SADC Course in Statistics

Joint distributions

(Session 05)

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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 0

2 1 1

allx ally

. f ( x, y ) for all x, y.

. f ( x, y ) or f ( x, y )dxdy .

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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:

X Xally

f ( x ) f ( x, y ) or f ( x ) f ( x, y )dy .

Y Yallx

f ( y ) f ( x, y ) or f ( y ) f ( x, y )dx .

The sums are for the discrete cases while the integrals are for the continuous cases.

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• The conditional probability mass/density function of X given Y = y is defined as

• Notice that the above definition resembles very closely to the definition of conditional probability.

.)(

),()|(| yf

yxfyxf

YYX

Conditional Distributions

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Independent random variables• Random variables X and Y are said to be

independent if and only if

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

)()(),( yfxfyxf YX

).()|(| xfyxf XYX

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Back to Example 1

Outcomes of die (X) fY(y)

Coin outcomes(Y) 1 2 3 4 5 6

H 1/2

T 1/2

fX(x) 1/6 1/6 1/6 1/6 1/6 1/6 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?

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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

Sexual Behaviour(X)

Condom Use (Y)

Only with regularpartners

Only with casual

partners

Total

Never 617 439 1056

Sometimes 92 64 156

Always 53 133 186

Total 762 636 1398

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Joint and Marginal Probabilities

Sexual Behaviour(X)

Condom Use (Y)

Only with regularpartners

Only with casual

partners

fY(y)

Never 0.441 0.314 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

Sexual Behaviour(X)

Condom Use (Y)

Only with regularpartners

Only with casual

partners

Never

Sometimes

Always

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Practical work follows to ensure learning objectives

are achieved…