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Random vectors

Establish the correlation matrix for W, X, Y, Z. ,

Solution. The means of the variables are given by

1 1 3 2 16
µ W = 2 × = , µ X = 4 × + 2 × = ,
4 2 13 13 13
1 6 3
µ Y = 1 × , µ Z = 2 × = .
2 52 13
2
2
2
The variances, computed from σ = Var(U) = E(U ) − (E(U)) , where U = W, X, Y or Z,
U
are
( ) 2 ( ) 2
1 1 3 3 2 16 472
2
σ 2 = 4 × − = ; σ = 16 × + 4 × − = ,
W
4 2 4 X 13 13 13 169
( ) 2 ( ) 2
1 1 1 6 3 69
2 2
σ = 1 × − = , σ = 4 × − = .
Z
Y
2 2 4 52 13 169
The covariances are found by first calculating E(WX) etc. and then forming E(WX)−µ W µ X
etc.
2 2 8 8 1 16
E[WX] = 2 · 4 · + 2 · 2 · = , Cov[W, X] = − · = 0,
52 52 13 13 2 13
1 1 1 1 1 1
E[WY ] = 2 · 1 · = , Cov[W, Y ] = − · = ,
4 2 2 2 2 4
1 3 3
E[WZ] = 0, Cov[W, Z] = 0 − · = − ,
2 13 26
6 4 8 8 6 1
E[XY ] = 4 · 1 · + 2 · 1 · = , Cov[X, Y ] = − · = 0,
52 52 13 13 13 2
6 12 12 16 3 108
E[XZ] = 4 · 2 · = , Cov[X, Z] = − · = ,
52 13 13 13 13 169
1 3 3
E[Y Z] = 0, Cov[Y Z] = 0 − · = − .
2 13 26
The correlations Corr[W, X] and Corr[X, Y ] are clearly zero; the remainder are given by
( ) −1/2
1 3 1
Corr[W, Y ] = · = 0.577;
4 4 4
( ) −1/2
3 3 69
Corr[W, Z] = − · = −0.209;
26 4 169
( ) −1/2
108 472 69
Corr[X, Z] = · = 0.598;
169 169 169
( ) −1/2
3 1 69
Corr[Y, Z] = − · = −0.361.
26 4 169
Finally, then, we can write down the correlation matrix:
 
1 0 0.58 −0.21
 0 1 0 0.6 
ρ =   .
 0.58 0 1 −0.36 
−0.21 0.6 −0.36 1

As would be expected, X is uncorrelated with either W or Y, colour and face-value being two
independent characteristics. Positive correlations are to be expected between W and Y and
between X and Z; both correlations are fairly strong. Moderate anticorrelations exist between
Z and both W and Y, reflecting the fact that it is impossible for W and Y to be positive if Z is
positive.

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