Unformatted text preview: Bivariate Discrete Distributions
Let X and Y be two discrete random variables defined on a sample
space S of an experiment.
The joint probability mass function p(x, y) is defined for each pair of
numbers (x, y) by In this class the pairs of numbers can be created as a product set and
show as the margins in a two way table. The table cells can contain the
joint probabilities.
p(x,y)
x
1
2 2
.05
.25 y
4
.15
.05 6
.20
.30 Such joint probabilities must satisfy twp conditions: Check sum of p(x, y) in table above: .05 + .15 + .20 + .25 + .05 + .30 = 1 Probabilities can be calculated for any set A of the (x, y) pairs. Example A = { (x,y) : x + y ≤ 5} = { (1,2), (1,4), ( 2,2) }
p(x,y)
x
1
2 y
4
.15
.05 2
.05
.25 6
.20
.30 P(A) = .05 + .15 + .25 = .45 Probabilities can be calculated for individual random variables X (or Y) Example
p(x,y)
x
1
2 2
.05
.25 y
4
.15
.05 6
.20
.30 p(x)
.4
.6 Doing this for all possible values of X (1 and 2 below) yields the
marginal probability mass function for X.
p(x,y)
x
1
2 y
4
.15
.05 2
.05
.25 6
.20
.30 p(x)
.4
.6 The label marginal simply indicates the context of a joint distribution.
A marginal pmf is like any other probability mass function. We can
compute its expected value, variance and standard deviation. 1
2 .4
.6 .4
1.2
1.6 1
4 = .4
.6 .4
2.4
= 2.8 We obtain the marginal pmf for Y similarly.
p(x,y)
x
1
2 2
.05
.25
.30 y
4
.15
.05
.20
p(y) 6
.20
.30
.50 p(x)
.4
.6
1.00 We can then compute the expected value, variance and standard
deviation for y. The calculations can be done just as easily in a row
format but have been put in a column format show the similarity to the
x calculations above. 2
4
6 .3
.2
.5 .6
.8
3.0
4.4 4
16
36 = .3
.2
.5 1.2
3.2
18.0
= 22.4 Conditional Probability Mass Functions
Knowing the outcome for X or Y limits the possible pairs of outcomes.
Suppose we know that y=4.
P(x,y)
X
1
2 2
.05
.25
.30 y
4
.15
.05
.20 6
.20
.30
.50 x
.4
.6
1.00 The conditional distribution of Xy=4 is another probability mass
function. The probabilities for the possible values of X must sum to 1,
so we divide the joint probabilities in the column by the marginal
probability P(Y=4) = .20. 1
2 .15/.20 =.75
.05/.20 =.25
1.00 We could compute the expected value, variance and standard deviation
of this probability mass function. However this time the task is to
compare marginal pmf and conditional pmf for X. 1
2 .4
.6
1.0 . 75
.25
1.00 The pmfs are different. Thus knowing what Y is tells us something
about pmf of X and hence X and Y are not independent. Similarly if the expected values for the marginal and conditional
distributions of X differ then X and Y are not independent. 1
2 .4
.6 .4
1.2
1.6 1
2 .75
.25 .75
.50
= 1.25 Independence requires that
The joint probabilities are the product of the margin probabilities.
An example is below. The top left probability is .12 = .4 x .3. All the
bivariate cell probabilities are the product of the respect margin
probabilities.
p(x,y)
X
1
2 2
.12
.18
.3 y
4
.08
.12
.2
p(y) 6
.20
.30
.5 p(x)
.4
.6
1.00 Expected Value of a Function of Random Variables, X and Y,
For Bivariate Probability Mass Function. Let = XY. Find E(XY) using the first bivariate pmf from above.
p(x,y)
x
1
2 xy
x
1
2
xy p(x,y)
x
1
2 2
.05
.25 y
4
.15
.05 6
.20
.30 2
1 2=2
2 2=4 y
4
4
8 6
6
12 2
2 .05 = .1
4 .25 = 1.0 y
4
4 .15 = .6
8 .05 = .4 6
6 .20 = 1.2
12 .30 = 3.6 E(XY) = .1 + .6 + 1.2 +
1.0 + .4 + 3.6 = 6.9 Covariance of X and Y
Cov(X,Y) = =
So
The pmf being used is still:
p(x,y)
x
1
2 2
.05
.25 Y
4
.15
.05 Previous results were E(XY) = 6.9 , E(X) =
So
– 6
.20
.30
= 1.6 and E(Y) = = 4.4. Correlation of X and Y
This is denoted as Cor(X,Y), Previous results were and defined by 0 . ...
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This note was uploaded on 12/20/2011 for the course STAT 344 taught by Professor Staff during the Spring '08 term at George Mason.
 Spring '08
 Staff
 Probability

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