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.