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Chapter 8
The Independence Problem
In some problems a random sample of n
pairs
of observations (X, Y) are observed from a
bivariate population.
Our main interest is in whether or not there is a statistical relationship between these two
variables. In particular we are interested in finding whether these two variables are independent
of each other and if they are
not
independent, then we want to find the extent and the type of
dependence between them.
Data:
We obtain n pairs of observations, (X
1
, Y
1
), (X
2
, Y
2
), …, (X
n
, Y
n
), one pair of
observations from each subject in the population.
Assumption
A.
The n bivariate observations (X
1
, Y
1
), (X
2
, Y
2
), …, (X
n
, Y
n
) are
a random sample
from
a continuous bivariate population. That is, (X, Y) pairs are mutually independent and
identically distributed (iid) according to some continuous bivariate population. Thus we
are
assuming independence
between the observed pairs.
8.1 Test of Independence based on Signs (Kendall)
Hypotheses:
Let F
X, Y
be the joint distribution function of the common bivariate population.
Let F
X
and F
Y
be the marginal distributions functions of the population of X’s and the population
of Y’s, respectively [though not necessarily the same].
Then, the hypotheses of interest are
Ho: F
X, Y
(x, y) = F
X
(x)×F
Y
(y) for all (x, y) pairs.
Ha: F
X, Y
(x, y) ≠ F
X
(x)×F
Y
(y) for at least one (x, y) pair.
Type of dependence we are interested in this section is Kendal’s population correlation
coefficient, defined as [See next page]
2
1
2
1
2
(
) (
) 0
1
P
X
X
Y
Y
Note that (X
2
– X
1
)×(Y
2
– Y
1
) > 0, if {(X
2
– X
1
) > 0
and
(Y
2
– Y
1
) > 0}
or
if {(X
2
– X
1
) < 0
and
(Y
2
– Y
1
) < 0}.
Furthermore, these two events are mutually exclusive and each has probability ½ if X and Y are
independent. Hence,
if X and Y are independent variables
P((X
2
– X
1
)×(Y
2
– Y
1
) > 0) = P({(X
2
– X
1
) > 0
and
(Y
2
– Y
1
) > 0})
+ P({(X
2
– X
1
) < 0
and
(Y
2
– Y
1
) < 0}.
= P[ (X
1
– X
2
) > 0] × P[ (Y
1
– Y
2
) > 0]
+ P[ (X
1
– X
2
) < 0] × P[ (Y
1
– Y
2
) < 0]
= ½ × ½ + ½ × ½ = ¼ + ¼ = ½ .
Thus,
τ = 2×( ½ + ½ ) – 1 = 0 if X and Y are independent random variables.
So we conclude
that the
test of independence
is the same as
Ho: τ = 0 vs. Ha: τ ≠ 0
(or τ < 0 or τ > 0).
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View Full DocumentSome new concepts:
Two pairs of observations (X
i
, Y
i
) and (X
j
, Y
j
), i ≠ j, are said to be
a
concordant pair
if
(X
i
– X
j
)×(Y
i
– Y
j
) > 0 and
a discordant pair
if
(X
i
– X
j
)×(Y
i
– Y
j
) < 0.
From these, we infer that,
There is a
positive association
between X and Y if there are many more concordant pairs
than discordant pairs. Similarly,
If the discordant pairs dominate then we infer
negative association.
Test Statistic
is
τ
= P(Concordant pair) – P(Discordant Pair)
= 2×P(Concordant pair) – 1
= 2×P[ (X
i
– X
j
)×(Y
i
– Y
j
) > 0] – 1.
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 Summer '11
 YESILCAY

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