Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

132 Multivariate Distributions

For instance, in a simplified notation that appears to be clear,

μ 1 = E(X)=

∂M(0,0)

∂t 1

μ 2 = E(Y)=

∂M(0,0)

∂t 2

σ^21 = E(X^2 )−μ^21 =

∂^2 M(0,0)

∂t^21

−μ^21

σ^22 = E(Y^2 )−μ^22 =

∂^2 M(0,0)

∂t^22

−μ^22

E[(X−μ 1 )(Y−μ 2 )] =

∂^2 M(0,0)

∂t 1 ∂t 2

−μ 1 μ 2 , (2.5.10)

and from these we can compute the correlation coefficientρ.
It is fairly obvious that the results of equations (2.5.10) hold ifXandY are
random variables of the discrete type. Thus the correlation coefficients may be com-
puted by using the mgf of the joint distribution if that function is readily available.
An illustrative example follows.

Example 2.5.6(Example 2.1.10, Continued).In Example 2.1.10, we considered
the joint density
f(x, y)=

{

e−y 0 <x<y<∞

0elsewhere,

and showed that the mgf was

M(t 1 ,t 2 )=

1

(1−t 1 −t 2 )(1−t 2 )

,

fort 1 +t 2 <1andt 2 <1. For this distribution, equations (2.5.10) become

μ 1 =1,μ 2 =2

σ^21 =1,σ^22 = 2 (2.5.11)

E[(X−μ 1 )(Y−μ 2 )] = 1.

Verification of (2.5.11) is left as an exercise; see Exercise 2.5.5. If, momentarily, we
accept these results, the correlation coefficient ofXandYisρ=1/

√

2.

EXERCISES

2.5.1.Let the random variablesXandYhave the joint pmf

(a)p(x, y)=^13 ,(x, y)=(0,0),(1,1),(2,2), zero elsewhere.

(b)p(x, y)=^13 ,(x, y)=(0,2),(1,1),(2,0), zero elsewhere.

(c)p(x, y)=^13 ,(x, y)=(0,0),(1,1),(2,0), zero elsewhere.