Lecture 18: Sampling distributions - University of Wisconsin–Madison
嚜燉ecture 18: Sampling distributions
In many applications, the population is one or several normal
distributions (or approximately).
We now study properties of some important statistics based on a
random sample from a normal distribution.
If X1 , ..., Xn is a random sample from N(?, 考 2 ), then the joint pdf is
!
1 n
1
2
xi ﹋ R, i = 1, ..., n
exp ? 2 ﹉ (xi ? ?) ,
2考 i=1
(2羽)n/2 考 n
Theorem 5.3.1.
Let X1 , ..., Xn be a random sample from N(?, 考 2 ) and let X? and S 2 be
the sample mean and sample variance. Then
a. X? and S 2 are independent random variables;
b. X? ‵ N(?, 考 2 /n);
c. (n ? 1)S 2 /考 2 has the chi-square distribution with n ? 1 degrees of
beamer-tu-logo
freedom.
UW-Madison (Statistics)
Stat 609 Lecture 18
2015
1 / 18
Proof.
We have already established property b (Chapter 4).
To prove property a, it is enough to show the independence of Z? and
SZ2 , the sample mean and variance based on Zi = (Xi ? ?)/考 ‵ N(0, 1),
i = 1, ..., n, because we can apply Theorem 4.6.12 and
考2 n
X? = 考 Z? ? ? and S 2 =
﹉ (Zi ? Z? )2 = 考 2 SZ2
n ? 1 i=1
Consider the transformation
Y1 = Z? ,
Yi = Zi ? Z? ,
i = 2, ..., n,
Then
Z1 = Y1 ? (Y2 + ﹞ ﹞ ﹞ + Yn ),
Zi = Yi + Y1 ,
i = 2, ..., n,
and
? (Z1 , ..., Zn )
1
=
? (Y1 , ..., Yn )
n
Since the joint pdf of Z1 , ..., Zn is
!
1
1 n 2
exp ? ﹉ zi
zi ﹋ R, i = 1, ..., n,
2 i=1
(2羽)n/2
UW-Madison (Statistics)
Stat 609 Lecture 18
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2015
2 / 18
the joint pdf of (Y1 , ..., Yn ) is
?
!2 ?
!
n
n
n
1
1
exp ??
y1 ? ﹉ yi ? exp ? ﹉ (yi + y1 )2
2
2 i=2
(2羽)n/2
i=2
?
?
!2 ??
n
n
n
n
1
yi ﹋ R
=
exp ? y12 exp ?? ? ﹉ yi2 + ﹉ yi ??
n/2
i
=
1, ..., n.
2
2 i=2
(2羽)
i=2
Since the first exp factor involves y1 only and the second exp factor
involves y2 , ..., yn , we conclude (Theorem 4.6.11) that Y1 is
independent of (Y2 , ..., Yn ).
Since
n
n
Z1 ? Z? = ? ﹉ (Zi ? Z? ) = ? ﹉ Yi
i=2
and Zi ? Z? = Yi ,
i = 2, ..., n,
i=2
we have
SZ2
1 n
1
=
﹉ (Zi ? Z? )2 = n ? 1
n ? 1 i=1
UW-Madison (Statistics)
Stat 609 Lecture 18
n
﹉ Yi
i=2
!2
+
1 n 2
﹉ Yi
n ? 1 i=2
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2015
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which is a function of (Y2 , ..., Yn ).
Hence, Z? and SZ2 are independent by Theorem 4.6.12.
This proves a.
Finally, we prove c (the proof in the textbook can be simplified).
Note that
n
n
n
i=1
i=1
i=1
(n ?1)S 2 = ﹉ (Xi ? X? )2 = ﹉ (Xi ? ? + ? ? X? )2 = ﹉ (Xi ? ?)2 +n(? ? X? )2
Then
2
n
n
Xi ? ? 2
(n ? 1)S 2
X? ? ?
=
+
=
n
﹉ 考
﹉ Zi2
考
考2
i=1
i=1
Since Zi ‵ N(0, 1) and Z1 , ..., Zn are independent, we have previously
shown that
each Zi2 ‵ chi-square with degree of freedom 1,
the sum ﹉ni=1 Zi2 ‵ chi-square with degrees of freedom n, and its
mgf is (1 ? 2t)?n/2 , t < 1/2,
﹟
n(X? ? ?)/考 ‵ N(0, 1) and hence n[(X? ? ?)/考 ]2 ‵ chi-square
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with degree of freedom 1.
UW-Madison (Statistics)
Stat 609 Lecture 18
2015
4 / 18
The left hand side of the previous expression is a sum of two
independent random variables and, hence, if f (t) is the mgf of
(n ? 1)S 2 /考 2 , then the mgf of the sum on the left hand side is
(1 ? 2t)?1/2 f (t)
Since the right hand side of the previous expression has mgf
(1 ? 2t)?n/2 , we must have
f (t) = (1 ? 2t)?n/2 /(1 ? 2t)?1/2 = (1 ? 2t)?(n?1)/2
t < 1/2
This is the mgf of the chi-square with degrees of freedom n ? 1, and
the result follows.
The independence of X? and S 2 can be established in other ways.
t-distribution
Let X1 , ..., Xn be a random sample from N(?, 考 2 ).
Using the result in Chapter 4 about a ratio of independent normal and
chi-square random variables, the ratio
﹟
X? ? ?
(X? ? ?)/(考 / n)
﹟ =p
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S/ n
[(n ? 1)S 2 /考 2 ]/(n ? 1)
UW-Madison (Statistics)
Stat 609 Lecture 18
2015
5 / 18
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