(i) Let (X1, Y1), . . . , (Xn, Yn) be a sample from the bivariate normal distribution (5.73), and let S2 1 = (Xi − X¯)2, S12 = (Xi − X¯)(Yi − Y¯), S2 2 = (Yi − Y¯)2.

Then (S2 1 , S12, S2 2 ) are independently distributed of (X¯ , Y¯), and their joint distribution is the same as that of (

n−1 i=1 X

i 2, n−1 i=1 X

iY 

i , n−1 i=1 Y 

i 2), where

(X

i, Y 

i ),i = 1,..., n − 1, are a sample from the distribution (5.73) with

ξ = η = 0.

(ii) Let X1,..., Xm and Y1,..., Ym be two samples from N(0, 1). Then the joint density of S2 1 = X2 i , S12 = XiYi , S2 2 = Y 2 i is 1

4π(m − 1)

(s2 1 s2 2 − s2 12)

1 2 (m−3) exp

−1 2

(s2 1 + s2 2 )

for s2 12 ≤ s2 1 s2 2 , and zero elsewhere.

(iii) The joint density of the statistics (S2 1 , S12, S2 2 ) of part (i) is

(s2 1 s2 2 − s2 12)

1 2 (n−4)

4π(n − 2)



στ

1 − ρ2

n−1 exp

− 1 2(1 − ρ2)

 s2 1

σ2 − 2ρs12

στ +

s2 2

τ 2

 (5.84)

for s2 12 ≤ s2 1 s2 2 and zero elsewhere.

[(i): Make an orthogonal transformation from X1,..., Xn to X

1,..., X

n such that X

n = √nX¯ , and apply the same orthogonal transformation also to Y1,..., Yn. Then Y 

n = √nY¯, n−1 i=1 X

iY 

i = n i=1

(Xi − X¯)(Yi − Y¯), n−1 i=1 X

i 2 = n i=1

(Xi − X¯)

2

, n−1 i=1 Y 

i 2 = n i=1

(Yi − Y¯)

2

.

The pairs of variables (X

1, Y 

1), . . . , (X

n, Y 

n) are independent, each with a bivariate normal distribution with the same variances and correlation as those of (X, Y ) and with means E(X

i) − E(Y 

i ) = 0 for i = 1,..., n − 1.

(ii): Consider first the joint distribution of S12 = xiYi and S2 2 = Y 2 i given x1 ..., xm. Letting Z1 = S12/

x 2 i and making an orthogonal transformation from Y1,..., Ym to Z1,..., Zm so that S2 2 = m i=1 Z2 i , the variables Z1 and m i=2 Z2 i =

S2 2 − Z2 1 are independently distributed as N(0, 1) and χ2 m−1, respectively. From this the joint conditional density of S12 = s1Z1 and S2 2 is obtained by a simple transformation of variables. Since the conditional distribution depends on the x’s only through s2 1 , the joint density of S2 1 , S12, S2 2 is found by multiplying the above conditional density by the marginal one of S2 1 , which is χ2 m. The proof is completed through use of the identity

 1 2 (m − 1)

 1 2m 
= √π(m − 1)
2m−2 .
(iii): If(X
, Y 
) = (X
1, Y 
1;...; X
m, Y 
m)is a sample from a bivariate normal distribution with ξ = η = 0, then T = (
X
i 2, X
iY 
i , Y 
i 2) is sufficient for θ(σ, ρ, τ ), and the density of T is obtained from that given in part (ii) for θ0 = (1, 0, 1) through the identity (Problem 3.41 (i))
pT θ (t) = pT θ0 (t)
p X
,Y 
θ (x
, y
)
p X
,Y 
θ0 (x
, y
)
.
The result now follows from part (i) with m = n − 1.]