Back to probabilities we have f y y 1 y 2 da f x x 1

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Back to probabilities. We have f Y ( y 1 , y 2 ) dA = f X ( x 1 , x 2 ) dx 1 dx 2 or f Y ( y 1 , y 2 ) | J | dx 1 dx 2 = f X ( x 1 , x 2 ) dx 1 dx 2 or f Y ( y 1 , y 2 ) = | J | - 1 f X ( x 1 , x 2 ) or, in general for an invertible function g , f Y ( y ) = | J | - 1 f X ( g - 1 ( y )) Example 11 Box-Muller transformation. Let X 1 ∼ U (0 , 1) and X 2 ∼ U (0 , 1) (indepen- dent). Let Y 1 = p - 2 ln X 1 cos 2 πX 2 Y 2 = p - 2 ln X 1 sin 2 πX 2 Then Y 1 ∼ N (0 , 1) Y 2 ∼ N (0 , 1) . 2 Many-to-one mappings Let Y = g ( X 1 , X 2 ) where X 1 and X 2 are jointly distributed random variables. For a given value of y , the inverse may form a curve in ( x 1 , x 2 ) space. Let A y denote the region in the X 1 X 2 plane such that g ( X 1 , X 2 ) y . (This may not be a connected region.) Then { Y y } = { g ( X 1 , X 2 ) y } = { ( X 1 , X 2 ) A y } so F Y ( y ) = Z Z A y f X 1 X 2 ( x 1 , x 2 ) dx 1 dx 2 Let Δ A y denote the region of the X 1 X 2 plane such that y < g ( x 1 , x 2 ) y + dy . then f Y ( y ) dy = Z Z Δ A y f X 1 X 2 ( x 1 , x 2 ) dx 1 dx 2 Example 12 Let Z = X + Y . The region in the xy plane such that x + y z is the part of plane below the line x + y = z . We have F Z ( z ) = Z -∞ Z z - y -∞ f XY ( x, y ) dxdy
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ECE 6010: Lecture 4 – Change of Variables 6 Differentiating this with respect to z we have f Z ( z ) = Z -∞ f XY ( z - y, y ) dy. Independence... 2 Example 13 Let Z = X/Y . The region of the plane such that x/y z can be determined as follows. Fix z. For y > 0 we want the region where x yz , and for y < 0 we want the region where x yz . We obtain F Z ( z ) = Z 0 Z yz -∞ f XY ( x, y ) dxdy + Z 0 -∞ Z zy f XY ( x, y ) dxdy. To get the density: The region Δ A z is the triangular sector bounded by the lines x = yz and x = y ( z + dz ) . The coordinates of a point are x = zy and y. The area of a differential is | y | dydz . We obtain f Z ( z ) dz = Z -∞ f XY ( zy, y ) | y | dydz Then cancel dz : f Z ( z ) = Z -∞ f XY ( zy, y ) | y | dy. 2 Example 14 Let Z = X 2 + Y 2 . The region A z is the circle x 2 + y 2 z 2 . If f XY ( x, y ) = 1 2 πσ 2 e - ( x 2 + y 2 ) / 2 σ 2 we have F Z ( z ) = 1 2 πσ 2 Z z 0 re - r 2 / 2 σ 2 dr = 1 - e - z 2 / 2 σ 2 z > 0 . so f Z ( z ) = z σ 2 e - z 2 / 2 σ 2 , z > 0 . This is a Rayleigh distribution.
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