Download Analysis of Two Random Variables X(t) and X(t+d) in a given interval and more Exercises Probability and Statistics in PDF only on Docsity!
ECE863 2
6.5 a) Within the interval
[ ]
t ∈ 0,1, X(t) can take only two possible values:
X t ( ) = + g t( ) = + 1 or X t( ) = − g t( ) = − 1
Both of these values are equally likely since A takes on ± 1 with equal
probability.
( ) ( ) [ ]
P X t 1 P X t 1 t 0,
[ ]
P X t = 0 = 1 ∀ t ∉ 0,
b) For
[ ]
t ∈ 0,
( ) ( ) ( ) ( )
X
m t = 1 P X t = 1 + − 1 P X t = − 1
( ) [ ]
X
⇒ m t = 0 ∀ t ∈ 0,
Also,
( ) [ ]
X
⇒ m t = 0 ∀ t ∉ 0,
X
m t = 0 ∀t
6.5 c) Here, we have two random variables:
X t and ( )
X t +d
The joint pmf of
X t and ( )
X t + d depends on the time instances (t) and
(t + d).
ECE863 3
( ) ( ) [ ]
When t and t + d ∈ 0,1, then both X(t) and X(t + d) have the same
value (either both of them = +1 if A = 1 OR both of them = -1 if A = -1).
( ) ( )
( ) ( )
P X t 1, X t d 1
P X t 1, X t d 1
[ ]
( ) [ ]
t 0,
and t d 0,
( ) [ ] ( ) [ ]
When t ∈ 0,1 and t + d ∉ 0,1, then
( ) ( )
( ) ( )
P X t 1, X t d 0
P X t 1, X t d 0
[ ]
( ) [ ]
t 0,
t d 0,
( )
[ ]
When both t and t + d ∉ 0,1, then
( ) ( )
P X t = 0, X t + d = 0 = 1
ECE863 5
6.9 a) For a given value of t, At and B are independent (since A and B are
independent).
Therefore, for:
Z t = At +B
the pdf
( )
( )
( )
Z t A t B
f Z f a f b
′
where
A t At
( )
( )
B Z t A t
f z f u f z u du
∞
′
− ∞
Since
A ( t)
A
t
( )
( u)
A t A
1 u
f f
t t
′
( )
Z t A B
1 u
f f f z u du
t t
∞
− ∞
ECE863 6
6.9 b) ( )
[ ]
Z
E Z t =m t
[ ] [ ]
= t E A +E B
A B
= t m +m
z 1 2 z 1 2 Z 1 Z 2
C t , t = R t , t −m t m t
Z 1 2 1 2
R t , t = E Z t Z t
( )( )
1 2
= E At + B At +B
( )
2 2
A 1 2 B 1 2
E A t t m m t t E B
( ) ( ) ( )( )
A A Z 1 2 Z 1 2 1 B 2 B
⇒ C t , t = R t , t − m t + m m t +m
( ) ( ) ( )
2 2
Z 1 2 A B 1 2 A1 2 B 1 2
= R t , t − m m t + t + m t t +m t t
⇒ ( ) [ ] [ ]
Z 1 2 1 2
C t , t = VAR A t t +VAR B
ECE863 8
6.18 a) ( )
[ ]
E Z t = 0 (Since
X
m and
Y
m are zeros)
Z 1 2 1 2
C t , t E Z t Z t
( )
( )
1 1 1 2 2 2 2 2
E X t Cos t Y t Sin t X t Cos t Y t Sin t
= ω + ω ω + ω
Since
X t and
Y t independent, and
X Y
m = m = 0
1 2 2 1
⇒ E X t Y t = E X t Y t = 0
Z 1 2 1 2 1 2 1 2 1 2
⇒ C t , t = E X t X t Cos ωt Cos ωt + E Y t Y t Sin ωt Sin ωt
Z 1 2 1 2 1 2 1 2 1 2
⇒ C t t = C t , t Cos ωt Cos ωt + C t , t Sin ω t Sin ωt
( ) ( ) ( ) ( )
Z 1 2 1 2 1 2
C t , t = C t , t Cos t − t ω
6.18 b) Z(t) is a Gaussian random process with zero mean and variance
( )
2
Z t
σ =C t, t
( )
( ( )) ( )
2
z t /2C t,t
Z t
f z e
2 C t, t
−
π
ECE863 9
n
Y takes on two possible values: 1 and 0
( ) ( )
n n n
P Y 1 P I 1 and I is not erased
( )
n n n
P I is not erased I 1 P I 1
( )
= 1 − α p
Therefore,
n
Y is a Bernoulli process with a parameter: ( )
Y
p = 1 − α p
'
n
⇒ S is a binomial process with ( )
n k
' k
n Y Y
n
P S k p 1 p
k
−
Since the binomial process is an “iid sum” process (sum of Bernoulli
random variables with the same parameter), then
'
n
⇒ S has independent and stationary increments
6.25 a)
[ ] [ ]
n 1 2 n
E M E X X X nE X
n n
[ ] [ ]
n X
E M = E X =m
( )( ) n x n x M 1 2 1 2
C n ,n E M m M m
( ) ( )
1
n 1 x n 2 x
1 2
2
E S n m S n m
n n
( )( ) M 1 2 n 1 x n 2 x
1 2
1 2
C n ,n E S n m S n m
n n
S 1 2
1 2
C n n
n n
2
M 1 2 1 2 X
1 2
C n ,n min n n
n n
[ ] ( )
2
n M X
VAR M C n,n
n
= = σ
ECE863 11
Using the stationary increment property of S n
S S S
n n k k
f u f u
−
( ( ) )
( ( ) )
S S S S
n n n k k
f nx, n k y f nx f n k y nx
both ( )
S
n
f u and ( )
S
k
f u are Gaussian pdf functions since they represent
the pdf of a sum of Gaussian iid random variables.
