Transcript Lecture 05

5. Functions of a Random Variable
Let X be a r.v defined on the model (, F , P ), and suppose
g(x) is a function of the variable x. Define
Y  g ( X ).
(5-1)
Is Y necessarily a r.v? If so what is its PDF FY ( y ), pdf fY ( y ) ?
Clearly if Y is a r.v, then for every Borel set B, the set of for
Y ( ) must
B
which
belong to F. Given that X is a r.v, this is
gis1 (also
B) a Borel set, i.e., if g(x) is a Borel
assured if
function. In that case if X is a r.v, so is Y, and for every Borel
set B
P(Y  B)  P( X  g 1 ( B)).
(5-2)
1
In particular
FY ( y )  P(Y ( )  y )  Pg ( X ( ))  y   PX ( )  g 1 ( , y ]. (5-3)
Thus the distribution function as well of the density
function of Y can be determined in terms of that of X. To
obtain the distribution function of Y, we must determine the
Borel set on the x-axis such that X ( )  g 1 ( y ) for every
given y, and the probability of that set. At this point, we
shall consider some of the following functions to illustrate
the technical details.
aX  b
X nis
1
X
log X
X2
Y  g( X )
|X |
X
eX
| X | U ( x)
2
Example 5.1: Y  aX  b
Solution: Suppose a  0.
(5-4)
y b

 y b
FY ( y )  P Y ( )  y   PaX ( )  b  y   P X ( ) 

F

 . (5-5)
X
a 

 a 
and
1  y b
fY ( y )  f X 
.
a  a 
(5-6)
On the other hand if a  0, then

FY ( y )  P Y ( )  y   P aX ( )  b  y   P X ( ) 

 y b
 1  FX 
,
 a 
and hence
fY ( y )  
y b

a 
(5-7)
1  y b
fX 
.
a  a 
(5-8)
3
From (5-6) and (5-8), we obtain (for all a)
fY ( y ) 
1
 y b
fX 
.
|a |  a 
(5-9)
Example 5.2: Y  X 2 .
(5-10)
FY ( y )  PY ( )  y   PX 2 ( )  y .
(5-11)
If y  0, then the event  X 2 ( )  y  , and hence
FY ( y )  0, y  0.
For y  0, from Fig. 5.1, the event
is equivalent to {x1  X ( )  x2}.
(5-12)
{Y ( )  y}  { X 2 ( )  y}
Y  X2
y
x1
Fig. 5.1
x2
X
4
Hence
FY ( y )  Px1  X ( )  x2   FX ( x2 )  FX ( x1 )
 FX ( y )  FX (  y ),
y  0.
(5-13)
By direct differentiation, we get


 1
f ( y )  f X ( y ) ,

fY ( y )   2 y X

0,

If
f X (x)
y  0,
otherwise .
(5-14)
represents an even function, then (5-14) reduces to
fY ( y ) 
1
fX
y
 y  U ( y ).
(5-15)
In particular if X  N (0,1), so that
f X ( x) 
1
 x2 / 2
e
,
2
(5-16)
5
and substituting this into (5-14) or (5-15), we obtain the
p.d.f of Y  X 2 to be
fY ( y ) 
1
e  y / 2U ( y ).
2y
(5-17)
On comparing this with (3-36), we notice that (5-17)
represents a Chi-square r.v with n = 1, since (1 / 2)   .
Thus, if X is a Gaussian r.v with   0, then Y  X 2
represents a Chi-square r.v with one degree of freedom
(n = 1).
Example 5.3: Let
 X  c,

