Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

94 Multivariate Distributions

Likewise if (X 1 ,X 2 ) is discrete, thenE(Y)existsif

∑

x 1

∑

x 2

|g(x 1 ,x 2 )|pX 1 ,X 2 (x 1 ,x 2 )<∞.

Then
E(Y)=

∑

x 1

∑

x 2

g(x 1 ,x 2 )pX 1 ,X 2 (x 1 ,x 2 ). (2.1.11)

We can now show thatEis a linear operator.

Theorem 2.1.1.Let(X 1 ,X 2 )be a random vector. LetY 1 =g 1 (X 1 ,X 2 )andY 2 =
g 2 (X 1 ,X 2 )be random variables whose expectations exist. Then for all real numbers
k 1 andk 2 ,
E(k 1 Y 1 +k 2 Y 2 )=k 1 E(Y 1 )+k 2 E(Y 2 ). (2.1.12)

Proof:We prove it for the continuous case. The existence of the expected value of
k 1 Y 1 +k 2 Y 2 follows directly from the triangle inequality and linearity of integrals;
i.e.,

∫∞

−∞

∫∞

−∞

|k 1 g 1 (x 1 ,x 2 )+k 2 g 1 (x 1 ,x 2 )|fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2

≤|k 1 |

∫∞

−∞

∫∞

−∞

|g 1 (x 1 ,x 2 )|fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2

+|k 2 |

∫∞

−∞

∫∞

−∞

|g 2 (x 1 ,x 2 )|fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2 <∞.

By once again using linearity of the integral, we have

E(k 1 Y 1 +k 2 Y 2 )=

∫∞

−∞

∫∞

−∞

[k 1 g 1 (x 1 ,x 2 )+k 2 g 2 (x 1 ,x 2 )]fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2

= k 1

∫∞

−∞

∫∞

−∞

g 1 (x 1 ,x 2 )fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2

+k 2

∫∞

−∞

∫∞

−∞

g 2 (x 1 ,x 2 )fX 1 ,X 2 (x 1 ,x 2 )dx 1 dx 2

= k 1 E(Y 1 )+k 2 E(Y 2 ),

i.e., the desired result.

We also note that the expected value of any functiong(X 2 )ofX 2 can be found
in two ways:

E(g(X 2 )) =

∫∞

−∞

∫∞

−∞

g(x 2 )f(x 1 ,x 2 )dx 1 dx 2 =

∫∞

−∞

g(x 2 )fX 2 (x 2 )dx 2 ,

the latter single integral being obtained from the double integral by integrating on

x 1 first. The following example illustrates these ideas.