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Module 13

Transformations of Absolutely Continuous

Random Variables

() Module 13 Transformations of Absolutely Continuous Random Variables

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X: an A.C. r.v. with d.f. FX() and p.d.f. fX();

FX(x) = x

fX(t)dt, x R;

For

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We will assume throughout that fX() is continuous everywhere except

at (possibly) finite number of points (say, x1, x2, . . . , xn), where it hasjump discontinuities. In that case FX() is differentiable everywhereexcept at discontinuity points{x1, . . . , xn} offX(). Moreover

fX(t) =

FX

(t), ift / {

x1

, . . . , xn}

0, otherwise.

Support

SX = {x R :FX(x+) FX(x )>0, >0} .

() Module 13 Transformations of Absolutely Continuous Random Variables

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g :R

R

: a given function;

Then Y =g(X) is a r.v.;

Goal: To find the probability distribution (i.e., d.f. FY() and/or,p.d.f./p.m.f. fY()) ofY =g(X);

Remark 1: We have seen that when X is discrete, Y =g(X) is also

discrete. When X is A.C., Y =g(X) may not be A.C. (or evencontinuous) as the following example illustrates.

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Example 1: Let Xbe an A.C. r.v. with p.d.f.

fX(x) =

12 , if 1< x

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P({

Y = 0}

) = P({

0

X

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Result 1: Suppose SX =k

i=1Si,X, where{Si,X, i= 1, . . . , k} is acollection of disjoint intervals and in Si,X (i= 1, ..., k), g :Si,X

R is

strictly monotone with inverse function g1i (y) such that ddy

g1i (y) iscontinuous. Let g(Si,X) = {g(x) :x Si,X}, i= 1, ..., k. Then the r.v.Y =g(X) is A.C. with p.d.f.

fY(y) =

ki=1

fX

g1i (y) ddyg1i (y)

Ig(Si,X)(y),

where, for a set A, IA() denotes its indicator function, i.e.,

IA(y) =

1, ify A0, otherwise

.

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Corollary 1: Suppose that g :SX R is strictly monotone with inversefunction g1(y) such that d

dyg1(y) is continuous. Let

g(SX) = {g(x) :x SX}. Then Y =g(X) is of A.C. type with p.d.f.

fY(y) =fX(g1(y))

ddyg1(y) Ig(SX)(y).

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Example 2: Let X be an A.C. r.v. with p.d.f.

fX(x) =

|x|2 , if 1

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S1,X = [1, 0) S2,X = [0, 2]g11 (y) =

y g12 (y) =

y

ddy

g11 (y) = 12y

ddy

g12 (y) = 12y

g(S1,X) = (0, 1] g(S2,X) = [0, 4)y g(S1,X) 0< y 1 y g(S2,X) 0 y 4fX(g

11 (y))

ddyg11 (y) fX(g12 (y))

ddyg12 (y)

=fX(

y)

12y I(0,1](y) =fX(

y)

12y I[0,4](y)

Thus a p.d.f. ofY is

fY(y) =2

i=1

fX(g1i (y))

d

dyg1i (y)

Ig(Si,X)(y)

=

12 , if 0

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FY(y) =P({Y y}) = y

fY(t)dt=

0, ify

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(b) For y

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For y 4, FY(y) = 1.

Thus the d.f. of Y is

FY(y) =

0, ify

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ClearlySY = [0, 4], FY() is differentiable everywhere except at points 0, 1and 4. Let

g(y) = FY(y), ify / {0, 1, 4}

0, otherwise

=

12 , if 0

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Take home problem

Let X be a r.v. with p.d.f.

fX(x) = ex, ifx>0

0, otherwise .

Let Y = 2X+ 3.

(a) Find the p.d.f. of Y and hence find the d.f. of Y;

(b) Find the d.f. of Y and hence find the p.d.f. of Y.

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Abstract of Next Module

X: a r.v. associated with a random experiment

E;

Each time the random experiment is performed we get a value ofX;

Question: If the random experiment is performed infinitely what is themean (or expectation) of observed values ofX org(X), for some functionreal-valued function g()?In the discrete case, the relative frequency interpretation of probabilitysuggests that we take

E(g(X)) = limN

xSX g(x) Number of times we get {X =x}

N

=xSX

g(x) limN

frequency of{X =x}N

=xSX

g(x)P({X =x}) =xSX

g(x)fX(x).

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