5.3.1 Properties of the sample mean and variance
Lemma 5.3.2 (Facts about chi-squared random variables)
We use the notation χ2p to denote a chi-squared random variable with p degrees of freedom.
(a) If Z is a N(0, 1) random variable, then Z2 ∼ χ21; that is, the square of a standard normal random variable is a chi-squared random variable.
(b) If X1, . . . , Xn are independent and Xi ∼ χ2p, then X1+ · · · + Xn ∼ χ2p1+···+pn; that is, independent chi-squared variables add to a chi-squared variables, and the degrees of freedom also add.
Proof: . Part (a) can be established based on the density formula for variable transforma- tions. Part (b) can be established with the moment generating function. ¤
Theorem 5.3.1 Let X1, . . . , Xn be a random sample from a N(µ, σ2) distribution, and let X = (1/n)¯ Pn
i=1Xi and S2 = [1/(n − 1)]Pn
i=1(Xi− ¯X)2. Then (a) ¯X and S2 are independent random variables.
(b) ¯X has a N(µ, σ2/n) distribution.
(c) (n − 1)S2/σ2 has a chi-squared distribution with n − 1 degrees of freedom.
Proof: Without loss of generality, we assume that µ = 0 and σ = 1. Parts (a) and (c) are proved as follows.
S2 = 1 n − 1
Xn i=1
(Xi− ¯X)2 = 1
n − 1[(X1− ¯X)2+ Xn
i=2
(Xi− ¯X)2)]
= 1
n − 1[¡Xn
i=2
(Xi− ¯X)¢2 +
Xn i=2
(Xi− ¯X)2)]
The last equality follows from the fact Pn
i=1(Xi − ¯X) = 0. Thus, S2 can be written as a function only of (X1− ¯X, . . . , Xn− ¯X). We will now show that these random variables are independent of ¯X. The joint pdf of the sample X1, . . . , Xn is given by
f (x1, . . . , xn) = 1
(2π)n/2e−(1/2)Pni=1x2i, −∞ < xi < ∞.
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Make the transformation
y1 = ¯x, y2 = x2 − ¯x,
...
yn = xn− ¯x.
This is a linear transformation with a Jacobian equal to 1/n. We have f (y1, . . . , yn) = n
(2π)n/2e−(1/2)(y1−Pni=2yi)2e−(1/2)Pni=2(yi+y1)2, −∞ < yi < ∞
=£ ( n
2π)1/2e(−ny12)/2¤£ n1/2
(2π)(n−1)/2e−(1/2)[Pni=2yi2+(Pni=2yi)2]¤
, −∞ < yi < ∞.
Hence, Y1 is independent of Y2, . . . , Yn, and ¯X is independent of S2. Since
¯ xn+1 =
Pn+1
i=1 xi
n + 1 = xn+1+ n¯xn
n + 1 = ¯xn+ 1
n + 1(xn+1− ¯xn), we have
nSn+12 = Xn+1
i=1
(xi− ¯xn+1)2 = Xn+1
i=1
[(xi− ¯xn) − 1
n + 1(xn+1− ¯xn)]2
= Xn+1
i=1
[(xi− ¯xn)2− 2(xi− ¯xn)(xn+1− ¯xn
n + 1 ) + 1
(n + 1)2(xn+1− ¯xn)2]
= Xn
i=1
(xi− ¯xn)2 + (xn+1− ¯xn)2− 2(xn+1− ¯xn)2
n + 1 + (n + 1)
(n + 1)2(xn+1− ¯xn)2
= (n − 1)S2+ n
n + 1(xn+1− ¯xn)2.
Now consider n = 2, S22 = 12(X2− X1)2. Since (X2− X1)/√
2 ∼ N(0, 1), part (a) of Lemma 5.3.2 shows that S22 ∼ χ21. Proceeding with the induction, we assume that for n = k, (k − 1)Sk2 ∼ χ2k−1. For n = k + 1, we have
kSk+12 = (k − 1)Sk2 + k
k + 1(Xk+1− ¯Xk)2.
Since Sk2 is independent of Xk+1 and ¯Xk, and Xk+1− ¯Xk ∼ N(0,k+1k ), kSk+12 ∼ χ2k. ¤
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Lemma 5.3.3
Let Xj ∼ N(µj, σj2), j = 1, . . . , n, independent. For constants aij and brj (j = 1, . . . , n; i = 1, . . . , k; r = 1, . . . , m), where k + m ≤ n, define
Ui = Xn
j=1
aijXj, i = 1, . . . , k,
Vr = Xn
j=1
brjXj, r = 1, . . . , m.
(a) The random variables Ui and Vr are independent if and only if Cov(Ui, Vr) = 0. Fur- thermore, Cov(Ui, Vr) =Pn
j=1aijbrjσj2.
(b) The random vectors (U1, . . . , Uk) and (V1, . . . , Vm) are independent if and only if Ui is independent of Vr for all pairs i, r (i = 1, . . . , k; r = 1, . . . , m).
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