Page 149 - Probability and Statistical Inference
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126 3. Multivariate Random Variables
Also, X , ..., X form a collection of dependent random variables if and only
k
1
if these are not independent.
In order to show that X , ..., X are dependent, one simply needs
1
k
to show the existence of a particular set of values x , ..., x ,
1
k
with x ∈ χ , i = 1, ..., k such that
i i
Example 3.5.1 (Example 3.2.4 Continued) We immediately note that P(X
= 1 ∩ X = 1) = .12, P(X = 1) = .32 and P(X = 1) = .55. But, P(X = 1
1
2
2
1
1)P(X = 1) = .176 which is different from P(X = 1 ∩ X = 1). We have
1
2
2
shown that when x = 1, x = 1 are chosen, then, f(x ,x ) ≠ f (x )f (x . Thus,
1
1
2
2
1
2
1
2
the two random variables X , X are dependent. !
1 2
Example 3.5.2 (Example 3.3.1 Continued) From (3.3.9)-(3.3.11) we have
2
f(x , x ) = 6(1 x ) for 0 < x < x < 1, f (x ) = 18x (1 x ) (1 x ) for 0
1 2 2 1 2 i i 2 1 2
< x , x < 1. For x = .4, x = .5, we have f(x , x ) = 3, but f (x ) = 1.62
1 2 1 2 1 2 i i
≠ f(x , x ). Thus, the two random variables X , x are dependent. !
1 2 1 2
Example 3.5.3 Consider two random variables X , X having their joint
1
2
pdf given by
First, one can easily show that f (x ) = 2x for 0 < x < 1, i = 1, 2. Since one
i
i
i
i
obviously has f(x , x ) = f (x ) f (x ) for all 0 < x , x < 1, it follows that X , X 2
1
2
2
2
1
1
1
2
1
are independent. !
It is important to understand, for example, that in order for
the k random variables X , X , ..., X to be independent, their
2
k
1
joint and the marginal pmfs or the pdfs must satisfy (3.5.1).
Look at the next example for the case in point.
Example 3.5.4 It is easy to construct examples of random variables X ,
1
X , X such that (i) X and X are independent, (ii) X and X are independent,
3
3
1
2
2
2
and (iii) X and X are also independent, but (iv) X , X , X together are depen-
1
1
3
3
2
dent. Suppose that we start with the non-negative integrable functions f (x )
i i
for 0 < x < 1 which are not identically equal to unity such that we have
i
, i = 1, 2, 3. Let us then define