ST2334: Chapter 3 - Two dimensional random variables and condition probability distribution

Imagine picking a student in a crowd

Maybe there are two characteristics that we are interested in.

- CAP

- Modules taking

This leads to two dimensional random variables

Because of 2 dimension, we talk about marginal and conditional probability distributions.

The x is a certain value and the y is a certain value

P(X =x) marginal

p(X = x | Y = y ) conditional

Given this, how does it affect the other value

Marginal and conditional will not be the same unless they are independent. (Chap 2)

Chap 2 covers for one variables but for chapter 3, covers 2 variables

Definition

Let E be experiment and S is sample space 

Let X and Y be two functions each assigning a real number to each s where s is variables in S

This is like in a graph, it is denoted as s(X) and s(Y)

This is a two dimension random variables or random vector

X and Y takes certain variables and refer to the same sample space

Range space

Rx,y =  {(x,y) | x = X(s), y = Y(s), s is part of S}

Random variables takes certain values and is random. It is associated with the probability function. 

A function maps one value to another value in S

Definition 2

Let X1, X2 ... Xn be n functions each assigning a real number to every outcome.

We call (X1...Xn) an n dimensional random variable

Definition 3

1. (X,Y) is two dimensional discrete random variable if the possible values of X(s), Y(s) are finite or countable infinite

This means that x and y can be represented as xi , yj where i = 1,2,3... j = 1,2,3..

We use two different subscripts

2. If (X, Y) is Continuous random variable if the possible values assume all values in region of Euclidean plane R^2

e.g 1

Tv set to be service

X represent the age to nearest year of set and Y the number of defective components in the set

For any set, there is a function 

X(s) where 0 ,1,2,3...

Y(s) where  = 1,2,3...

The range for x is 0 to infinite and

the range for y is 0 to n

Where n is the total number of components in the tv set

e.g 2

Fast food restaurant operates drive up and walk up

X is proportion of time for drive up in use (at least one customer is being served or waiting)

Y is walk up window is in used

Rx, 0 < = X <= 1

Ry, 0 < = Y <=1

The range must be between 1 and 0 because it is proportional of time use

This is continuous because it takes all values within the unit square

Definition

Let (X,Y) be a 2 dim discrete random variable define on the sample space of an experiment.

We associate a number fxy (Xi, Yi) representing P(X= xi, Y =yi)

and it satisfy the following conditions:

1. fxy >= 0

2. The sum of fxy = 1

The function is defined for all pairs of values

e.g 1

Find value of k so that the function 

fx,y = kxy for x = 1,2,3 and y = 1,2,3

can serve as a joint probability function

There are 3 possible values for each variables, thus there are a possible 9 combinations (3*3)

We need to sum up all the probabilities, it will give us one

1k + 2k  +3k + 2k + 4k+ 6k+ 3k + 6k + 9k = 1

k = 1/36

How do we know when the f(x,y) is 0?

When the values are outside the range of the function (ie, x and y are wrong numbers)

e.g 2

A company has lines A and B which produce 5 and 3 machines respectively

Assume number of machiens produce is random variable

X = Line A

Y = Line B

What is the prob that more machine are produced by line A than by line B on a given day

We are finding:

P(X > Y)

We can find the answer by summing all probabilities where X is greater than Y

We do not bother what is the value of Y, just care only where X is greater than Y

Marginal probability is the sum of all the X when Y is a certain similiar number

ie when Y is 1, sum X1,X2,X3...

What is the conditional probability of X given Y when Y = 0 

F(X | Y = 0) where X has to be defined as a specific number

e.g 3

There are 9 exe, 4 are married, 3 are never married and 2 are divorce

3 of the exe are to be randomly selected for promotion

X is number of married 

Y are never married

Find joint probability function of X and Y

X = 0 ,1,2,3

Y = 0,1,2,3

We cannot have X = 0 and Y = 0 at the same time

The number to select 3 out of 9 is 9C3

The number of ways to select x exe from 4 married, y exe from 3 never married and the rest from divorce is

4Cx * 3Cy * 2C(3-x-7)

such that 1< = x+y < = 3

//this means that x and y must at least contribute 1

//x + y also cannot be bigger than 3 because the max is 3 people to be selected

f(x,y) = P(X = x , Y = y )

= 4Cx * 3Cy * 2C(3-x-7) / 9C3

such that 1< = x+y < = 3 holds true.

What if I add in a variable z to represent the divorce?

