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4. Functions of Random Variables and Sampling Distribution 199

where –∞ < y , ..., y < ∞. In (4.4.8), the terms involving y , ..., y factorize
1
n
1
n
and none of the variables’ domains involve the other variables. Thus it is clear
(Theorem 3.5.3) that the random variables Y , ..., Y are independently dis-
n
1
tributed. Using such a factorization, we also observe that Y is distributed as
1
whereas Y , ..., Y are iid N(0, σ ). !
2
2 n
The following result plays a crucial role in many statistical analyses. Inci-
dentallY, one should note that sometimes a population is also referred to as a
universe in the statistical literature. The result we are about to mention is also
indispensable in much of the distribution theory. These points will become
clear in the sequel.
2
Theorem 4.4.2 (Joint Sampling Distribution of and S from a
Normal Universe) Suppose that X , ..., X are iid N(µ, σ ) random vari-
2
n
1
ables, n ≥ 2. Define the sample mean , and the sample
variance Then, we have:
(i) The sample mean is distributed independently of the sample
variance S ;
2
(ii) The sampling distribution of is and that of (n – 1)S /
2
σ is Chi-square with (n – 1) degrees of freedom.
2
Proof (i) Using the Helmert transformation from (4.4.6), we can rewrite
= n –1/2 Y . Next, observe that
1
using the same Helmert variables. It is clear that is a function of Y alone,
1
whereas from (4.4.9) we note that S depends functionallY on (Y , ..., Y )
2
n
2
only. But, since Y , ..., Y are all independent, we conclude that is distrib-
n
1
2
uted independentlY of S . Refer to the Theorem 3.5.2, part (ii) as needed.
(ii) Recall that = n –1/2 Y where Y is distributed as N( µ, σ ) and so
2
1
1
the sampling distribution of follows immediately. From (4.4.9) again, it is
clear that which is the sum of (n – 1) indepen-
dent Chi-square random variables each having one degree of freedom. Hence,
using the reproductive property of independent Chi-square variables (Theo-
rem 4.3.2, part (iii)), it follows that (n – 1)S /σ has a Chi-square distribution
2
2
with (n – 1) degrees of freedom. ¢
Remark 4.4.1 Let us reconsider the setup in the Example 4.4.9 and Theo-
rem 4.4.2. It is important to note that the sample variance S is the average of
2
and each Helmert variable independently contributes one de-
gree of freedom toward the total (n – 1) degrees of freedom. Having n
observations X , ..., X , the decomposition in (4.4.9) shows how ex-
n
1
actly S can be split up into (n – 1) independent and identically distributed
2