need to a large m. As the xn value is dependent on the (5xn-1+7)value, such that the value it can be is between 0 to m. Thus, if we want to create large group of random number, it is better to have large m such that the random value generated will not be repeated.
Example:
For xn = (2xn-1+1) mod 3 where x0=2, x1 = 5 mod 3 = 2
Notice that, with the small value m, the random number generated repeated itself is faster than when the value is large enough.

For many years the “rand” function in Matlab used this algorithm with these parameters A=7^5=16807, b=0, m=2^31 -1=2147483647 – recommended in a 1988 paper by Park and Miller (Important part is that m should be large)

### Inverse Transform Method

This method is useful for generating types of distribution other than uniform distribution, such as exponential distribution and normal distribution. Exponential distribution has the property that generated numbers are frequently close to 0. Normal distribution has the property that generated numbers are frequently close to its mean.

Theorem:
Take U ~ U[0,1] and let x=F-1(U)
Then x has distribution function F(.)
Where F(x) = P(X<=x) cdf; F-1(U) denotes the inverse function of F(.) Or that F(x)=U -> x=F-1(U)

Proof of the theorem:
F(x)=P(X<=x)
=P(F-1(U)<=x)
=P(F(F-1(U))<=F(x)) #Applying F, which is monotonic, to both sides
=P(U<=F(x))
=F(x) #Because Pr(U<=y)= y,since U is uniform on the unit interval

F(.) is a monotonic function, which is a function that strictly increasing or decreasing.

Example: $f(x) = \lambda e^{-\lambda x}$
$F(x)= \int_0^x f(x) dx$
$= \int_0^x \lambda e ^{-\lambda x}\ dx$
$= \frac{\lambda}{-\lambda}\, e^{-\lambda x}\, | \underset{0}{x}$
$= -e^{\lambda x} + e^0$
$=1 - e^{- \lambda x}$

y=1-e-λx;
1-y=e-λx;
x=-ln(1-y)/λ;
y=-ln(1-x)/λ;
F-1(x)=-ln(1-x)/λ;

Step 1: Draw U ~U[0,1];
Step 2: x=-ln(1-U)/ λ;

Example 2: Given a CDF of X: F(x) = x5, transform U~U[0,1]. Sol: Let y=x5, solve for x => x=y(1/5) =>F-1(x) = x(1/5) Hence, to obtain a value of x from F(x), we first set u as an uniform distribution, then obtain the inverse function of F(x), and set x= u(1/5)

In Matlab, you can use functions: "who" to see what variables you have defined "clear all" to clear all variables you have defined "close all" to close all figures

MatLab for Inverse Transform Method:

>>u=rand(1,1000)
>>hist(u)       #will generate a fairly uniform diagram


#let λ=2 in this example; however, you can make another value for λ
>>x=(-log(1-u))/2;
>>size(x)       #1000 in size
>>figure
>>hist(x)       #exponential


Limitations:
1. This method is flawed since not all functions are invertible nor monotonic.
2. It may be impractical since some CDF's and/or integrals are not easy to compute.

### Probability Distribution Function Tool in MATLAB

>>disttool