-> that is 3 dimension already, the formula does not hold true and it is very complicated

x+ y + z must be equal to 3

This must always hold:

1. fxy is greater or equal to 0

2. Integrating fx(x,y) dxdy = 1

When integrating, treat y as an constant.  Integrate with respect to y first.'
Does it matter if we integrate x or y first?
-> There might be some.. but it doesnt matter for this module (LOL)🤣
But if we take the right order, it may make the integration easier. Try to alternate who to integrate first.

e.g 1
P(X + Y >=1)

<Insert 3-23>

Look at the boundary of the pdf
Drawing it out, we can plot the area of the graph that we are looking for using the PDF

Looking at the probability,  we realised that we want the area of the graph that is above the line
x+y =1
However, there is no prob density function for this.

If integrating x first before y,
Integrate x first -> x^2 + (xy)/3 dx with range of 1-x to 0
Because y = 1 - x, we will integrate dy from a range of 2 to 1- x (2 is taken from pdf)


But the easier method is to find the area of the triangle (Under the graph x + y =1)
and take the complement of that

e.g2
Given a pdf
12/13 x ( x+ y)  for 0<=x<=1, 1<=y <=2
0                           otherwise

Define A = 0 < x< 1/2, 1< y <2

By integrating x first and y second,
x -> limit of 1/2 to 0
Ensure there is no more x after this
y-> limit of 2 to 1

But what if we integrate y first??
y-> limit of 1 to 2 
x-> limit of 1/2 to 0
In this case , because it is a nice shape, there is no need to change

Marginal and Conditional Probability Distributions

What if I want X only or Y only. 

I also want to know what is X | Y is given some value

Definition

Discrete

fx(x) = the sum of f(x,y) with the same x value

fy(y) = sum of all fxy)x, y) with fix y

Continuous

fx(x) = integrate of dxy with respect to dy

fy(y) = integrate of dxy with respect to dx

IF the value x or y falls out of the range, f(x) / f(y) = 0

Definition

Define as the ratio of the joint distribution / marginal probability

Marginal must always be greater than 0.

fY|X (y|x) = fxy(x,y) / fx(x)

In this case, only one is the random variable

We already fix the random variable x to a certain number, thus it is not random anymore

This expression will show the density of y for a given x

1. The conditional pdf satisfy

a) For fixed y,

 fx|y(X|Y) >= 0

and for fixed x,

fY|X(y|x) >= 0

b) The sum of fx|y(X|Y) = 1

The sum of fY|X(y|x) = 1

fxy(x,y) = fY|X(y|x) * fx(x)

e.g1

Suppose 3 toss of a fair coins

Let X be number of head on last flip and Y the total number of heads for 3 tosses

(HHH HHT HTH .... are sample points)

_ _ _, there are 2 outcomes for each _ 

Therefore, 2* 2 *2 = 8

Find the conditional distribution of Y given X = 1

Constructing the table will help us see better.

<Insert 47>

Independent Random variable

x and y are independent when the condition does not rely on the other.

If the pdf is given x^y -> this is dependent as y^x means a different thing
But if the pdf is given x*y -> independent as y*x is the same as x*y

Proving independence

We can prove by showing fxy(x,y) = fx(x) * fy(y)

To get fx(x)/fy(y):

e.g 2
fxy = xy/36
for x = 1,2,3
and
y = 1,2,3
Are X and Y independent?

For joint PDF, check if its greater or equal to 0

If we sub in x and y, we know that it is greater than 0.
We must then check if f(1,1) + f(1,2) + ... f(3,3) = 1
This is a double summation. 

Solution:
fx(X) =
= x/36 * (summation of y from 1 to 3)
= x/36 *(1+2+3) = 1/6 x for x = 1,2,3

fy(Y) =
1/36 * (Summation of x from 1 to 3)
= 1/4 * y

fxy(xy) = 1/36 * xy = fx(x) fy(y) =  x/6 * y *6
for all x,y =1,2,3

Therefore, it is independent

e.g 3
X and Y are 2 independent rv with
fx(x) = e^-x for x>=0 and
fy(y) = e^-y for y>=0

what is fxy(x,y)?
(How to get the joint pdf)
Given this 2, it is not possible without fx(x)* fy|x(y|x) or fy(y) * fx|y(x|y)
but because it is independent... we can just multiply

Solution
fx,y(x,y) = fx(x)fy(y) { e^-x * e^ -y for x >+ 0 and y >= 0
                                   {  0        otherwise

Expectation

Correlation Coefficient