Y  g ( X )   0,
 X  c,

X  c,
 c  X  c,
X  c.
6
In this case
P(Y  0)  P(c  X ( )  c)  FX (c)  FX (c).
For y  0, we have x  c, and
Y ( )  X ( )  c
FY ( y )  P Y ( )  y   P( X ( )  c  y )
Similarly
(5-18)
so that
 P  X ( )  y  c   FX ( y  c), y  0.
y  0, if x  c, and Y ( )  X ( )  c so
(5-19)
that
FY ( y )  P Y ( )  y   P( X ( )  c  y )
 P  X ( )  y  c   FX ( y  c),
Thus
 f X ( y  c),
fY ( y )  
 f X ( y  c),
g( X )
y  0,
c
y  0.
FY (x)
X
x
(a)
(5-20)
(5-21)
FX (x)
c
y  0.
(b)
Fig. 5.2
(c)
y
7
Example 5.4: Half-wave rectifier
Y  g ( X );
 x,
g ( x)  
0,
x  0,
x  0.
(5-22)
Y
In this case
P(Y  0)  P( X ( )  0)  FX (0).
X
(5-23)
Fig. 5.3
and for y  0, since Y  X ,
FY ( y)  PY ( )  y   P X ( )  y   FX ( y).
(5-24)
y  0,
 f X ( y )U ( y ).
y  0,
(5-25)
Thus
 f X ( y ),
fY ( y )  
 0,
8
Note: As a general approach, given Y  g ( X ), first sketch
the graph y  g ( x ), and determine the range space of y.
Suppose a  y  b is the range space of y  g ( x ).
Then clearly for y  a, FY ( y)  0, and for y  b, FY ( y)  1, so
that FY ( y) can be nonzero only in a  y  b. Next, determine
whether there are discontinuities in the range space of y. If
so evaluate PY ( )  yi  at these discontinuities. In the
continuous region of y, use the basic approach
FY ( y)  Pg ( X ( ))  y 
and determine appropriate events in terms of the r.v X for
every y. Finally, we must have FY ( y ) for    y  , and
obtain
fY ( y ) 
dFY ( y )
in  a  y  b.
dy
9
However, if Y  g ( X ) is a continuous function, it is easy to
establish a direct procedure to obtain fY ( y). A continuos
function g(x) with g (x ) nonzero at all but a finite number
of points, has only a finite number of maxima and minima,
and it eventually becomes monotonic as | x | . Consider a
specific y on the y-axis, and a positive increment  y as
shown in Fig. 5.4
g ( x)
y  y
y
x1 x1  x1
x2  x2 x2
x3 x3  x3
x
Fig. 5.4
fY ( y ) for Y  g ( X ), where g () is of continuous type.
10
Using (3-28) we can write
Py  Y ( )  y  y 

y  y
y
fY (u )du  fY ( y )  y.
(5-26)
But the event y  Y ( )  y  y can be expressed in terms
of X ( ) as well. To see this, referring back to Fig. 5.4, we
notice that the equation y  g (x) has three solutions x1, x2 , x3
(for the specific y chosen there). As a result
when y  Y ( )  y  y, the r.v X could be in any one of the
three mutually exclusive intervals
{x1  X ( )  x1  x1}, {x2  x2  X ( )  x2} or {x3  X ( )  x3  x3} .
Hence the probability of the event in (5-26) is the sum of
the probability of the above three events, i.e.,
Py  Y ( )  y  y  P{x1  X ( )  x1  x1}
 P{x2  x2  X ( )  x2 }  P{x3  X ( )  x3  x3} .(5-27) 11
For small y, xi , making use of the approximation in (5-26),
we get
fY ( y )y  f X ( x1 )x1  f X ( x2 )( x2 )  f X ( x3 )x3.
(5-28)
In this case, x1  0, x2  0 and x3  0, so that (5-28) can be
rewritten as
fY ( y )   f X ( xi )
i
| xi |
1

f X ( xi )
y
i y / xi
(5-29)
and as y  0, (5-29) can be expressed as
fY ( y )  
i
1
1
f X ( xi )  
f X ( xi ).
dy / dx x
i g ( xi )
(5-30)
i
The summation index i in (5-30) depends on y, and for every
y the equation y  g ( xi ) must be solved to obtain the total
number of solutions at every y, and the actual solutions x1, x2 ,
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all in terms of y.
For example, if Y  X 2 , then for all y  0, x1   y and x2   y
represent the two solutions for each y. Notice that the
solutions xi are all in terms of y so that the right side of (5-30)
2
Y

