TRANSFORMATION OF FUNCTION OF A RANDOM

VARIABLE

UNIVARIATE TRANSFORMATIONS

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TRANSFORMATION OF RANDOM VARIABLES

• If X is an rv with cdf F(x), then Y=g(X) is also an rv.

• If we write y=g(x), the function g(x) defines a mapping from the original sample space of X, S, to a new sample space, , the sample space of the rv Y.

g(x): S

TRANSFORMATION OF RANDOM VARIABLES

• Let y=g(x) define a 1-to-1 transformation. That is, the equation y=g(x) can be solved uniquely:

• Ex: Y=X-1 X=Y+1 1-to-1

• Ex: Y=X² X=± sqrt(Y) not 1-to-1

• When transformation is not 1-to-1, find disjoint partitions of S for which transformation is 1-to-1.

3

)y(gx 1

4

TRANSFORMATION OF RANDOM VARIABLES

If X is a discrete r.v. then S is countable. The sample space for Y=g(X) is ={y:y=g(x),x S}, also countable. The pmf for Y is

1 1

Yx g y x g y

f y P Y y P X x f x

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Example

• Let X~GEO(p). That is,• Find the p.m.f. of Y=X-1• Solution: X=Y+1

• P.m.f. of the number of failures before the first success• Recall: X~GEO(p) is the p.m.f. of number of Bernoulli

trials required to get the first success

,...3,2,1xfor)p1(p)x(f 1x

,...2,1,0yfor)p1(p)1y(f)y(f yXY

Example

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• Let X be an rv with pmf

1/5, 2

1/ 6, 1

1/5, 0

1/15, 1

11/30, 2

x

x

p x x

x

x

Let Y=X2. S ={2, 1,0,1,2} ={0,1,4}1/5, 0

( ) 7 /30, 1

17 /30, 4

y

p y y

y

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FUNCTIONS OF CONTINUOUS RANDOM VARIABLE

• Let X be an rv of the continuous type with pdf f. Let y=g(x) be differentiable for all x and non-zero. Then, Y=g(X) is also an rv of the continuous type with pdf given by

..0

|)(|))(()(

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wo

yforygdy

dygf

yh

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FUNCTIONS OF CONTINUOUS RANDOM VARIABLE

• Example: Let X have the density

1, 0 1

0, otherwise

xf x

Let Y=eX.X=g1 (y)=log Y dx=(1/y)dy.

11. ,0 log 1

1, 1

0, otherwise

h y yy

y eyh y

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FUNCTIONS OF CONTINUOUS RANDOM VARIABLE

• Example: Let X have the density

2 / 21, .

2xf x e x

Let Y=X2. Find the pdf of Y.

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THE PROBABILITY INTEGRAL TRANSFORMATION

• Let X have continuous cdf FX(x) and define the rv Y as Y=FX(x). Then, Y is uniformly distributed on (0,1), that is,

P(Y y) = y, 0<y<1.

• This is very commonly used, especially in random number generation procedures.

Example 1

• Generate random numbers from X~ Exp(1/λ) if you only have numbers from Uniform(0,1).

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Example 2

• Generate random numbers from the distribution of X(1)=min(X1,X2,…,Xn) if X~ Exp(1/λ) if you only have numbers from Uniform(0,1).

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Example 3

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• Generate random numbers from the following distribution:

CDF method

• Example: Let

Consider . What is the p.d.f. of Y?

• Solution:

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0xfore1)x(F x2

XeY

1yfory2)y(Fdy

d)y(f

1yfory1)y(lnF

)ylnX(P)ye(P)yY(P)y(F

3YY

2X

XY

CDF method

• Example: Consider a continuous r.v. X, and Y=X². Find p.d.f. of Y.

• Solution:

15

)]y(f)y(f[y2

1

)y(dy

d)y(f)y(

dy

d)y(f)y(f

)y(F)y(F)yXy(P)yX(P)y(F

XX

XXY

XX2

Y

TRANSFORMATION OF FUNCTION OF TWO OR MORE

RANDOM VARIABLES

BIVARIATE TRANSFORMATIONS

DISCRETE CASE

• Let X1 and X2 be a bivariate random vector with a known probability distribution function. Consider a new bivariate random vector (U, V) defined by U=g1(X1, X2) and V=g2(X1, X2) where g1(X1, X2) and g2(X1, X2) are some functions of X1 and X2 .

