5. Functions of a Random Variable
Let X be a r.v defined on the model (Ω, 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 ξfor
which Y (ξ ) ∈ B must belong to F. Given that X is a r.v, this is
assured if g −1 ( B ) is also a Borel set, i.e., if g(x) is a Borel
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)
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In particular
FY ( y ) = P (Y (ξ ) ≤ y ) = P (g ( X (ξ )) ≤ y ) = P (X (ξ ) ≤ 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 (ξ ) ≤ 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
X2
Y = g( X )
log X
|X |
X
eX
| X | U ( x)
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Example 5.1: Y = aX + b
Solution: Suppose a > 0 .
FY ( y) = P(Y (ξ ) ≤ y ) = P(aX (ξ ) + b ≤ y ) = P X (ξ ) ≤
and
(5-4)
y −b
y −b
= FX
. (5-5)
a
a
y −b
1
fY ( y) = f X
.
a
a
On the other hand if a < 0 , then
FY ( y ) = P(Y (ξ ) ≤ y ) = P(aX (ξ ) + b ≤ y ) = P
= 1 − FX
and hence
y −b
,
a
fY ( y) = −
(5-6)
y −b
X (ξ ) >
a
(5-7)
y −b
1
fX
.
a
a
(5-8)
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From (5-6) and (5-8), we obtain (for all a)
fY ( y) =
y −b
1
fX
.
a
|a |
(5-9)
(5-10)
Example 5.2: Y = X 2 .
FY ( y ) = P (Y (ξ ) ≤ y ) = P (X 2 (ξ ) ≤ y ).
(5-11)
If y < 0 , then the event { X 2 (ξ ) ≤ y}= φ , and hence
(5-12)
FY ( y ) = 0, y < 0.
For y > 0 , from Fig. 5.1, the event
is equivalent to { x1 < X (ξ ) ≤ x 2 }.
{Y (ξ ) ≤ y } = { X 2 (ξ ) ≤ y }
Y = X2
y
x1
Fig. 5.1
x2
X
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Hence
FY ( y ) = P ( x1 < X (ξ ) ≤ x 2 ) = F X ( x 2 ) − F X ( x1 )
= FX ( y ) − FX ( −
y ),
(5-13)
y > 0.
By direct differentiation, we get
1
fY ( y ) =
2
(
f
y
X
( y ) + f X (−
)
y) ,
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
fX (x) =
1
− x2 /2
e
,
2π
(5-16)
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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 ).
2π y
(5-17)
On comparing this with (3-36), we notice that (5-17)
represents a Chi-square r.v with n = 1, since Γ (1 / 2 ) = π .
Thus, if X is a Gaussian r.v with µ = 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.
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In this case
P (Y = 0) = P ( − c < X (ξ ) ≤ c ) = FX ( c ) − FX ( − c ).
For y > 0 , we have x > c , and
Y (ξ ) = X (ξ ) − c
FY ( y ) = P (Y (ξ ) ≤ y ) = P ( X (ξ ) − c ≤ y )
Similarly
= P ( X (ξ ) ≤ y + c ) = FX ( y + c ), y > 0 .
y < 0 , if x < − c , and Y ( ξ ) = X ( ξ ) + c
FY ( y ) = P (Y (ξ ) ≤ y ) = P ( X (ξ ) + c ≤ y )
= P ( X (ξ ) ≤ y − c ) = FX ( y − c ),
Thus
f X ( y + c ),
y < 0.
g( X )
c
(a)
(5-19)
so that
(5-20)
(5-21)
y < 0.
FX (x )
−c
so that
y > 0,
f Y ( y ) = [ FX ( c ) − FX ( − c )]δ ( y ),
f X ( y − c ),
(5-18)
FY ( y )
X
(b)
Fig. 5.2
x
(c)
y
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Example 5.4: Half-wave rectifier
Y = g ( X );
g ( x) =
x,
x > 0,
0,
x ≤ 0.
(5-22)
Y
In this case
P (Y = 0) = P ( X (ξ ) ≤ 0) = FX ( 0).
(5-23)
X
Fig. 5.3
and for y > 0 , since Y = X ,
FY ( y ) = P(Y (ξ ) ≤ y ) = P( X (ξ ) ≤ y ) = FX ( y ).
(5-24)
Thus
f X ( y ),
y > 0,
f Y ( y ) = FX (0)δ ( y ) y = 0,
0,
= f X ( y )U ( y ) + FX (0)δ ( y ).
(5-25)
y < 0,
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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 (ξ ) = yi ) at these discontinuities. In the
continuous region of y, use the basic approach
FY ( y ) = P (g ( X (ξ )) ≤ 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 )
dy
in a < y < b.
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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
x 1 x1 + ∆x1
x 2 + ∆x 2 x 2
x 3 x3 + ∆x3
x
Fig. 5.4
fY (y)
for Y = g(X ), where g (⋅) is of continuous type.