X
is only a function of y. Referring back to the example
(Example 5.2) here for each y  0, there are two solutions
given by x1   y and x2   y . ( fY ( y )  0 for y  0 ).
YX
Moreover
2
dy
 2 x so that
dx
dy
dx
y
2 y
x  xi
and using (5-30) we get

x1

Fig. 5.5
 1
f X ( y )  f X ( y ) ,
y  0,

fY ( y )   2 y

0,
otherwise ,

which agrees with (5-14).
X
x2
(5-31)
13
Example 5.5:
1
Y  .
X
(5-32)
Find fY ( y).
Solution: Here for every y,
dy
1
  2 so that
dx
x
x1  1 / y
dy
dx
is the only solution, and

x  x1
1
2

y
,
2
1/ y
and substituting this into (5-30), we obtain
fY ( y ) 
1
1
f
.
X
2

y
 y
(5-33)
In particular, suppose X is a Cauchy r.v as in (3-38) with
parameter  so that
f X ( x) 
 /
,    x  .
2
2
 x
(5-34)
In that case from (5-33), Y  1/ X has the p.d.f
fY ( y ) 
1
 /
(1 /  ) / 

,
2
2
2
2
2
y   (1 / y )
(1 /  )  y
   y  .
(5-35)
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But (5-35) represents the p.d.f of a Cauchy r.v with
parameter 1/ . Thus if X  C ( ), then 1 / X  C (1 /  ).
Example 5.6: Suppose f X ( x)  2 x /  2 , 0  x   , and Y  sin X .
Determine fY ( y).
Solution: Since X has zero probability of falling outside the
interval (0,  ), y  sin x has zero probability of falling outside
the interval (0,1). Clearly fY ( y )  0 outside this interval. For
any 0  y  1, from Fig.5.6(b), the equation y  sin x has an
infinite number of solutions , x1, x2 , x3 ,, where x1  sin 1 y
is the principal solution. Moreover, using the symmetry we
also get x2    x1 etc. Further,
dy
 cos x  1  sin 2 x  1  y 2
dx
so that
dy
dx
 1  y2 .
x  xi
15
f X ( x)
x3

(a)
x
y  sin x
y
x1
x1
x2

x3
x
(b)
Fig. 5.6
Using this in (5-30), we obtain for 0  y  1,
fY ( y ) 


i  
1
1 y
2
f X ( xi ).
(5-36)
But from Fig. 5.6(a), 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).
16
Thus (Fig. 5.7)
fY ( y ) 

1
1  y2
 f X ( x1 ) 
2( x1    x1 )

2
1 y
2
f X ( x2 )  
2

,

  1  y 2

0,

fY ( y )
 2 x1 2 x2 
 2  2 
2
 
1 y  
1
2
0  y  1,
otherwise.
(5-37)

y
1
Fig. 5.7
Example 5.7: Let Y  tan X where X  U   / 2,  / 2 .
Determine fY ( y).
Solution: As x moves from   / 2,  / 2 , y moves from  ,   .
From Fig.5.8(b), the function Y  tan X is one-to-one
for   / 2  x   / 2 . For any y, x1  tan 1 y is the principal
solution. Further
dy d tan x

 sec 2 x  1  tan 2 x  1  y 2
dx dx
17
so that using (5-30)
fY ( y ) 
1
1/ 
f X ( x1 ) 
,
2
| dy / dx |x  x1
1 y
   y  ,
(5-38)
which represents a Cauchy density function with parameter
equal to unity (Fig. 5.9).
f X (x )
 / 2
 /2
x
fY ( y ) 
1
1  y2
(a)
y  tan x
 / 2
y
y
x1  / 2
(b)
Fig. 5.8
x
Fig. 5.9
18
Functions of a discrete-type r.v
Suppose X is a discrete-type r.v with
P( X  xi )  pi , x  x1, x2 ,, xi ,
(5-39)
and Y  g ( X ). Clearly Y is also of discrete-type, and
when x  xi , yi  g ( xi ), and for those yi
P(Y  yi )  P( X  xi )  pi , y  y1, y2 ,, yi ,
(5-40)
Example 5.8: Suppose X  P( ), so that
P( X  k )  e 
k
k!
, k  0,1,2,
(5-41)
Define Y  X 2  1. Find the p.m.f of Y.
Solution: X takes the values 0,1,2,, k , so that Y only
takes the value 1,3,, k 2  1, and
19
P(Y  k 2  1)  P( X  k )
so that for j  k 2  1


j 1

P(Y  j )  P X  j  1  e
, j  1,3,, k 2  1,. (5-42)
( j  1)!
20