Page 265 - Elements of Distribution Theory

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052184472Xc08 CUNY148/Severini May 24, 2005 17:54

8.5 Sampling Distributions 251

¯
2
Then X has a normal distribution with mean 0 and variance 1/n, (n − 1)S has a chi-
¯
2
squared distribution with n − 1 degrees of freedom, and X and S are independent.
Proof. Let m 0 denote a 1 × n vector of all ones, let m = m 0 /n and let X = (X 1 ,..., X n ).
¯
Then X = mX and
T
T
T
2
T
T
T
T
(n − 1)S = (X − nm mX) (X − nm mX) = X (I n − nm m) (I n − nm m)X.
¯
T
The marginal distribution of X follows from Theorem 8.1. Note that mm = 1/n. Hence,
2
T
T
T
T
T
T
T
(I − nm m) (I − nm m) = I − 2nm m + n m (mm )m = I − nm m
2
so that, by Theorem 8.6, (n − 1)S has a chi-squared distribution with degrees of freedom
T T
equal to the trace of I − nm m. Note that each diagonal element of I − nm m is 1 − 1/n =
(n − 1)/n so that the trace is n − 1.
¯
T
T
T
T
T
Finally, (I − nm m)m = m − nm (mm ) = 0so that, by Theorem 8.7, X and S are
independent.
√
¯
The distribution of the ratio nX/S now follows immediately from the definition of the
¯ 2
2
t-distribution given in Example 7.10. The distribution of nX /S is also easily determined.
Corollary 8.2. Let X 1 ,..., X n denote independent, identically distributed standard normal
random variables. Let
n
1
¯
X = X j
n
j=1
and let
n

¯ 2
2
S = (X j − X) /(n − 1).
j=1
Then
√
¯
(i) nX/S has a t-distribution with n − 1 degrees of freedom.
¯ 2
2
(ii) nX /S has a F-distribution with (1, n − 1) degrees of freedom.
¯
¯
Proof. From Theorem 8.9, X and S are independent; √ nX has a standard normal dis-
2
¯ 2
tribution, nX has a chi-squared distribution with 1 degree of freedom, and (n − 1)S has
a chi-squared distribution with n − 1degrees of freedom. The results now follow easily
from the definitions of the t- and F-distributions, given in Examples 7.10 and 7.11,
respectively.
¯
2
The statistics X and S considered in Theorem 8.9 may be interpreted as follows. Define
the vector m 0 as in the proof of Theorem 8.9. Then the projection of a random vector X
¯
onto the space spanned by m 0 is Xm 0 . The statistic
n

2 ¯ 2
(n − 1)S = (X j − X)
j=1