Download Iverse Transform-Probability and Statistics-Assignment Solution and more Exercises Probability and Statistics in PDF only on Docsity!
1
ECE 863 – Analysis of Stochastic Systems
Fall 2000
Solutions for Homework Set
Problem 1
[ ] [ ]
n n
n i i
i 1 i 1
E S E X E X n
= =
= = = μ
n n n
n k j k
k 1 j 1 k 1
j k
VAR S VAR X COV X , X
= = =
≠
The above terms can be organized in an (nxn) matrix as follows:
2 2
2 2 2
2
2 2
K
σ ρσ
ρσ σ ρσ
ρσ
ρσ σ
VAR(S
n
) is the sum of all terms inside the above matrix.
This leads to:
VAR(S
n
) = n σ
2
2
Problem 2
5.4 a) By Eqn. 5.7, we have
w w ( )w
Z X Y
w w w e e e
−α −β − α +β
2
b) The above characteristic function is for a Cauch RV with parameter (α+β).
Therefore, taking the inverse transform leads to:
( )
Z
2 2
f z
z
α + β
α + β +
Therefore, Z is also Cauchy.
Problem 3
5.8 a) ( )
( )
j aX bY w jawX jbwY
Z
w E e E e E e
( )
X Y
= Φ aw Φ bw
b) [ ] ( ) ( ) ( ) ( ) ( )
' ' '
Z X Y Y X
w 0 w 0 w 0
E Z w aw a bw bw b aw
j j j
= = =
where
'
Z
Φ w is the derivative of the characteristic function with respect to w.
Since ( )
Y Y X X
w 0 w 0
bw and aw
= =
Φ = Φ = Φ = Φ = , then the above
lead to:
[ ] [ ] [ ]
E Z = aE X +bE Y
2 "
Z
w 0
E Z w
=
( ) ( ) ( ) ( ) ( ) ( )
' ' " 2 " 2
X Y X X Y Y
w 0
aw a bw 2 aw a bw b aw bw b
=
[ ] [ ]
2 2 2 2
a E X b E Y 2abE X E Y
( )
[ ] [ ]
( [ ]) ( [ ]) [ ] [ ]
2
2
2 2
2 2 2 2
2 2
VAR(Z) E[Z ] E[Z]
a E X b E Y 2abE X E Y
a E X b E Y 2abE X E Y
( ) [ ] ( ) ( )
2
2 2 2
VAR Z E Z E Z a VAR X b VAR Y
4
Problem 7
5.12 a) Note first that
[ ]
n
k
k 1
E S/N n E X nE X
=
thus
[ ] [ ] [ ] [ ]
E S E E S/N E NE X E N E X
Now for the variance, we use: ( ) [ ] ( )
2
2
VAR S E S E S
2 2
E S E E S /N
First, let's find the conditional expectation:
n n n n
2
i j i j
i 1 j 1 i 1 j 1
E S N n E X X E XX
= = = =
since
2
i j
E XX E X
if i = j and [ ]
2
i j
E XX E X
if i ≠ j, then
( ) [ ]
2 2 2
E S N n nE X n n 1 E X
Therefore, ( ) [ ]
2 2 2
E S E NE X N N 1 E X
[ ] [ ] [ ] [ ]
2 2
2 2
E N E X E N E X E N E X
Then
( ) [ ]
2 2
VAR S E S E S
[ ] [ ] [ ] [ ] [ ] [ ]
2 2 2 2 2 2
E N E X E N E X E N E X E N E X
[ ]
( ) ( ) ( [ ])
2
n
VAR S ( ) = E N VAR X +VAR N E X
5
b) We start (again) by using conditional expectation:
1
n
X
i
n S X X
i n 1
X
E z /N n E z E z E z G z
= ¹
Therefore,
S S
E z E E z N
N
X
E G z
( )
N
w G z
X
E w
=
( )
N X
=G G z
Problem 8
5.14 This problem can be modeled as the random sum of random variables:
N
i
i 1
S X
=
Here, X i
is a Bernoulli random variable with a probability of “success” is (ε)
(in this case success = error):
P[ X
i
= 1 ] = ε and P[ X
i
= 0 ] =1− ε
Moreover, the number of elements in the sum is a binomial random variable N with
parameters (n) and (p):
k n k
n
P[N k] p ( p)
k
−
For the random variable N, the probability of “success” is (p). In this case, “success” means
“ a packet is transmitted”. Therefore:
th
i
N message transmissions in n trials
and X indicator function of error in i transmission
Consequently, S measures the number of errors in a random number (N) of trails
(transmissions)
7
Problem 9
n
n
n
2 2 2
VAR S
VAR S
n
P S
n n
− μ > ε ≤ =
ε ε
2 2
2 2
n 2 n 1
0 as n
n
σ + − ρσ
ε
Therefore, the Week Law of Large Numbers holds in this case.
Problem 10
5.21 a) Let's look at both sides of the equality that we are trying to prove:
n n
2 2 2 2
j j j n
j 1 j 1
LHS X 2 X X 2 nM n
= =
= − μ + μ = − μ + μ
n
2
2 2
j n j n n
j 1
RHS X 2M X M n M
=
= − + + − μ
n
2 2 2 2
j n n n n n
j 1
X 2M nM nM nM 2n M n
=
= − + + − μ + μ
n
2 2
n j
j 1
X 2n M n LHS
=
= − μ + μ =
b)
n n
2 2
2
j n j n
j 1 j 1
from part a
E k X M kE X n M
= =
− = − μ − − μ
n
2
2
j n
j 1
n
k E X knE M
knVAR(X) knVAR(M )
=
= − μ − − μ
2 2
2
n
kn kn since VAR M
n n
σ σ
= σ − =
2
= k n − 1 σ
8
Problem 10 – Continued:
c) If
k
n 1
then
2 2
n
E V
= σ
d) if
k
n
= then
n
2
2 2 2
j n
j 1
E X M 1
n n n
=
− = − σ = σ − σ
Therefore, the expected value of the estimator does not equal to the parameter
(which is the variance in this case) that we are trying to estimate.
Problem 11
5.28 Using the fact that the time between arrivals is an exponential RV with mean
and variance
2
; the time of the nth arrival is:
n n 1
S = X + …+X
where
[ ] [ ]
n i
E S = nE X = n/
and [ ] [ ]
n i
VAR S = nVAR X =n/
650
more than 650
P messages in P S 60
60 seconds
Since S
650
is a random variable with mean (65) and variance (6.5), then the above
probability is equivalent to the following:
650
S 65
P
From the central limit theorem, this probability converges to a Gaussian.
