Transformations Involving Joint Distributions

Transformations Involving Joint Distributions

Statistics 110 Summer 2006

Copyright c 2006 by Mark E. Irwin

Transformations Involving Joint Distributions

Want to look at problems like

? If X and Y are iid N (0, 2), what is the distribution of

?

Z

=

X2

+

Y

2

Gamma(1,

1 2

)

? U = X/Y C(0, 1)

? V = X - Y N (0, 22)

? What is the joint distribution of U = X + Y and V = X/Y if X Gamma(, ) and Y Gamma(, ) and X and Y are independent.

Approaches:

1.

CDF

approach

fZ(z) =

d dz

FZ

(z)

2.

Analogue to fY (y) = fX(g-1(y))

d dy

g-1(y)

(Density transformation)

Transformations Involving Joint Distributions

1

CDF approach: Let X1, X2, . . . , Xn have density fX1,X2,...,Xn(x1, x2, . . . , xn) and let Z = g(X1, X2, . . . , Xn). Let Az = {(x1, x2, . . . , xn) : g(x1, x2, . . . , xn) z}

FZ(z) = P [Z z] Then just differentiate this to get the density Example: Let Z = Y - X. Then Az = {(x, y) : y - x z} = {(x, y) : y x + z}

x+z

FZ(z) =

fX,Y (x, y)dydx

- -

=

fX,Y (x, y)dxdy

- y-z

Transformations Involving Joint Distributions

2

Making the change of variables x = y - u in the second form gives

z

FZ(z) =

fX,Y (y - u, y)dudy

- -

z

=

fX,Y (y - u, y)dydu

- -

Differentiating this gives the result

fZ(z) =

fX,Y (x, x + z)dx

-

by the change of variables x = y - z, the alternative form is derived

fZ(z) =

fX,Y (y - z, y)dy

-

Transformations Involving Joint Distributions

3

For example, let X and Y be independent N (0, 1) variables. Then the density of Z = Y - X is

fZ(z) =

1 exp 2

-

x2 2

1 exp 2

-

(x

+ 2

z)2

dx

= 1 2

1 exp 2

-2x2

+

2xz 2

+

z2

dx

= 1 1 exp 2 2

-2(x

+

z 2

)2

2

+

z2 2

dx

= 1 e-z2/4 2 exp

2 2

2

-2(x

+ 2

z 2

)2

dx

= 1 e-z2/4 2 2

so Z N (0, 2)

Transformations Involving Joint Distributions

4

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