Remember also,
[ ]
( )
2 2
n
VAR S = n σ = n Since σ = 1. And the same for
2
k
VAR S = k σ =k
⇒ using ( )
2
u /2n
Sn
u nx
f u e
2 n
−
=
π
and ( )
( )
2
u /2k
S
k
u n k y nx
f u e
2 k
−
= + −
π
( )
( )
( ) ( )
2
2
n k y nx /2k nx /2n
M M n n k
f n n k e e
2 kn
− + − −
π
6.33 a) ( )
k
t
t
P N t k e
k!
−λ
λ
t
P N t 0 e
−λ
b) The desired probability is:
t t
P N t 0 P N t 1 e te
−λ −λ
= + = = + λ
ECE863 12
6.46 First we derive the cdf for Y(t):
a)
( )
Y t
F y = P Y t ≤ y = P X t + μt ≤y
= P X t ≤ y − μt
( )
( )
X t
= F y − μt
( )
( )
Y t Y t
d
f y F y
dy
( )
( )
( )
Y t X t
f y = f y − μt
Since
( )
2
x /2 t
X t
f x e
2 t
− α
πα
( )
( )
2
y t /2 t
Y t
f y e
2 t
− − μ α
πα
6.46 b) Similar to the above expression
( )
( )
( )
Y t X t
f y = f y − μt , it can be shown:
( ) ( )
( ) ( )
( ) ( ( )) ( )
1 2 1 2 Y t Y t s X t X t s
f y , y f y t y t s
= − μ − μ +
Since the Wiener Process X(t) has independent and stationary
increments
( ) ( )
( )
( )
( )
( ( ) )
1 2 X t 1 2 1 Y t Y t s X t s t
f y , y f y t f y t s y t
⇒ = − μ − μ + − −μ
( )
( )
( )
( )
X t 1 X s 2 1
= f y − μt f y − y − μs
( ) ( )
2 2
y t /2 t y y s /2 s
1 2 1
1 2 Y t Y t s
e e
f y , y
2 t 2 s
− −μ α − − − μ α
πα πα
ECE863 14
6.50 Since
[ ]
X 1 2 1 2
C t , t = VAR ζ Cos 2 tπ Cos 2 tπ
X 1 2
⇒ C t , t does not depend on the difference
1 2
t −t
⇒ X is not wide sense stationary
⇒ X is not stationary
6.53 a) ( )
[ ]
[ ]
[ ]
E X t = E A cos ωt + E B sin ωt
( ) ( )( )
X 1 2 1 1 2 2
C t , t = E Acos ωt + Bsin ωt Acos ωt + Bsin ωt
2
1 2 1 2 1 2
E A cos t cos t E AB sin t cos t cos t sin t
= ω ω + ω ω + ω ω
2
1 2
E B sin t sin t
Since A & B are zero-mean iid RVs
[ ] [ ] [ ]
⇒ E AB = E A E B = 0
and
2 2
E A E B
( ) ( ( ))
2
X 1 2 1 2
C t , t E A cos t t
⇒ = ω −
⇒ X t is wide sense stationary
ECE863 15
6.56 a) We need to evaluate ( )
Y
m t and
Y 1 2
C t , t
[ ] ( )
Y
m t E X t aE X t s
X X
= m − a m (Since X is W.S.S.
X X
m t = m )
( )
Y X
⇒ m t = 1 −a m
[ ]
⇒ E Y t does not change with time
( )
2 2
Y 1 2 1 2 X
C t , t E Y t Y t 1 a m
Therefore, if
1 2
E Y t Y t
is a function of
1 2
t − t , then Y(t) is a
W.S.S. process.
( ) ( ) ( ( ) ( )) ( ( ) ( ))
1 2 1 1 2 2
E Y t Y t E X t aX t s X t aX t s
( ) ( ) ( ) ( )
2
X 1 2 X 1 2 X 1 2 X 1 2
= C t , t − a C t + s, t + C t , t + s + a C t + s, t +s
Since
X 1 2 X 1 2
C t , t = C t − t (X(t) is W.S.S.)
( ) ( ) ( ) ( ) ( )
1 2 X 1 2 X 1 2 X 1 2
⇒ E Y t Y t = C t − t − a C t − t + s + C t − t −s
2
X 1 2
( )
2
Y 1 2 X 1 2 X 1 2 X 1 2
⇒ C t , t = 1 + a C t − t − a C t − t + s + C t − t −s
( )
2 2
X
− 1 −a m
⇒ Y t ( ) is W.S.S. since
Y 1 2 Y 1 2
C t , t = C t −t
and ( )
( )
Y X
m t = constant = 1 −a m