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Using (3-28) we can write
P {y < Y (ξ ) ≤ y + ∆ y } =
y + ∆y
y
f Y ( u ) du ≈ f Y ( y ) ⋅ ∆ y .
(5-26)
But the event {y < Y (ξ ) ≤ y + ∆ y } can be expressed in terms
of X (ξ ) 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 (ξ ) ≤ y + ∆ y }, the r.v X could be in any one of the
three mutually exclusive intervals
{x1 < X (ξ ) ≤ x1 + ∆x1}, {x2 + ∆x2 < X (ξ ) ≤ x2} or {x3 < X (ξ ) ≤ 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 (ξ ) ≤ y + ∆ y } = P { x1 < X (ξ ) ≤ x1 + ∆ x1 }
+ P { x 2 + ∆ x 2 < X (ξ ) ≤ x 2 } + P { x 3 < X (ξ ) ≤ x 3 + ∆ x 3 } .(5-27) 11
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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 |
=
∆y
i
1
f X ( xi )
∆y / ∆xi
(5-29)
and as ∆y → 0, (5-29) can be expressed as
fY ( y ) =
i
1
f X ( xi ) =
dy / dx x
i
i
1
f X ( xi ).
g ′( xi )
(5-30)
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.
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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
dy
= 2 x so that
=2 y
dx
dx x = xi
y
and using (5-30) we get
1
fY ( y ) = 2
(
f
y
x1
X
)
( y ) + f X (− y ) ,
0,
which agrees with (5-14).
Fig. 5.5
y > 0,
otherwise ,
X
x2
(5-31)
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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
1
1
fY ( y ) = 2 f X
.
y
y
(5-33)
In particular, suppose X is a Cauchy r.v as in (3-39) with
parameter α so that
α /π
f X ( x) = 2
, − ∞ < x < +∞.
2
α +x
(5-34)
In that case from (5-33), Y = 1 / X has the p.d.f
fY ( y ) =
1
(1 / α ) / π
α /π
=
,
2
2
2
2
2
y α + (1 / y )
(1 / α ) + y
− ∞ < y < +∞ .
(5-35)
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But (5-35) represents the p.d.f of a Cauchy r.v with
parameter 1 / α . Thus if X ∼ C (α ), then 1 / X ∼ C (1 / α ).
Example 5.6: Suppose f X ( x ) = 2 x / π 2 , 0 < x < π , and Y = sin X .
Determine fY ( y ).
Solution: Since X has zero probability of falling outside the
interval (0,π ), 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 = π − x1 etc. Further,
dy
= cos x = 1 − sin 2 x = 1 − y 2
dx
so that
dy
= 1 − y2 .
dx x = xi
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fX (x)
x3
π
(a)
x
y = sin x
y
x −1
x1
x2
π
x3
x
(b)
Fig. 5.6
Using this in (5-30), we obtain for 0 < y < 1,
fY ( y ) =
+∞
i = −∞
i ≠0
1
1− y
2
f X ( x i ).
(5-36)
But from Fig. 5.6(a), in this case f X ( x−1 ) = f X ( x3 ) = f X ( x4 ) =
(Except for f X ( x1 ) and f X ( x 2 ) the rest are all zeros).
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Thus (Fig. 5.7)
fY ( y ) =
=
1
1 − y2
( f X ( x1 ) +
2 ( x 1 + π − x1 )
π
2
1− y
2
f X ( x2 ) ) =
2
= π 1 − y2
0,
,
1
1 − y2
2 x1
π2
0 < y < 1,
otherwise.
+
fY ( y )
2 x2
π2
2
(5-37)
π
Fig. 5.7
y
1
Example 5.7: Let Y = tan X where X ∼ U (− π / 2, π / 2 ) .
Determine fY ( y ).
Solution: As x moves from (−π / 2, π / 2) , y moves from (− ∞ , + ∞ ) .
From Fig.5.8(b), the function Y = tan X is one-to-one
for −π / 2< x <π / 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
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so that using (5-30)
fY ( y ) =
1
1/π
f X ( x1 ) =
,
2
| dy / dx |x = x1
1+ y
(5-38)
− ∞ < y < +∞ ,
which represents a Cauchy density function with parameter
equal to unity (Fig. 5.9).
f X (x )
−π /2
π /2
x
fY ( y ) =
(a)
y = tan x
−π /2
y
y
x1 π / 2
(b)
Fig. 5.8
1
1 + y2
x
Fig. 5.9
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Functions of a discrete-type r.v
Suppose X is a discrete-type r.v with
P ( X = xi ) = pi , x = x1 , x2 ,
(5-39)
, xi ,
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 (λ ), so that
P ( X = k ) = e −λ
λk
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 ,
takes the value 1, 2, 5, , k 2 + 1,
and
(5-41)
so that Y only
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P(Y = k 2 + 1) = P( X = k )
so that for j = k 2 + 1
(
P(Y = j ) = P X =
)
j − 1 = e− λ
λ
j −1
( j − 1)!
, j = 1, 2,5,
, k 2 + 1,
. (5-42)
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