Lecture 24: Sample Variance S2 and its Properties, Lecture notes of Statistics

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Lecture 24: The Sample Variance S^2 The squared

variation

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Suppose we have n numbers x 1 , x 2 ,... , xn. Then their squared variation

sv = sv(x 1 , x 2 ,... , xn) sv(x 1 , x 2 ,... , xn) =

∑n

i= 1

(xi − x)^2

Their mean (average) squared variation msv or σ^2 n (denoted σ^2 and called the

“population variance on page 33 of our text) is given by

msv = σ^2 n =

n

sv =

n

∑^ n

i= 1

(xi − x)^2

Here x is the average

n

∑^ n

i= 1

xi.

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The Shortcut Formula for the Squared Variation

Theorem

sv(x 1 , x 2 ,... , xn) =

∑^ n

i= 1

x^2 i −

n

∑^ n

i= 1

xi )^2 (∗)

Proof

Note since x =

n

∑^ n

i= 1

xi we have

∑n

i= 1

xi = nx

Now

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Proof (Cont.)

∑^ n

i= 1

x i^2 − 2 x(nx) + nx^2

∑^ n

i= 1

x i^2 − 2 nx^2 + nx^2

∑^ n

i= 1

x i^2 − nx^2

∑^ n

i= 1

x i^2 − n

∑^ n

i= 1

xi

n

2

∑^ n

i= 1

x i^2 −n

( n

i= 1

xi

n^2

∑^ n

i= 1

x i^2 −

n

∑^ n

i= 1

xi

2

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Let met give a conceptual proof of the theorem the way a professorial mathematician would prove the theorem.

Definition

A polynomial p(x 1 , x 2 ,... , xn) is symmetric, if it is unchanged by permuting the

variables.

Examples 3

p(x, y, z) = x^2 + y^2 + z^2 is symmetric

p(x, y, z) = xy + z^2 is not symmetric

Theorem

Any symmetric polynomial pin x 1 , x 2 ,... , xn can be rewritten as a polynomial in

the power sums

∑n

i= 1

xik that is

p(x 1 ,... , xn) = q

xi ,

x 12 ,... ,

x i`

if deg p = `.

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Bottom Line

sv =

∑n

i= 1

(xi − x)^2 is a symmetric polynomial in x 1 , x 2 ,... , xn so there exist a and

b with

sv(x 1 , x 2 ,... , Xn) = a

∑^ n

i= 1

x i^2 + b

∑^ n

i= 1

xi

2

This is true for all x 1 ,... , xn (an “identify”) so we just choose x 1 ,... , xn cleverly to

get a and b.

First choose x 1 = 1, x 2 = −1, x 3 =... = xn = 0 so

∑n

i= 1

xi = 0 and

∑n

i= 1

x i^2 = 2

since x = 0

(∗∗) becomes

2 = a 2 + b( 0 ) so a = 1

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In Which We Return to Statistics

Estimating the Population Variance We have seen that X is a good (the best)

estimator of the population mean-μ, in particular it was an unbiased estimator.

How do we estimate the population variance?

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Answer - use the Sample variance s^2 to estimate the population variance σ^2

The reason is that if we take the associated sample variance random variable

S^2 =

n − 1

n∑− 1

i= 1

(Xi − X)^2

then we have

Amazing Theorem

Why do you need

n − 1

? We will see.

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Corollary

Suppose X 1 , X 2 ,... , Xn is a random sample from a population of mean μ and

variance σ^2. Then

(i) E(X i^2 ) = μ^2 + σ^2

(ii) E(T 0 ) = n^2 μ^2 + nσ^2

Proof.

(i) E(Xi ) = μ and V(Y ) = σ^2 so plug into (#)

(ii) E(T 0 ) = nμ and V(T 0 ) = nσ^2

so plug into (#)

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We can now prove (b)

E(S^2 ) = E

n − 1

∑^ n

i= 1

X i^2 −

n(n − 1 )

Xi )^2

since E is linear

n − 1

∑^ n

i= 1

E(X i^2 ) −

n(n − 1 )

E(T 02 )

by (i) and (ii)

n − 1

∑^ n

i= 1

(μ^2 + σ^2 ) −

n − 1

n

(n^2 μ^2 + nσ^2 )

n − 1

[

nμ^2 + nσ^2 −

n

(n^2 μ^2 + nσ^2 )

]

n − 1

[

nμ

2 + nσ 2 −

nμ

2 − σ 2 ]

n − 1

[

(n − 1 )σ^2

]

= σ^2

Amazing - you need

n − 1

not

n