(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

4πΓ(m − 1) (s 2

1s 2

2 − s 2

12)

1 2 (m−3) exp

−1 2

(s 2

1 + s 2

2)



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

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

(s2 1s2 2 − s2 12)

1 2 (n−4)

4πΓ(n − 2) 

στ1 − ρ2

n−1 exp

− 1 2(1 − ρ2)

s2 1

σ2 − 2ρs12

στ + s2 2

τ 2



(5.84)

for s2 12 ≤ s2 1s2 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/

x2 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 Γ
#
1 2 (m − 1)$
Γ
1 2m 
= √πΓ(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 3.39 (i)]
pT θ (t) = pT θ0 (t)
pX
,Y

θ (x

, y

)
pX
,Y

θ0 (x
, y
)
.
The result now follows from part (i) with m = n − 1.]