Estocasticos

EE 5375/7375 Random Processes

September 9, 2003
Homework #2 Solutions

Problem 1. textbook problem 3.12
Let U be a uniform random variable in the interval [-1,1]. Find the probabilities (a) P (U > 0) (b) P (U < 5)
(c) P (|U | < 1/3) (d) P (1/3 < U < 1/2) (e) P (|U | ≥ 3/4).
(a) The pdf for U is fU (u) = 1/2 for −1 ≤ u ≤ 1. P (U > 0) =
1
(b) P (U < 5) = −1 1 du = 1.
2
(c) P (|U | <1/3) = P (−1/3 < U < 1/3) =
1/2 1
du =
1 /3 2
−3/4 1
1
2 du + 3/4
−1

(d) P (1/3 < U < 1/2) =
(e) P (|U | ≥ 3/4) =

1
12 .
1
2 du

=

1/3 1
du
−1/3 2

1
8

+

1
8

11
du
02

= 1.
2

1
= 3.

= 1.
4

Problem 2. textbook problem 3.15
A random variable Y has the PDF
FY (y ) =

0
1 − y −n

if y < 1
if y ≥ 1

where n is a positive integer. (a) Plot thePDF of Y . (b) Find the probability P (k < Y ≤ k + 1) for a
positive integer k .
(a) The PDF is shown in Fig. 1.
(b) P (k < Y ≤ k + 1) = FY (k + 1) − FY (k ) = 1 − (k + 1)−n − (1 − k −n ) = k −n − (k + 1)−n .
Problem 3. textbook problem 3.16
A continuous random variable X has the PDF

0
FX (x) = c(1 + sin(x))

1

if x ≤ −π/2
if −π/2 < x ≤ π/2
if π/2 ≤ x

(a) Find c. (b) Plot FX(x).
(a) A continuous random variable must have a smooth PDF without discontinuities. Therefore c must
satisfy (i) c(1 + sin(−π/2)) = 0 and (ii) c(1 + sin(π/2)) = 1. Condition (i) is satisfied for every c because
1 + sin(−π/2) = 0. Condition (ii) is c(1 + sin(π/2)) = 2c = 1, so c = 1/2.
(b) The PDF is shown in Fig. 2.
Problem 4. textbook problem 3.18
Let X be an exponential random variablewith parameter λ. (a) For d > 0 and a positive integer k , find the
probabilities P (X ≤ d), P (kd ≤ X ≤ (k + 1)d), and P (X > kd). (b) Segment the positive real line into 5
equiprobable disjoint intervals.
(a) The exponential pdf is fX (x) = λe−λx and PDF is FX (x) = 1 − e−λx for x ≥ 0. Hence
P (X ≤ d) = FX (d) = 1 − e−λd
P (kd ≤ X ≤ (k + 1)d) = FX ((k + 1)d) − FX (kd) = 1 − e−λ(k+1)d − (1 −e−λkd ) = e−λkd (1 − e−λd )
P (X > kd) = 1 − FX (kd) = e−λkd
1

(b) The problem is to find {X1 , X2 , X3 , X4 } where FX (X1 ) = 1/5, FX (X2 ) = 2/5, FX (X3 ) = 3/5, and
FX (X4 ) = 4/5. In general, FX (Xi ) = i/5 for i = 1, 2, 3, 4. These points will divide the real line into 5
equiprobable intervals. We have FX (Xi ) = 1 − e−λXi = i/5, therefore
1 − e−λXi =

i
5

i
= e−λXi
5
1
i
Xi= − ln 1 −
λ
5

1−

That is,
1
X1 = − ln
λ

4
5

1
, X2 = − ln
λ

3
5

1
, X3 = − ln
λ

2
5

1
, X4 = − ln
λ

1
5

Problem 5. textbook problem 3.19
A random variable X has the pdf
cx(1 − x)
0

fX (x) =

if 0 ≤ x ≤ 1
otherwise

(a) Find c. (b) Find P (1/2 ≤ X ≤ 3/4). (c) Find FX (x).
(a) The pdf must integrate to 1:
1

1

cx(1 − x)dx =

1

cx2dx =

cxdx −

0

0

0

so c = 6.
2
3/4
32
(b) P (1/2 ≤ X ≤ 3/4) = 1/2 6x(1 − x)dx = 3 4 − 3 1 − 2
2
x
x
x
(c) FX (x) = 0 6y (1 − y )dy = 0 6ydy − 0 6y 2 dy = 3x2 − 2x3 .

c
c
c
− = =1
23
6

33
4

+2

13
2

=

27
16

−

3
4

−

54
64

+

2
8

=

11
32 .

Problem 6. textbook problem 3.21
A random variable X has the pdf shown in Fig. P3.2.(a) Find fX (x). (b) Find the PDF of X . (c) Find b
such that P (|X | < b) = 1/2.
(a) The pdf has a symmetric triangular shape:
fX (x) =

c 1−
0

|x|
a

if |x| ≤ a
if |x| > a

The area of the triangle is c(2a)/2 but this must equal 1, so c = 1/a.
(b) The PDF is found by integration of the pdf. First, FX (x) = 0 for x < −a. Second, for −a ≤ x ≤ 0,
x

FX (x) =
−a

y
11
1
1+dy = +
a
a
2a

For 0 ≤ x ≤ a,
FX (x) =

1
+
2

x

x+

x2
2a

1
y
11
1−
dy = +
a
a
2a

x−

x2
2a

1
2
= FX (b) − FX (−b) =
2
a

b−

b2
2a

0

Finally, for all x > a, FX (x) = 1.
(c) We have
P (|X | < b) =

2

If we solve for b, then we find b = a(1 −

1
√ ).
2

Problem 7. textbook problem 3.27
Let X be the exponential random variable...