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Example
• A device containing two key components fails when and only when both
components fail. The lifetime, T1 and T2, of these components are
independent with a common density function given by
 e t
fT t   
 0
t 0
otherwise
• The cost, X, of operating the device until failure is 2T1 + T2. Find the
density function of X.
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Convolution
• Suppose X, Y jointly distributed random variables. We want to find the
probability / density function of Z=X+Y.
• Discrete case
X, Y have joint probability function pX,Y(x,y). Z = z whenever X = x and
Y = z – x. So the probability that Z = z is the sum over all x of these joint
probabilities. That is
pZ z    p X ,Y x, z  x .
x
• If X, Y independent then
pZ z    p X x  pY z  x .
x
This is known as the convolution of pX(x) and pY(y).
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Example
• Suppose X~ Poisson(λ1) independent of Y~ Poisson(λ2). Find the
distribution of X+Y.
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Convolution - Continuous case
• Suppose X, Y random variables with joint density function fX,Y(x,y). We want to
find the density function of Z=X+Y.
Can find distribution function of Z and differentiate. How?
The Cdf of Z can be found as follows:

FZ  z   P X  Y  z  
zx
  f x, y dydx
X ,Y
x  y 

z
  f x, v  x dvdx

X ,Y
x  v 
z


  f x, v  x dxdv.
X ,Y
v  x 

If
 f x, v  x dx
XY
is continuous at z then the density function of Z is given by
x 
• If X, Y independent then
f Z z  
f Z z  

 f x, z  xdx
XY
x 

 f x f z  xdx
X
Y
x
This is known as the convolution of fX(x) and fY(y).
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Example
• X, Y independent each having Exponential distribution with mean 1/λ. Find
the density for W=X+Y.
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Some Recalls on Normal Distribution
• If Z ~ N(0,1) the density of Z is
z

 1
 Z z   
e 2

 2
2
,   z  
• If X = σZ + μ then X ~ N(μ, σ2) and the density of X is
 x 
 1

2
f X x   
e 2
 2 2
2
,   x  
• If X ~ N(μ, σ2) then
Z
X 

~ N 0,1.
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More on Normal Distribution
• If X, Y independent standard normal random variables, find the density of
W=X+Y.
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In general,
•
If X1, X2,…, Xn i.i.d N(0,1) then X1+ X2+…+ Xn ~ N(0,n).
• If X 1 ~ N 1 , 12  , X 2 ~ N 2 ,  22 ,…, X n ~ N  n ,  n2  then


X 1  X 2    X n ~ N 1     n ,  12     n2 .
• If X1, X2,…, Xn i.i.d N(μ, σ2) then
Sn = X1+ X2+…+ Xn ~ N(nμ,
nσ2)
 2 
Sn
and X n 
.
~ N   ,
n
n 

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Sum of Independent χ2(1) random variables
• Recall: The Chi-Square density with 1 degree of freedom is the
Gamma(½ , ½) density.
• If X1, X2 i.i.d with distribution χ2(1). Find the density of Y = X1+ X2.
• In general, if X1, X2,…, Xn ~ χ2(1) independent then
X1+ X2+…+ Xn ~ χ2(n) = Gamma(n/2, ½).
• Recall: The Chi-Square density with parameter n is



f X x   



n/2
1
 
1
n
 2  e 2 x x 2 1
n
2
0
 
0 x
otherwise
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Cauchy Distribution
• The standard Cauchy distribution can be expressed as the ration of two
Standard Normal random variables.
• Suppose X, Y are independent Standard Normal random variables.
Let Z  Y . Want to find the density of Z.
X
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Change-of-Variables for Double Integrals
• Consider the transformation , u = f(x,y), v = g(x,y) and suppose we are
interested in evaluating  F x, y dAxy .
Dxy
• Why change variables?
In calculus: - to simplify the integrand.
- to simplify the region of integration.
In probability, want the density of a new random variable which is a
function of other random variables.
• Example: Suppose we are interested in finding P A   f X ,Y x, y dxdy
. .
A
Further, suppose T is a transformation with T(x,y) = (f(x,y),g(x,y)) = (u,v).
Then, P A   f U ,V u, v dudv.
T  A
• Question: how to get fU,V(u,v) from fX,Y(x,y) ?
• In order to derive the change-of-variable formula for double integral, we
need the formula which describe how areas are related under the
transformation T: R2  R2 defined by u = f(x,y), v = g(x,y).
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Jacobian
• Definition: The Jacobian Matrix of the transformation T is given by
 f

