Chapter 6, Section 6 Transformations of Variables

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Chapter 6, Section 6 Transformations of Variables
Method of Transformations Bivariate Distributions
? John J Currano, 12/05/2007
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• Method of Transformations Bivariate Cases
• There are two possible situations of interest
• U h (Y1, Y2)
• U1 h1(Y1, Y2), U2 h2(Y1, Y2).
• They are handled the same way in case 1, we let
  U1 U and U2 be either Y1 or Y2 and use case 2
  methods.

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Method of Transformations Bivariate Discrete
Case We are given (Y1, Y2) discrete with joint
probability function f (y1, y2) and support A.
We use f instead of p for the joint pf. If (y1,
y2) ? A, let u1 h1(y1, y2) and u2 h2(y1,
y2) and suppose that the function T (y1, y2)
? (u1, u2) defines a 1 1 function from A onto
the set B of images. Then there is an inverse
function T ?1 (u1, u2) ? (y1, y2) which
defines two functions from B to A y1 w1(u1,
u2) and y2 w2(u1, u2).
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• Method of Transformations Bivariate Discrete
  Case
• (Y1, Y2) discrete, joint probability function f
  (y1, y2), support A.
• T (y1, y2) ? (u1, u2), u1 h1(y1, y2),
  u2 h2(y1, y2)
• defines a 1 1 function from A onto the set B
  of images.
• T ?1 (u1, u2) ? (y1, y2), y1 w1(u1, u2),
  y2 w2(u1, u2).
• Then the random vector (U1, U2) ( h1(y1, y2),
  h2(y1, y2) )
• has probability function

