Transcript Joint distributions

Joint distributions
(Session 05)
SADC Course in Statistics
Learning Objectives
By the end of this session you will be able to
• describe what is meant by a joint
probability density function
• explain how marginal conditional
probability distribution functions can be
derived from the joint density function
• compute joint and marginal probabilities
corresponding to a two-way frequency
table
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Bivariate distributions
• In many applications one has to work with
two or more random variables at the same
time. To determine the health of a child one
needs to consider the age, weight, height
and other variables.
• A function f is a bivariate joint probability
mass /density function if
1.
f ( x, y )  0

2.
for all
 

 f ( x, y )  1
allx ally
x, y.
or

f ( x, y )dxdy  1.
 
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Example 1
• Consider a trial where a coin and a die are
tossed.
• How many outcomes are possible? Table
below shows the possibilities. We will return
to this table shortly.
Die outcomes
Coin outcomes
1
2
3
4
5
6
H






T






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Marginal Distributions
• Given the bivariate joint probability mass/
density function f, the marginal mass/
densities fX and fY are defined as:


f X ( x )   f ( x, y )
or
fX ( x ) 


allx
f ( x, y )dy .

ally
fY ( y )   f ( x, y )

or
fY ( y ) 

f ( x, y )dx .

The sums are for the discrete cases while
the integrals are for the continuous cases.
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Conditional Distributions
• The conditional probability mass/density
function of X given Y = y is defined as
f ( x, y )
f X |Y ( x | y ) 
.
fY ( y )
• Notice that the above definition resembles
very closely to the definition of conditional
probability.
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Independent random variables
• Random variables X and Y are said to be
independent if and only if
f ( x, y)  f X ( x) fY ( y)
that is, the joint mass/density function is
equal to the product of the marginal
mass/density functions.
• It follows that if X and Y are independent,
then
f X |Y ( x | y )  f X ( x).
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Back to Example 1
Note that the coin/die throwing trial
corresponds to independent outcomes
because what happens with the coin cannot
affect the die outcome. Below are the
marginal probabilities. Can you compute the
joint distribution?
Outcomes of die (X)
Coin outcomes(Y)
1
2
3
4
5
fY(y)
6
H
1/2
T
1/2
fX(x)
1/6 1/6 1/6 1/6 1/6 1/6
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Example 2 (Use of condoms)
• A cross-sectional survey on HIV and AIDS
was conducted in a major mining town in
South Africa in 2001.
• Among the issues investigated were sexual
behaviour and the use of condoms. A total of
2231 people between the ages of 13 to 59
provided responses.
• The sample consisted of migrant
mineworkers, sex workers and members of
the local community.
• The following are some results for men.
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Sexual behaviour and condom use
Condom
Use (Y)
Never
Sexual Behaviour(X)
Only with
Only with
regular
casual
partners
partners
617
439
Total
1056
Sometimes
92
64
156
Always
53
133
186
Total
762
636
1398
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Joint and Marginal Probabilities
Condom
Use (Y)
Never
Sexual Behaviour(X)
Only with
Only with
regular
casual
partners
partners
0.441
0.314
fY(y)
0.755
Sometimes
0.066
0.046
0.112
Always
0.038
0.095
0.133
fX(x)
0.545
0.455
1.000
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Class Exercise (Part I):
• Is condom use independent of sexual
behaviour in terms of type of sexual
partner?
• Use the definition of independence and
allow for sampling errors.
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Class Exercise (Part II):
• Calculate the conditional probability that a
man from the study area has casual
partners given that he always uses
condoms.
• To do this part of the exercise, it would be
helpful to first calculate conditional
probabilities of X, given each value for Y.
Note these down in the table below, and
then answer the question above.
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Conditional Probabilities of X given Y
Condom
Use (Y)
Sexual Behaviour(X)
Only with
Only with
regular
casual
partners
partners
Never
Sometimes
Always
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Practical work follows to
ensure learning objectives
are achieved…
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