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 .
2
DISCRETE CASE
• If B is any subset of 2, then (U,V)B iff
(X1,X2)A where
AU ,V  x1 , x2  : g1x1 , x2 , g 2 x1 , x2   B 
• 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
fU ,V u , v   Pr U  u ,V  v   Pr  X1 , X 2   AU ,V  

 x1 ,x2  AU ,V
f X 1 , X 2  x1 , x2 
3
2
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
Y1  g1  X 1 , X 2 , , X n  
Y2  g 2  X 1 , X 2 , , X n 
*


Yk  g k  X 1 , X 2 , , X n 
T
X Y
~
~
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CONTINUOUS CASE
• If the transformation T is one-to-one and onto,
than there is no problem of determining the
inverse transformation. An and Bk=n, then
T:AB. T-1(B)=A. It follows that there is a oneto-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
1
x1  g1  y1 , y2 , , yk  

x2  g 21  y1 , y2 , , yk  
* *


1
xn  g n  y1 , y2 , , yk  n 

1
g i / y i
• Assuming that the partial derivatives
exist at every point (y1, y2,…,yk=n)B. Under
these assumptions, we have the following
determinant J
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CONTINUOUS CASE
 g11
g11 



yn 
 y1
J  det 

 
1 
 g 1

g
n
 n 

yn 
 y1

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.
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CONTINUOUS CASE
• As a result, the function g is defined as
follows:
1 1
1


f
g
,
g
,

,
g
 X1,, X n 1 2
n | J |, for y1, y 2 ,, y n   B
gy , y ,, y  
1
2
n


0, otherwise
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Example
• Recall that I claimed: Let X1,X2,…,Xn be
independent rvs with Xi~Gamma(i, ). Then,
 X ~ Gamma   , 
n
i 1
i

n
i 1
i

• Prove this for n=2 (for simplicity).
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M.G.F. Method
• If X1,X2,…,Xn are independent random
variables
with MGFs Mxi (t), then the MGF
n
of Y   Xi is MY (t)  MX (t)...MX (t)
i 1
1
n
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Example
• Recall that I claimed:
~ Bin  n , p . Then,
independent
Let X
i
i
 X ~ Bin  n  n 
k
i 1
i
1
2
 n , p .
k
• Let’s prove this.
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Example
• Recall that I claimed: Let X1,X2,…,Xn be
independent rvs with Xi~Gamma(i, ). Then,

 X ~ Gamma   , 
n
i 1
i
n
i 1
i

• We proved this with transformation technique
for n=2.
• Now, prove this for general n.
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More Examples on Transformations
• Example 1:
• Recall that I claimed: If X~N( , 2), then
Z
X 

~ N (0,1)
• Let’s prove this.
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Example 2
• Recall that I claimed:
Let X be an rv with X~N(0, 1). Then,
X ~
2
2
1
Let’s prove this.
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Example 3
Recall that I claimed:
• If X and Y have independent N(0,1)
distribution, then Z=X/Y has a Cauchy
distribution with =0 and σ=1.
Recall the p.d.f. of Cauchy distribution:
f (x) 
1
 1  (
1
x 

)2
,  0
Let’s prove this claim.
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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:
X1 & X2 ~ Exp(1)
Y1=X1
Y2=X1+X2
X1 & X2 ~ Exp(1)
Y1=X1-X2
Y2=X1+X2
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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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