Download Functions of Random Variables

Functions of Random Variables
Often we have to consider random variables which are functions of other random
variables. Let X be a random variable and g (.) is a function. Then Y  g ( X ) is a
random variable. We are interested to find the pdf of Y . For example, suppose
X represents the random voltage input to a full-wave rectifier. Then the rectifier output
Y is given by Y  X . We have to find the probability description of the random variable
Y . We consider the following cases:
(a) X is a discrete random variable with probability mass function p X ( x)
The probability mass function of Y is given by
pY ( y )  P(Y  y)
 P( x | g ( x)  y)


P( X  x)

p X ( x)
x| g ( x )  y

x| g ( x )  y
(b) X is a continuous random variable with probability density function
y  g ( x) is one-to-one and monotonically increasing
The probability distribution function of Y is given by
f X ( x) and
FY ( y )  P Y  y
 P  g ( X )  y
 P  X  g 1 ( y )
 P( X  x)  x  g 1 ( y )
 FX ( x) x  g 1 ( y )
fY ( y ) 
dFY ( y )
dy

dFX ( x) 
dy  x  g 1 ( y )

dFX ( x) dx 
dx dy  x  g 1 ( y )
dFX ( x) 

 dx 
dy

dx  x  g 1 ( y )

fY ( y ) 
f X ( x) 
g ( x)  x  g 1 ( y )
f X ( x) f X ( x) 

dy
g ( x)  x  g 1 ( y )
dx
This is illustrated in Fig.
Example 1: Probability density function of a linear function of a random variable
Suppose Y  aX  b, a  0.
y b
dy
Then x 
and
a
a
dx
y b
fX (
)
f X ( x)
a
 fY ( y ) 

dy
a
dx
Example 2: Probability density function of the distribution function of a random
variable
Suppose the distribution function FX ( x) of a continuous random variable X is
monotonically increasing and one-to-one and define the random variable
Y  FX ( X ). Then, fY ( y)  1 0  y  1.
y  FX ( x)
Clearly 0  y  1
dy dFX ( x)

 f X ( x)
dx
dx
f ( x) f X ( x)
 fY ( y )  X

1
dy
f X ( x)
dx
 fY ( y )  1 0  y  1.
Remark
(1) The distribution given by fY ( y)  1 0  y  1 is called a uniform distribution over the
interval [0,1].
(2) The above result is particularly important in simulating a random variable with a
particular distribution function. We assumed FX ( x) to be one-to-one function for
invariability. However, the result is more general- the random variable defined by the
distribution function of any random variable is uniformly distributed over [0,1]. For
example, if X is a discrete RV,
FY ( y ) =P(Y  y )
 P( FX ( x)  y )
 P( X  FX1 ( y ))
 FX ( FX1 ( y ))
 y ( Assigning FX1 ( y ) to the left-most point of the interval for which FX ( x)  y ).
dF ( y )
 fY ( y )  Y
 1 0  y  1.
dy
Y  FX ( X )
y
x  FX1 ( y )
X
(c) X is a continuous random variable with probability density function
y  g ( x) has multiple solutions for x
Suppose for y  Y , y  g ( x) has solutions xi , i  1, 2,3,............., n . Then
f X ( x) and
n
fY ( y )  
i 1
f X ( x)
dy
dx x  xi
Proof:
Consider the plot of Y  g ( X ) . Suppose at a point y  g ( x) , we have three distinct roots
as shown. Consider the event  y  Y  y  dy . This event will be equivalent to union
events
x1  X  x1  dx1 ,x2  dx2  X  x2 
and
x3  X  x3  dx3
 P  y  Y  y  dy  P x1  X  x1  dx1  P x2  dx2  X  x2   P x3  X  x3  dx3
 fY ( y)dy  f X ( x1 )dx1  f X ( x2 )(dx2 )  f X ( x3 )dx3
Where the negative sign in  dx2 is used to account for positive probability.
Therefore, dividing by dy and taking the limit, we get
 dx 
 dx 
 dx 
fY ( y )  f X ( x1 )  1   f X ( x2 )   2   f X ( x3 )  3 
 dy 
 dy 
 dy 
 f X ( x1 )
3

i 1
dx
dx1
dx
 f X ( x2 ) 2  f X ( x3 ) 3
dy
dy
dy
f X ( xi )
dy
dx x  xi
In the above, we assumed y  g ( x) to have three roots. In general, if y  g ( x) has n
roots, then
n
fY ( y )  
i 1
f X ( xi )
dy
dx x  xi
Example 3: Probability density function of a linear function of a random variable
Suppose Y  aX  b, a  0.
y b
dy
Then x 
and
a
a
dx
y b
fX (
)
f X ( x)
a
 fY ( y ) 

dy
a
dx
Example 4: Probability density function of the output of a full-wave rectifier
Suppose Y  X ,
 a  X  a,
a0
Y
y
y
y
y  x has two solutions x1  y and x2   y and
 fY ( y ) 
f X ( x ) x  y

X
dy
 1 at each solution point.
dx
f X ( x)x  y
1
1
 f X ( y )  f X ( y )
Example 5: Probability density function of the output of a square-law device
Y  cX 2 , c  0
 y  cx 2
And
 x  
y
c
y0
dy
dy
 2cx so that
 2c y / c  2 cy
dx
dx
 fY ( y ) 
fX


y / c  fX
2 cy
= 0 otherwise

y /c

y0