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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 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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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
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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