To find the probability function of U1, sum over
all values of U2 to find the probability
function of U2, sum over all values of U1.
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Example. Let Y1 Poisson(?1) and Y2
Poisson(?2) be independent. Find the
probability function of U1 Y1 Y2 .
Solution. Introduce a random variable U2 so
that (y1, y2) ? (u1, u2) is a simple 1 1
transformation. U2 Y2 works well here
Since Y1 and Y2 are independent,
for y1 0, 1, 2, . . . and y2 0, 1, 2, . .
. .
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Example. Let Y1 Poisson(?1) and Y2
Poisson(?2) be independent. Find the
probability function of U1 Y1 Y2 .
for y1 0, 1, 2, . . . and y2 0, 1, 2, .
. .
for u1 0, 1, 2, . . . and u2 0, 1, 2, .
. . , u1 . Note. (u1, u2 ) ? B ?? 0
u2 u1 ? ? y2 y1 y2
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Example. Let Y1 Poisson(?1) and Y2
Poisson(?2) be independent. Find the
probability function of U1 Y1 Y2 .
for u1 0, 1, 2, . . . and u2 0, 1, 2, .
. . , u1
Then,
Thus, U1 Y1 Y2 Poisson(?1 ?2 ).
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Method of Transformations Bivariate Continuous
Case Suppose we are given (Y1, Y2) continuous
with joint pdf f (y1, y2) and joint support A
and a 1 1 transformation T A ? ??2. For
(y1, y2) ? A, let T ( (y1, y2) ) (u1, u2).
This defines two functions of y1 and y2 u1
h1(y1, y2) u2 h2(y1, y2) T is usually
specified by telling what h1 and h2 are.
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Method of Transformations Bivariate Continuous
Case
Example 1.
Example 2.
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Method of Transformations Bivariate Continuous
Case Given (Y1, Y2) f (y1, y2) continuous with
joint support A T A ? ??2, a 1 1
transformation T ( (y1, y2) ) ( h1(y1, y2),
h2(y1, y2) ) U1 h1(Y1, Y2) and U2 h2(Y1,
Y2). We want to find the joint pdf, g (u1, u2),
of (U1, U2).
Procedure. Let B T(A) Support of (U1,
U2). T A ? B is 1 1 ? T ?1 B ?
A exists and is 1 1 T ?1 can be found by
solving the equations ui hi(y1, y2) for y1
and y2 as in the examples on the previous
slide. This yields
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Method of Transformations Bivariate Continuous
Case Given (Y1, Y2) f (y1, y2) continuous with
joint support A
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Method of Transformations Bivariate Continuous
Case Given (Y1, Y2) f (y1, y2) continuous with
joint support A
Then a theorem from real analysis implies g (u1,
u2) f ( w1 (u1, u2), w2 (u1, u2) ) J where
J is the absolute value of the Jacobian, J,
provided that J is not identically zero on B
T(A).
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Example 1. f (y1, y2) 6y1, for 0 y1, 0 U1 Y1 Y2 . Solution. Let U2 Y2 , and let
T be the transformation from A Support of (Y1,
Y2) (y1, y2) 0 to B Support of (U1, U2) ?? Then T is
the transformation of Example 1 on slide 9, so
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U1 Y1 Y2 , U2 Y2 , T A (y1, y2)
0 How do we find B? There is a neat geometric
trick. Sketch the region A, the Support of (Y1,
Y2), labeling its boundaries with their
equations. Then substitute into the boundary
equations the expressions for y1 and y2 in terms
of u1 and u2 from T ?1 above these are the
equations for the boundaries of B.
y1 0 ? u1 u2 y2 0 ? u2 0 y1
y2 1 ? u1 1
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U1 Y1 Y2 , U2 Y2 , T A (y1, y2)
0 y1 0 ? u1 u2 y2 0 ? u2 0 y1
y2 1 ? u1 1
Thus, B (u1, u2) 0 , g(u1, u2) 6(u1 u2) for 0 0 elsewhere
? This example can also be done using the
Distribution ?? ?? Function Technique
try it and compare. ?
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Example 2. Given two independent gamma random
variables, Y1 Gamma(?, 1), and
Y2 Gamma(?, 1),
T is the transformation of Example 2 on slide
9. We wish to find the joint pdf of (U1, U2)
T(Y1, Y2) and the marginal distributions of both
U1 and U2. Let A joint support of (Y1, Y2)
(y1, y2) y1 0, y2 0 and B
joint support of (U1, U2) T(A). What is B?
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A joint support of (Y1, Y2) (y1, y2) y1
0, y2 0 B joint support of (U1, U2)
T(A). What is B?
The boundaries of A are the lines y1 0 and
y2 0, so the boundaries of B are
u1 u2 0 ? u1 0 or u2 0, and u1
u1 u2 0 ? u1 (1 u2) 0 ? u1
0 or (1 u2) 0 ? u1 0 or u2 1.
Thus, B (u1, u2) 0 0 .
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A support of (Y1, Y2) (y1, y2) y1 0,
y2 0 B support of (U1, U2) (u1, u2)
0 0
The Jacobian of the transformation, T, is
which is not identically zero on B.
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Y1 Gamma(?, 1), , and Y2 Gamma(?,
1), are independent,
A (y1, y2) y1 0, y2 0 , and B
(u1, u2) 0 0 . Since Y1
and Y2 are independent,
on B (u1, u2) 0 0 (and
0 elsewhere).
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Since u1 is positive on on B (u1, u2) 0 u2 0 , ? u1 u1 , and therefore
Since B (u1, u2) 0 0 is
a rectangle, g(u1, u2) factors into h1(u1)
h2(u2) and therefore U1 and U2 are
independent. Furthermore, the variable portions
of the marginal density functions of U1 and U2
are and , respectively,
so that U1 Gamma(? ? , 1 ) and U2
Beta(? , ? ).
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Summary. If Y1 and Y2 are two independent
gamma random variables with Y1 Gamma(?, 1)
and Y2 Gamma(?, 1), then U1 Y1 Y2
Gamma(? ? ) and
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Example 3. Suppose that Y1 and Y2 are
independent continuous random variables, so that
f (y1, y2) f1(y1) f2(y2). Then if U1 h1(Y1)
and U2 h2(Y2) where both h1 and h2 are 1
1 differentiable functions,
so U1 and U2 are independent. In fact,
U1 and U2 are independent even if h1 or h2 is not
1 1.