Page 226 - Probability and Statistical Inference

4. Functions of Random Variables and Sampling Distribution 203

W – W , ..., U = W – W n:n–1 , and realize that n(X – µ)/σ is same as
n:1
n:2
n:1
n:n
n
nU /σ, which has the standard exponential distribution. That is, n(X – µ)/σ
1
n:1
is distributed as the standard exponential distribution which is the same as
Gamma (1, 1). Also, recall that U , ..., U must be independent random vari-
1
n
ables. Next, one notes that

Now, it is clear that T is distributed independently of X because T function-
n:1
ally depends only on (U , ..., U ) whereas X functionally depends only on
n:1
n
2
U . But, U is independent of (U , ..., U ). Also, note that
n
2
1
1
where Z , ..., Z are iid random variables having the Chi-square distribution
nn
2n
with two degrees of freedom. Thus, using the reproductive property of inde-
pendent Chi-square variables (Theorem 4.3.2, part (iii)), we conclude that
has the Chi-square distribution with 2(n – 1) degrees of
freedom. !
Suppose that X , ..., X are iid random variables.
n
1
If their common distribution is negative exponential, then
X and are independent.
n:1
Remark 4.4.5 If we compare (4.4.20) with the representation given in
(4.4.9), it may appear that in principle, the basic essence of the Remark 4.4.1
holds in this case too. It indeed does, but only partially. One realizes fast that
in the present situation, one is forced to work with the spacings between the
successive order statistics. Thus, the decomposition of 2Tσ into unit inde-
–1
pendent components consists of (n – 1) random terms, each depending on the
sample size n. This is fundamentally different from what we had observed in
(4.4.9) and emphasized earlier in the Remark 4.4.2.
In the next example X and X are independent, but the
1
2
transformed variables Y and Y are dependent.
1 2
Example 4.4.13 (Example 4.4.5 Continued) Suppose that X and X are iid
2
1
standard exponential random variables. Thus,