# If x and y are independent then e xy e x e y which

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eimage of S under the linear transformation. Since S is a small set, R is also a small set. So P {(W, Z ) ∈ S } = P {(X, Y ) ∈ R} ≈ fX,Y (u, v )area(R). ) 1 Thus, fW,Z (α.β ) ≈ fX,Y (x, y ) area(R) = | det A| fX,Y (x, y ). This observation leads to the following area(S proposition: Proposition 4.7.1 Suppose W = A X , where Z Y det(A) = 0. Then W has joint pdf given by Z fW,Z (α, β ) = X Y 1 fX,Y | det A| has pdf fX,Y , and A is a matrix with A−1 α β . Example 4.7.2 Suppose X and Y have joint pdf fX,Y , and W = X − Y and Z = X + Y. Express the joint pdf of W and Z in terms of fX,Y . 146 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES Solution: We apply Proposition 4.7.1, using A= 1 −1 11 det(A) = 2 1 2 A−1 = 1 −2 1 2 1 2 . u α v to β such that α = u −α+β u 2 , or equivalently, v = That is, the linear transformation used in this example maps α+β 2 β = u + v. The inverse mapping is given by u = Proposition 4.7.1 yields: 1 fW,Z (α, β ) = fX,Y 2 and v = α + β −α + β , 2 2 − v and A−1 α β . , for all (α, β ) ∈ R2 . Example 4.7.3 Suppose X and Y are independent, continuous-type random variables. Find the joint pdf of W and Z, where W = X + Y and Z = Y. Also, ﬁnd the pdf of W. Solution: We again apply Proposition 4.7.1, th...
View Full Document

## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

Ask a homework question - tutors are online