Chapter 6, Section 3 Transformations of Variables

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Chapter 6, Section 3 Transformations of Variables
Method of Distribution Functions
? John J Currano, 04/20/2009
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• The problem under consideration is the following
  
• Given some information about one or more random
  variables (their cdfs, pdfs, mgfs), find the
  distribution of some function of the random
  variable(s).
• We consider three techniques
• The Method of Distribution Functions (pdf or
  cdf?? cdf ? pdf)
• The Method of Transformations (pdf ?? pdf)
• The Method of Moment-Generating Functions (mgf ?
  mgf)
• We start with the Method of Distributions, which
  is not usually used with discrete random
  variables.

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• Method of Distribution Functions Univariate
  Continuous Case
• Y is a random variable whose pdf, f (y), and/or
  cdf, F (y), is known
• U r (Y) for some function r
• G (u) and g (u) are the cdf and pdf,
  respectively, of U
• Then

which is F (b) F (a) if y r (y)
? u a, b , and g (u) G?(u).
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Example 1. Y Uniform(?1, 1), so
U Y 2. Let G(u) denote the cdf of U.
Then G(u) for u 0, G(u) for u
1, and for 0 lt u lt 1, G(u)
Then Support(U) (0, 1).
0
1
P (Y 2 ? u )
P (U ? u )
.
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Example 1. Y Uniform(?1, 1), so
U Y 2. For 0 lt u lt 1, G(u)
. But so that and therefore to
compute G(u), we can use the fact that Y is
uniform for 0 lt u lt 1.
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Example 1. Y Uniform(?1, 1), so
U Y 2.
Thus,
and
so,
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Example 2. (p. 307 6.2) Find the pdfs of U1 3Y
and U2 3 Y. Solution. This problem is
easier to do if we first find F(y)
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U1 3Y U2 3 Y
Then
and
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U1 3Y U2 3 Y
Also, G2(u) Pr(U2 u) Pr(3 Y u) Pr(Y
3 u) 1 Pr(Y 3 u)
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U1 3Y U2 3 Y
So G2(u)
and
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Method of Distribution Functions Bivariate
Continuous Case Example 3. (The procedure is
similar to the univariate case.)
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Example 6.3 (Text, pages 301-304) (Y1, Y2) is a
random sample from the U(0, 1) distribution,
Here the shape of R and the limits of integration
depend on u
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Example 4. (Y1, Y2) is uniformly distributed on
the region 0 lt y2 lt y1 lt 20. Find the
distribution of U Y1 Y2. Since the joint
support of (Y1, Y2) is a triangle with area 200,
Once again the shape of R and the limits of
integration depend on u
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U Y1 Y2 Case 1 For u ? 0, G(u) 0. Case
2 For 0 lt u ? 20,
Since we have a uniform distribution, it can also
be calculated geometrically.
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U Y1 Y2 Case 3 For 20 lt u ? 40,
Case 4 For u gt 40, G(u) 1.
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Example 4. (Y1, Y2) is uniformly distributed on
the region 0 lt y2 lt y1 lt 20. Find the
distribution of U Y1 Y2.
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Using geometry to calculate G(u) where U Y1
Y2
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• Simulation of Distributions
• Assume
• U Uniform(0, 1), an easy distribution to
  simulate Note FU(u) P(U u) u for 0
  lt u lt 1.
• X F(x) with F(x) strictly ? on Support(X)
  Thus F(x) has an inverse F ?1 (0, 1) ?
  Support(X)
• V F ?1 (U)
• Then V has distribution function F that
  is,
• V and X have the same distribution.
• Therefore, V can be used to simulate X.

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• Simulation of Distributions
• Proof. We are given
• U Uniform(0, 1), so P(U u) u for 0 lt u
  lt 1
• X F(x) with F(x) strictly ? on Support(X)
• V F ?1(U)
• Then, for v ? F ?1 ( (0, 1) ) Support(X),
• FV (v) P(V v) P( F ?1(U) v )
• P( F ( F ?1(U) ) F(v) ) since F is ?
• P( U F(v) )
• F(v) by (1)
• Thus the distribution functions of V and X are
  equal, so V X.
• Note. Example 6.5 on p. 306 proves this fact when
  X has an exponential distribution.

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Example 4. (p. 309 6.17bc) Y belongs to the
power family with for parameters ??, ? gt 0. (b)
For fixed ? and ?, find a transformation G(u) so
that G(U) has distribution function F, where U
Uniform(0, 1). Then Y G(U).
Solution. By the previous example, we may take G
F sinceon Support(Y) (0, ? ),
is strictly increasing. To find G F , solve
?1
?1
Thus, if U Uniform(0, 1), then
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(c) Suppose now that we wish to get a random
sample of size 5 from a power family
distribution with ? 2 and ? 4. We first
simulate a random sample of size 5 from
Uniform(0, 1). Suppose we obtain 0.2700, 0.6901,
0.1413, 0.1523, and 0.3609. We can use these
values, and the transformation to obtain a
sample of size 5 from Y
Sometimes we must evaluate G(u) by numerical
methods e.g., normal, ?? 2.