07jacobian - EECS 501 2-D EXAMPLE OF JACOBIAN METHOD Fall...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 501 2-D EXAMPLE OF JACOBIAN METHOD Fall 2001 Given: fx,y (X, Y ) = 6e−(3X +2Y ) for X, Y ≥ 0; 0 otherwise (2-D exponential). Goal: Compute fz,w (Z, W ) for transformation z = x + y and w = x/(x + y ). 1. Compute the inverse transformation of the given one: z = z (x, y ) = x + y w = w(x, y ) = x/(x + y ) →Inverse x = x(z, w) = zw y = y (z, w) = z (1 − w) since w = x/(x + y ) = x/z → x = zw and y = z − x = z − zw = z (1 − w). 2. Compute the Jacobian=determinant of the Jacobian matrix J: |J | = |det ∂z ∂x ∂w ∂x ∂z ∂y ∂w ∂y | = |det 1 y (x+y )2 1 −x (x+y )2 |=|− 1 x+y | = 1 x+y . since x, y ≥ 0 for this particular fx,y (X, Y ). 3. Substitute inverse transformation into fx,y (X, Y )/|J (X, Y )|: fz,w (Z, W ) = = fx,y (X,Y ) |J (X,Y )| |X =x(Z,W ),Y =y (Z,W ) fx,y (ZW,Z (1−W )) 1/(ZW +Z (1−W )) as defined above = 6Ze−(3ZW +2Z (1−W )) = 6Ze−Z (W +2) for Z ≥ 0 and 0 ≤ W ≤ 1 since x, y ≥ 0 → 0 ≤ w ≤ 1. 4. If desired, compute marginal pdfs for z and/or w: fz (Z ) = 6Ze−2Z fw ( W ) = ∞ 0 1 −ZW e dW 0 = 6(e−2Z − e−3Z ) for Z ≥ 0. 6Ze−Z (W +2) dZ = 6 (W +2)2 for 0 ≤ W ≤ 1. Check: Both marginal pdfs integrate to 1. Compare to exponential fx (X ), fy (Y ). Given: fx,y (X, Y ) = 9e−(3X +3Y ) for X, Y ≥ 0; 0 otherwise (2-D exponential). Goal: Compute fz,w (Z, W ) for transformation z = x + y and w = x/(x + y ). Now get fz,w (Z, W ) = 9Ze−3Z for Z ≥ 0 and 0 ≤ W ≤ 1. This is a 2nd -order Erlang or Gamma pdf in z ; a uniform pdf in w. Note that z and w are now independent random variables, unlike before. ...
View Full Document

This note was uploaded on 07/22/2011 for the course EECS 370 taught by Professor Lehman during the Spring '10 term at University of Florida.

Ask a homework question - tutors are online