4.6. The random variables X and Y are related by the equation,
Y = e
X
.
(a) If X is uniformly distributed, 0 X 5, find f
Y
(y) . Sketch this density function.
(b) If X is governed by the density function f
X
= ce
x
, 0 X 5, where c is a constant, find f
Y
(y) .
Sketch this density function.
Solution: (a) From the given information f
X
(x) = 1/5. Performing the density transformation (which is
valid for any density function of X):
x
1
= ln y
g(x) = exp(x)
g
(x) = exp(x)
f
Y
(y) =
f
X
(x
1
)
|g
(x
1
)|
=
1/5
exp(ln y)
=
1
5y
, 1 y 148.4
f
Y
(1) = 1/5
f
Y
(148.4) = 1/(5 · 148.4) = 1.3477 × 10
3
.
The density functions are shown in the two figures given next.
Density function f
X
(x) = 1/5, 0 x 5
Density function f
Y
(y) =
1
5y
, 1 y 148.4
(b) Suppose now that X is governed by the density function
f
X
= ce
x
, 0 X 5,
where c is a constant.
First find the value of c :
5
0
c exp(x)dx = 1 = c = 1.0068. Then,
f
Y
(y) =
f
X
(x
1
)
|g
(x
1
)|
=
1.0068 exp(ln y)
exp(ln y)
=
1.0068
y
2
, 1 y 148.4
f
Y
(1) = 1.0068
f
Y
(148.4) = 1.0068/ (148.4)
2
= 4. 5717 × 10
5
.
86
The density functions are shown in the next two figures.
Density Function, f
X
(x) = 1.0068e
x
, 0 x 5
Density Function, f
Y
(y) =
1.0068
y
2
, 1 y 148.5
87
4.7. Derive the probability density function for Y = |X|, given
f
X
(x) =
1
4
, 2 < x 0,
1
2
exp (x) , x > 0.
Sketch all density functions.
Solution: Performing the density transformation:
x
1
= y, x
2
= y
g(x) = X
g
(x) = 1
f
Y
(y) =
f
X
(x
1
)
|g
(x
1
)|
+
f
X
(x
2
)
|g
(x
2
)|
=
1
4
+
1
2
exp (y) +
1
4
+
1
2
exp (y)
=
1
2
+
1
2
(exp (y) + exp (y)) , y > 0.
The plots for the density function are shown in the next two figures.
f
X
(x) Density Function f
Y
(y) =
1
2
+
1
2
(exp (x) + exp (x)) , y > 0
88
4.8. Given Y = 3X 4 and f
X
(x) = 0.5, where 1 x 1, derive f
Y
(y) and sketch both density
functions.
Solution: For any density function f
X
(x) , the general transformation for one root is:
f
Y
(y) =
f
X
(x
1
)
|g
(x
1
)|
.
If g(x) = 3X 4, then g
(x) = 3 and
f
Y
(y) =
0.5
3
= 0.16667, 7 y 1.
The plots for the density functions are shown in the next two figures.
Density Function f
X
(x) =
1
2
, 1 x 1 Density Function f
Y
(y) =
1
6
, 7 x 1
89
4.9. Given Y = 3X 4 and f
X
(x) = N (0, 0.33), derive f
Y
(y) and plot both density functions.
Solution: As X is governed by the Gaussian density, the probability density function is given by
f
X
(x) =
1
σ
2π
exp
1
2
x µ
σ
2
, < x < .
In this case
x
1
= (y + 4)/3,
g(x) = 3x 4
g
(x) = 3
f
Y
(y) =
1
3
1
σ
2π
exp
(Y + 4)
3
µ
2
2σ
2
, over all x,
f
Y
(y) =
1
3
1
(0.33)
2π
exp
(Y + 4)
2
18
for µ = 0, σ = 1.
The density function plots are given in the next two figures.
Density Function f
X
(x) =
1
(1)
2π
exp
1
2
x
1
2
!
, −∞ 0 .
90

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