Transcript Two Random Variables

Two Random Variables
W&W, Chapter 5
Joint Distributions
So far we have been talking about the
probability of a single variable, or a variable
conditional on another.
We often want to determine the joint probability
of two variables, such as X and Y.
Suppose we are able to determine the following
information for education (X) and age (Y) for
all U.S. citizens based on the census.
Joint Distributions
Education (X)
Age (Y):
25-35
30
Age: 3555
45
Age: 55100
70
None
0
.01
.02
.05
Primary
1
.03
.06
.10
Secondary
2
.18
.21
.15
College
3
.07
.08
.04
Joint Distributions
Each cell is the relative frequency (f/N).
We can define the joint probability
distribution as:
p(x,y) = Pr(X=x and Y=y)
Example: what is the probability of getting
a 30 year old college graduate?
Joint Distributions
p(x,y) = Pr(X=3 and Y=30)
= .07
We can see that:
p(x) = y p(x,y)
p(x=1) = .03 + .06 + .10 = .19
Marginal Probability
We call this the marginal probability
because it is calculated by summing
across rows or columns and is thus
reported in the margins of the table.
We can calculate this for our entire table.
Marginal Probability Distribution
Education
(X)
Age (Y):
30
p(x)
None: 0
.01
.02
.05
.08
Primary: 1
.03
.06
.10
.19
Secondary: .18
2
.21
.15
.54
College: 3
.07
.08
.04
.19
p(y)
.29
.37
.34
1
45
70
Independence
Two random variables X and Y are
independent if the events (X=x) and
(Y=y) are independent, or:
p(x,y) = p(x)p(y) for all x and y
Note that this is similar to Event E is
independent of F if:
Pr(E and F) = Pr(E)Pr(F) Eq. 3-21
Example
Are education and age independent?
Start with the upper left hand cell:
p(x,y) = .01
p(x) = .08
p(y) = .29
We can see they are not independent
because (.08)(.29)=.0232, which is not
equal to .01.
Independence
In a table like this, if X and Y are
independent, then the rows of the table
p(x,y) will be proportional and so will the
columns (see Example 5-1, page 158).
Covariance
It is useful to know how two variables vary
together, or how they co-vary. We
begin with the familiar concept of
variance (E is expectation).
2 = E(x- )2 = (x- )2 p(x)
X,Y = Covariance of X and Y
= E(X - X)(Y - Y)
= (X - X)(Y - Y)p(x,y)
Covariance
Let’s calculate the covariance for education (X)
and age (Y).
First we need to calculate the mean for X and Y:
X = xp(x) = (0)(.08)+(1)(.19)+(2)(.54)+(3)(.19)=1.84
Y = yp(y) = (30)(.29)+(45)(.37)+(70)(.34)=49.15
Now calculate each value in the table minus its
mean (for X and Y), multiplied by the joint
probability!
Covariance
X,Y = (X - X)(Y - Y)p(x,y)
= (0-1.84)(30-49.15)(.01) +
(0-1.84)(45-49.15)(.02) + (0-1.84)(70-49.15)(.05) +
(1-1.84)(30-49.15)(.03) + (1-1.84)(45-49.15)(.06) +
(1-1.84)(70-49.15)(.10) + (2-1.84)(30-49.15)(.18) +
(2-1.84)(45-49.15)(.21) + (2-1.84)(70-49.15)(.15) +
(3-1.84)(30-49.15)(.07) + (3-1.84)(45-49.15)(.08) +
(3-1.84)(70-49.15)(.04) = -3.636
Covariance
The covariance is negative, which tells us that
as age increases, education decreases (and
vice versa).
It is negative because when one variable is
above its mean, the other is below its mean
on average.
We can calculate covariance alternatively as
X,Y
= E(XY) - X Y
= (xy)p(x,y) - X Y
Covariance and Independence
If X and Y are independent, then they are
uncorrelated, or their covariance is zero:
X,Y = 0
The value for covariance depends on the units
in which X and Y are measured. If X, for
example, were measured in inches instead of
feet, each X deviation and hence X,Y itself
would increase by 12 times.
Correlation
We can calculate the correlation instead:
 = X,Y
X Y
Correlation is independent of the scale it is
measured in, and is always bounded:
-1    1
Correlation
A perfect positive correlation (=1); all x,y coordinate
points will fall on a straight line with positive slope.
A perfect negative correlation (=-1); all x,y coordinate
points will fall on a straight line with negative slope.
A correlation of zero indicates no relationship between
X and Y (or independence!).
Positive correlations (as X increases, Y increases)
Negative correlations (as X increases, Y decreases)
Example of Correlation
Calculate the correlation between
education and age:
 = X,Y = -3.636
X Y
 = -0.2743
(.8212)(16.14)
Interpretation
There is a weak, negative correlation
between education and age, which
means that older people have less
education.
Later on we will learn how to conduct a
hypothesis test to determine if  is
significantly different from zero.