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DISCRETE CASE• If B is any subset of 2, then (U,V)B iff

(X1,X2)A where

• Then, Pr(U,V)B=Pr(X1,X2)A and probability distribution of (U,V) is completely determined by the probability distribution of (X1,X2). Then, the joint pmf of (U,V) is

221221121 Bx,xg,x,xg:x,xA V,U

V,UAx,x

X,XV,UV,U x,xfAX,XPrvV,uUPrv,uf21

21 212118

EXAMPLE

• Let X1 and X2 be independent Poisson distribution random variables with parameters 1 and 2. Find the distribution of U=X1+X2.

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CONTINUOUS CASE• Let X=(X1, X2, …, Xn) have a continuous joint

distribution for which its joint pdf is f, and consider the joint pdf of new random variables Y1, Y2,…, Yk defined as

*

X,,X,XgY

X,,X,XgY

X,,X,XgY

nkk

n

n

21

2122

2111

YX~

T

~

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CONTINUOUS CASE

• If the transformation T is one-to-one and onto, then there is no problem of determining the inverse transformation. An and Bk=n, then T:AB. T-1(B)=A. It follows that there is a one-to-one correspondence between the points (y1, y2,…,yk) in B and the points (x1, x2,…,xn) in A. Therefore, for (y1, y2,…,yk)B we can invert the equation in (*) and obtain new equation as follows:

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CONTINUOUS CASE

• Assuming that the partial derivatives exist at every point (y1, y2,…,yk=n)B. Under these assumptions, we have the following determinant J

**

y,,y,ygx

y,,y,ygx

y,,y,ygx

nknn

k

k

211

211

22

211

11

ii y/g 1

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CONTINUOUS CASE

called as the Jacobian of the transformation specified by (**). Then, the joint pdf of Y1, Y2,…,Yk can be obtained by using the change of variable technique of multiple variables.

n

nn

n

y

g

y

g

y

g

y

g

detJ1

1

1

11

1

11

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CONTINUOUS CASE

• As a result, the function g is defined as follows:

otherwise,0

By,,y,y|,J|g,,g,gfy,,y,yg n21

1n

12

11nX,,1X

n21

for

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Example

• Recall that I claimed: Let X1,X2,…,Xn be independent rvs with Xi~Gamma(i, ). Then,

• Prove this for n=2 (for simplicity).

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1 1

~ ,n n

i ii i

X Gamma

M.G.F. Method

• If X1,X2,…,Xn are independent random variables with MGFs Mxi (t), then the MGF of is

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n

1iiXY )t(M)...t(M)t(M nX1XY

Example

• Recall that I claimed: Let X1,X2,…,Xn be independent rvs with Xi~Gamma(i, ). Then,

• We proved this with transformation technique for n=2.

• Now, prove this for general n.27

1 1

~ ,n n

i ii i

X Gamma

Example

• Recall that I claimed:

• Let’s prove this.

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Let ~ , . ,independent

i iX Bin n p Then

1 21

~ , .k

i ki

X Bin n n n p

More Examples on Transformations

• Example 1:

• Recall the relationship:

If , then X~N( , 2)

• Let’s prove this.

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~ (0,1)X

Z N

Example 2

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• Recall that I claimed:

Let X be an rv with X~N(0, 1). Then, 2 2

1~X

Let’s prove this.

Example 3

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• If X and Y have independent N(0,1) distribution, then Z=X/Y has a Cauchy distribution with =0 and σ=1.

Recall that I claimed:

0,)

x(1

11)x(f

2

Recall the p.d.f. of Cauchy distribution:

Let’s prove this claim.

Example 4

• See Examples 6.3.12 and 6.3.13 in Bain and Engelhardt (pages 207 & 208 in 2nd edition). This is an example of two different transformations:

• In Example 6.3.12: In Example 6.3.13:

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X1 & X2 ~ Exp(1)Y1=X1

Y2=X1+X2

X1 & X2 ~ Exp(1)Y1=X1-X2

Y2=X1+X2

Example 5

• Let X1 and X2 are independent with N(μ1,σ²1) and N(μ2,σ²2), respectively. Find the p.d.f. of Y=X1-X2.

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Example 6

• Let X~N( , 2) and Y=exp(X). Find the p.d.f. of Y.

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