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

Get *Probabilistic Models for Dynamical Systems, 2nd Edition* now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.