PROBABILITY AND STATISTICS
FOR ENGINEERING
Functions of a Random Variable
Hossein Sameti
Department of Computer Engineering
Sharif University of Technology
Functions of a Random Variable
X:
g(X):


a r.v defined on the model (, F , P ),
a function of the variable x, Y  g ( X ).
Is Y necessarily a r.v?
If so what is its PDF
FY ( y), pdf fY ( y ) ?
2
Functions of a Random Variable
 With Y ( )  B , In particular
FY ( y )  P (Y ( )  y )  P  g ( X ( ))  y 
 For a specific 𝑦, the values of 𝑥 such that 𝑔(𝑥) ≤ 𝑦 form a set on the
𝑥 axis denoted by 𝑅𝑦 . Clearly, g(X ξ ) ≤ 𝑦 if 𝑋(𝜉) is a number in the
set 𝑅𝑦 . Hence
𝐹𝑌 𝑦 = 𝑃(𝑋𝜖𝑅𝑦 )
3
 For 𝑔(𝑋) to be a random variable, the function 𝑔(𝑥) must
have these properties:
– Its domain must include the range of the random variable 𝑋.
– It must be a Borel function, that is, for every 𝑦, the set 𝑅𝑦 such
that 𝑔 𝑥 ≤ 𝑦 must consist of the union and intersection of a
countable number of intervals. Only then {𝑌 ≤ 𝑦} is an event.
– The events {𝑔 𝑋 = ±∞} must have zero probability.
4
Example: Y  aX  b
 For a  0 :
 For a  0,
FY ( y )  PY ( )  y 
FY ( y )  PY ( )  y 
 PaX ( )  b  y 
 PaX ( )  b  y 
y b

 P X ( ) 

a 

 y b
 1  FX 
 ,
 a 
y b

 P X ( ) 

a 

 y b
 FX 
.
 a 
And fY ( y ) 
1  y b
fX 
.
a  a 
We conclude:
and hence
fY ( y ) 
fY ( y )  
1
 y b
fX 
.
|a |  a 
1  y b
fX 
.
a  a 
5
Example: Y  X 2
FY ( y )  PY ( )  y   PX 2 ( )  y .
y  0,
 For
X
2
( )  y  ,
FY ( y )  0, y  0.
 For
y  0,
Y  X2
y
according to the figure,
2
 the event {Y ( )  y}  { X ( )  y}
is equivalent to {x1  X ( )  x2 }.
x1
x2
X
6
Example: Y  X 2
- continued
FY ( y )  Px1  X ( )  x2   FX ( x2 )  FX ( x1 )
 FX ( y )  FX (  y ),

y  0.

 1
f ( y )  f X ( y ) ,
y  0,

fY ( y )   2 y X

0,
otherwise .
 If f X (x ) represents an even function,
fY ( y ) 
1
fX
y
 y  U ( y ).
7
Example: Y  X 2
- continued
 If X ~ N (0,1), so that f X ( x ) 
1
 x2 / 2
e
,
2
 We obtain the p.d.f of Y  X 2 to be
fY ( y ) 
1
e  y / 2U ( y ).
2y
 which represents a Chi-square r.v with n = 1, since (1 / 2)   .
 Thus, if X is a Gaussian r.v with   0, then
Y  X2
represents a
Chi-square r.v with one degree of freedom (n = 1).
8
Example
 For:
X  c,
 X  c,

Y  g ( X )   0,
 c  X  c,
 X  c,
X  c.

g( X )
c
X
c
 We have P(Y  0)  P(c  X ( )  c)  FX (c)  FX (c).
 For y  0, we have x  c, and Y ( )  X ( )  c so that
FY ( y )  P Y ( )  y   P( X ( )  c  y )
 P X ( )  y  c   FX ( y  c),
 For y  0, we have x  c, and
y  0.
Y ( )  X ( )  c
so that
FY ( y )  P Y ( )  y   P( X ( )  c  y )
 P X ( )  y  c   FX ( y  c),
y  0.
9
Example – continued
 Thus,
 f X ( y  c ), y  0,

fY ( y )  [ FX ( c )  FX ( c)] ( y ),
 f ( y  c ), y  0.
 X
FX (x)
FY ( y )
g( X )
c
c
X
x
y
10
Example: Half-wave Rectifier
 Consider Y  g ( X ); g ( x )   x, x  0,
Y
0, x  0.
X
 In this case P(Y  0)  P( X ( )  0)  FX (0).
 For
y  0,
 Thus,
Y  X,
FY ( y )  PY ( )  y   P X ( )  y   FX ( y).
since
y  0,
 f X ( y ),

fY ( y )   FX (0) ( y ) y  0,

0,
y  0,

 f X ( y )U ( y )  FX (0) ( y ).
11
Continuous Functions of a Random Variable





A continuous function g(x)
g (x ) nonzero at all but a finite number of points
has only a finite number of maxima and minima
eventually becomes monotonic as | x | .
Consider a specific y on the y-axis, and a positive increment
y
g (x )
y  y
y
x1 x1  x1
x2  x2 x2
x3 x3  x3
x
12
Continuous Functions of a Random Variable