x
J T  x, y   
 g
 x

f 

y    u , v  


g    x, y  
y 
• The Jacobian of a transformation T is the determinant of the Jacobian
matrix.
• In words: the Jacobian of a transformation T describes the extent to which
T increases or decreases area.
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Change-of-Variable Theorem in 2-dimentions
• Let x = f(u,v) and y = g(u,v) be a 1-1 mapping of the region Auv onto Axy
with f, g having continuous partials derivatives and det(J(u,v)) ≠ 0 on Auv.
If F(x,y) is continuous on Axy then
 F x, y dxdy   F  f u, v , g u, v  J u, v  dudv
Axy
where
x
J u, v   u
y
u
Auv
x
v  1
y J x, y 
v
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Example
• Evaluate  xydxdy where Axy is bounded by y = x, y = ex, xy = 2 and xy = 3.
Axy
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Change-of-Variable for Joint Distributions
• Theorem
Let X and Y be jointly continuous random variables with joint density
function fX,Y(x,y) and let DXY = {(x,y): fX,Y(x,y) >0}. If the mapping T given
by T(x,y) = (u(x,y),v(x,y)) maps DXY onto DUV. Then U, V are jointly
continuous random variable with joint density function given by
 f xu , v , y u , v  J u , v 
fU ,V u , v    X ,Y
0

if u, v  DU ,V
otherwise
where J(u,v) is the Jacobian of T-1 given by
x
J u, v   u
y
u
x
v
y
v
assuming derivatives exists and are continuous at all points in DUV .
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Example
• Let X, Y have joint density function given by
e  x y
f X ,Y x, y   
 0
Find the density function of U 
if x, y  0
otherwise
X
.
X Y
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Example
• Show that the integral over the Standard Normal distribution is 1.
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Density of Quotient
• Suppose X, Y are independent continuous random variables and we are
Y
interested in the density of
Z .
X
y
• Can define the following transformation z  , w  x .
x
• The inverse transformation is x = w, y = wz. The Jacobian of the inverse
transformation is given by
x
J w, z   w
y
w
x
z  1 0  w
y z w
z
• Apply 2-D change-of-variable theorem for densities to get
fW ,Z w, z   f X ,Y w, wz  w  f X w fY wz  w
• The density for Z is then given by

f Z z    f X w fY wz  w dw

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Example
• Suppose X, Y are independent N(0,1). The density of Z 
week 9
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is
X
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Example – F distribution
X /n
.
• Suppose X ~ (n) independent of Y ~ (m). Find the density of Z 
Y /m
χ2
χ2
• This is the Density for a random variable with an F-distribution with
parameters n and m (often called degrees of freedom). Z ~ F(n,m).
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Example – t distribution
• Suppose Z ~ N(0,1) independent of X ~ χ2(n). Find the density of T 
Z
X
.
n
• This is the Density for a random variable with a t-distribution with
parameter n (often called degrees of freedom). T ~ t(n)
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Some Recalls on Beta Distribution
• If X has Beta(α,β) distribution where α > 0 and β > 0 are positive
parameters the density function of X is
      1
 1
x 1  x 
0  x 1

f X x      


0
otherwise
• If α = β = 1, then X ~ Uniform(0,1).
• If α = β = ½ , then the density of X is
1


f X x    x1  x 

0

for 0  x  1
otherwise
• Depending on the values of α and β, density can look like:
• If X ~ Beta(α,β) then E  X  

 
and V  X  
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
.
2
        1
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Derivation of Beta Distribution
• Let X1, X2 be independent χ2(1) random variables. We want the density of
X1
X1  X 2
• Can define the following transformation
Y1 
X1
X1  X 2
,
Y2  X 1  X 2
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