Download Examples on Transformations of Random Variables

Examples on Transformations of Random Variables
1. Let X ∼ U ([−π, π]). Find the distribution of the random variable Y =
cos X.
The density of X is given by
fX (x) =
½
1
2π
0
if x ∈ [−π, π]
otherwise
Method 1: Note that the range of random variable Y is [−1, 1]. There
are two solutions to the equation y = cos x for x ∈ [−π, π], one in [−π, 0]
and the other in [0, π]. Hence, the density of Y = cos X is given by
¯ ¯
X
¯ dx ¯
−1
fX (cos y) ¯¯ ¯¯
fY (y) =
dy
cos x=y
¯
¯
X
¯
¯
1
−1
¯
¯
fX (cos y) ¯
=
¯
−1
−
sin
(cos
y)
cos x=y
½ 2
1
if cos−1 y ∈ [0, π]
2π sin (cos−1 y)
=
0
otherwise
Thus,
½
1
π sin (cos−1 y)
if y ∈ [−1, 1]
(1)
otherwise
p
If we further use the fact that sin(cos−1 (y)) = 1 − y 2 we obtain the
following expression for the PDF:
(
√1
if y ∈ [−1, 1]
π 1−y 2
fY (y) =
(2)
0
otherwise
fY (y) =
0
Method 2: The CDF of X is
FX (x) =
x+π
x
1
x−a
=
=
+
b−a
2π
2π 2
The sets that are equivalent to the event {Y ≤ y} are {X < − cos−1 (y)}
and {X > cos−1 (y)}. The CDF of Y is given by
FY (y)
=
P (Y ≤ y) = P (cos X ≤ y)

 0
P (X ≤ − cos−1 y) + P (X ≥ cos−1 y)
=

1

y ∈ [−∞, −1)
 0
−1
cos
y
=
if y ∈ [−1, 1]
1− π

1
otherwise
1
y ∈ [−∞, −1)
if y ∈ [−1, 1]
otherwise
assuming that cos−1 y ≥ 0. The probability density function of Y is
obtained as the derivative of this CDF expression.
2. Square law : Let X ∼ U ([−1, 1]). Find the distribution of the random
variable Y = X 2 .
The PDF of X is given by
fX (x) =
½
1
2
0
if x ∈ [−1, 1]
otherwise
Method 1: Note that the range of random variable Y is [0, 1]. There are
two solutions to the equation y = x2 . Hence, the density of Y = X 2 is
given by
¯ ¯
X
¯ dx ¯
fX (x) ¯¯ ¯¯
fY (y) =
dy
x2 =y
¯
¯
¯
¯
1 ¯¯ 1 ¯¯ 1 ¯¯ 1 ¯¯
=
√ +
√
2 ¯ −2 y ¯ 2 ¯ 2 y ¯
½ 1
√
if y ∈ [0, 1]
2 y
=
(3)
0
otherwise
Method 2: The CDF of X is
FX (x) =
x
2
The CDF of Y is given by
FY (y)
P (Y ≤ y) = P (X 2 ≤ y)

y ∈ [−∞, 0)
 0
√
√
P (− y ≤ X ≤ y) if y ∈ [0, 1]
=

1
otherwise

y ∈ [−∞, −1)
 0√
√
√
− y
y
=
− 2 = y if y ∈ [−1, 1]
 2
1
otherwise
=
Hence, the density of Y is given by (2).
3. Full-wave rectifier: Let X ∼ U ([−1, 1]). Find the distribution of the
Y = Xu(X).
The density of X is given by
fX (x) =
½
1
2
0
if x ∈ [−1, 1]
otherwise
Note that this transformation is not differentiable so the gradient method
is not applicable in this case. Instead let us look at the set equivalence
method.
2
– For y < 0, the event {Y ≤ y} does not have a solution on the real
line and hence reduces to a null event. Consequently the probability
of this event is 0.
– For y = 0, the event {Xu(X) ≤ 0} has only one solution X = 0 and
the probability of this event is FX (0+ ) − FX (0− ).
– For y > 0, the event that {Xu(X) ≤ y} reduces to the event {X ≤ y}
and the probability of this event is just FX (y).
The CDF of the transformed random variable can then be summarized as:

y<0
 0
FX (0+ ) − FX (0− ) y = 0
FY (y) =

FX (y)
y>0
Since the input distribution is uniform this reduces to:
½
0
y≤0
FY (y) =
FX (y) y > 0
3