For fY ( y ) Y  g (X ) where g () is of continuous type,
Py  Y ( )  y  y  
y  y
y
 y  g (x ) has three solutions
x1 , x2 , x3
fY (u )du  fY ( y )  y.
when
y  Y ( )  y  y,
 X could be in any one of three disjoint intervals:
g ( x)
{x1  X ( )  x1  x1},
y  y
{x2  x2  X ( )  x2 }
or {x3  X ( )  x3  x3} .
y
x1 x1  x1
x2  x2 x2
x
x3 x3  x3
So,
Py  Y ( )  y  y  P{x1  X ( )  x1  x1}
 P{x2  x2  X ( )  x2 }  P{x3  X ( )  x3  x3} .
13
Continuous Functions of a Random Variable
 For small y, xi , we get
fY ( y )y  f X ( x1 )x1  f X ( x2 )( x2 )  f X ( x3 )x3.
 Here, x1  0, x2  0 and x3  0, so
| xi |
1
fY ( y )   f X ( xi )

f X ( xi )
y
i
i y / xi
 As y  0,
1
1
fY ( y )  
f X ( xi )  
f X ( xi ).
i dy / dx x
i g ( xi )
i
 If the solutions are all in terms of y, the right side is only a function of y.
14
Example
YX
2
Revisited
 For all y  0, x1   y and x2   y
Y  X2
y
 fY ( y )  0 for y  0 .
x1
 Moreover
dy
 2 x so that
dx
 and using fY ( y )  
i
X
x2
dy
2 y
dx x  xi
1
f X ( xi ) ,

g ( xi )


 1
f ( y )  f X ( y ) ,
y  0,

fY ( y )   2 y X

0,
otherwise ,

which agrees with previous solution.
15
1
Example: Y 
X
 Find
fY ( y ).
 Here for every
y, x1  1 / y
is the only solution, and
dy
1
  2 so that
dx
x
 and substituting this into
dy
dx
fY ( y )  
i

x  x1
1
2

y
,
2
1/ y
1
f X ( xi ) , we obtain
g ( xi )
1
1
fY ( y )  2 f X  .
y
 y
16
Example
 Suppose f X ( x)  2 x /  2 , 0  x   , and Y  sin X . Determine fY ( y ).
 X has zero probability of falling outside the interval (0,  ).
 y  sin x has zero probability of falling outside (0,1).
 fY ( y)  0 outside this interval.
 For any 0  y  1, the equation y  sin x
solutions
, x1 , x2 , x3 ,,
where x1  sin 1
 using the symmetry we also get
 so that
x2    x1
has an infinite number of
y is the principal solution.
etc. Further,
dy
 cos x  1  sin 2 x  1  y 2
dx
dy
dx
 1  y2 .
x  xi
17
Example – continued
f X ( x)
(a)
y  sin x
x3

x
y
x1
x1
x2

x3
x
(b)
 for 0  y  1,
fY ( y ) 


i 
i 0
1
1 y
2
f X ( xi ).
 In this case f X ( x1 )  f X ( x3 )  f X ( x4 )    0
 (Except for f X ( x1 ) and f X ( x2 ) the rest are all zeros).
18
Example – continued
 Thus,
fY ( y ) 
1
1 y
 f X ( x1 )  f X ( x2 )  
2
 2 x1 2 x2 
 2  2 
2
 
1 y  
1
2

, 0  y  1,
2( x1    x1 ) 

  1  y 2
 2 1  y2

0,
otherwise.
fY ( y )
2

1
y
19
Example:
Y  tan X , X    / 2,  / 2 .
 As x moves in   / 2,  / 2 , y moves in  ,   .
 The function Y  tan X is one-to-one for   / 2  x   / 2 .
f X (x )
1
 For any y, x1  tan y is the principal solution.
 / 2
 /2
dy d tan x

 sec 2 x  1  tan 2 x  1  y 2
dx dx
y  tan x
 / 2
1
1/ 
fY ( y ) 
f X ( x1 ) 
,
2
| dy / dx | x  x1
1 y
   y  
x
y
x
x1  / 2
(Cauchy density function with parameter equal to unity)
fY ( y ) 
1
1  y2
y
20
Functions of a Discrete-type R.V
 Suppose X is a discrete-type r.v with
P( X  xi )  pi , x  x1, x2 ,, xi ,
 and Y  g ( X ).
 Clearly Y is also of discrete-type, and when x  xi , yi  g ( xi ),
and for those
yi s,
P(Y  yi )  P( X  xi )  pi , y  y1, y2 ,, yi ,
21
Example
 Suppose X ~ P( ), so that
P( X  k )  e

k
k!
,
k  0,1,2,
 Define Y  X 2  1. Find the p.m.f of Y.
 Solution: X takes the values 0,1,2,, k ,
 Y only takes the values 1, 2, 5, , k 2  1,
P(Y  k 2  1)  P( X  k )
 so that for j  k 2  1

P(Y  j )  P X 
j 1

j  1  e
, j  1, 2,5, , k 2  1, .
( j  1)!

22