Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

130 Multivariate Distributions

which is the desired result.

Note that if the variance, Equation (2.5.9), is denoted byk(x), thenE[k(X)] =
σ^22 (1−ρ^2 )≥0. Accordingly,ρ^2 ≤1, or− 1 ≤ρ≤1. This verifies Theorem 2.5.1
for the special case of linear conditional means.
As a corollary to Theorem 2.5.3, suppose that the variance, Equation (2.5.9), is
positive but not a function ofx; that is, the variance is a constantk>0. Now ifk
is multiplied byf 1 (x) and integrated onx, the result isk,sothatk=σ^22 (1−ρ^2 ).
Thus, in this case, the variance of each conditional distribution ofY,givenX=x,is
σ 22 (1−ρ^2 ). Ifρ= 0, the variance of each conditional distribution ofY,givenX=x,
isσ 22 , the variance of the marginal distribution ofY. On the other hand, ifρ^2 is near
1, the variance of each conditional distribution ofY,givenX=x, is relatively small,
and there is a high concentration of the probability for this conditional distribution
near the meanE(Y|x)=μ 2 +ρ(σ 2 /σ 1 )(x−μ 1 ). Similar comments can be made
aboutE(X|y) if it is linear. In particular,E(X|y)=μ 1 +ρ(σ 1 /σ 2 )(y−μ 2 )and
E[Var(X|Y)] =σ^21 (1−ρ^2 ).

Example 2.5.4. Let the random variablesXandY have the linear conditional
meansE(Y|x)=4x+3 andE(X|y)= 161 y−3. In accordance with the general
formulas for the linear conditional means, we see thatE(Y|x)=μ 2 ifx=μ 1 and
E(X|y)=μ 1 ify=μ 2. Accordingly, in this special case, we haveμ 2 =4μ 1 +3
andμ 1 = 161 μ 2 −3sothatμ 1 =−^154 andμ 2 =−12. The general formulas for the
linear conditional means also show that the product of the coefficients ofxandy,
respectively, is equal toρ^2 and that the quotient of these coefficients is equal to
σ 22 /σ 12 .Hereρ^2 =4( 161 )=^14 withρ=^12 (not−^12 ), andσ^22 /σ^21 = 64. Thus, from the
two linear conditional means, we are able to find the values ofμ 1 ,μ 2 ,ρ,andσ 2 /σ 1 ,
but not the values ofσ 1 andσ 2.

y

x

• h (0, 0)
  
  
  • a
  
  
  
  

a

E(Y|x) = bx

h

Figure 2.5.1:Illustration for Example 2.5.5.