Lecture 18: Sampling distributions - University of Wisconsin–Madison

Lecture 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(?, ¦Ò 2 ), then the joint pdf is

!

1 n

1

2

xi ¡Ê R, i = 1, ..., n

exp ? 2 ¡Æ (xi ? ?) ,

2¦Ò i=1

(2¦Ð)n/2 ¦Ò n

Theorem 5.3.1.

Let X1 , ..., Xn be a random sample from N(?, ¦Ò 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(?, ¦Ò 2 /n);

c. (n ? 1)S 2 /¦Ò 2 has the chi-square distribution with n ? 1 degrees of

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freedom.

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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 ? ?)/¦Ò ¡« N(0, 1),

i = 1, ..., n, because we can apply Theorem 4.6.12 and

¦Ò2 n

X? = ¦Ò Z? ? ? and S 2 =

¡Æ (Zi ? Z? )2 = ¦Ò 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 ? ¡Æ zi

zi ¡Ê R, i = 1, ..., n,

2 i=1

(2¦Ð)n/2

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the joint pdf of (Y1 , ..., Yn ) is

?

!2 ?

!

n

n

n

1

1

exp ??

y1 ? ¡Æ yi ? exp ? ¡Æ (yi + y1 )2

2

2 i=2

(2¦Ð)n/2

i=2

?

?

!2 ??





n

n

n

n

1

yi ¡Ê R

=

exp ? y12 exp ?? ? ¡Æ yi2 + ¡Æ yi ??

n/2

i

=

1, ..., n.

2

2 i=2

(2¦Ð)

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? = ? ¡Æ (Zi ? Z? ) = ? ¡Æ Yi

i=2

and Zi ? Z? = Yi ,

i = 2, ..., n,

i=2

we have

SZ2

1 n

1

=

¡Æ (Zi ? Z? )2 = n ? 1

n ? 1 i=1

UW-Madison (Statistics)

Stat 609 Lecture 18

n

¡Æ Yi

i=2

!2

+

1 n 2

¡Æ Yi

n ? 1 i=2

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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 = ¡Æ (Xi ? X? )2 = ¡Æ (Xi ? ? + ? ? X? )2 = ¡Æ (Xi ? ?)2 +n(? ? X? )2

Then

2



n 

n

Xi ? ? 2

(n ? 1)S 2

X? ? ?

=

+

=

n

¡Æ ¦Ò

¡Æ Zi2

¦Ò

¦Ò2

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 ¡Æni=1 Zi2 ¡« chi-square with degrees of freedom n, and its

mgf is (1 ? 2t)?n/2 , t < 1/2,

¡Ì

n(X? ? ?)/¦Ò ¡« N(0, 1) and hence n[(X? ? ?)/¦Ò ]2 ¡« chi-square

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with degree of freedom 1.

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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 /¦Ò 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(?, ¦Ò 2 ).

Using the result in Chapter 4 about a ratio of independent normal and

chi-square random variables, the ratio

¡Ì

X? ? ?

(X? ? ?)/(¦Ò / n)

¡Ì =p

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S/ n

[(n ? 1)S 2 /¦Ò 2 ]/(n ? 1)

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