Statistics Exam
Telecommunication Engineering
June 24th , 2013
Solution
Questions
1h 30m
Q1. (2p) A communication system is composed of the components: A, B, C, D and E, where each
component works independently. The system is shown in Figure 1:
B
A
C
in
out
D
E
Figure 1: Diagram of the system.
Let P (A) = P (B) = P (D) = P (E) = 0.8 and P (C) = 0.9, the probabilities that each component works
properly in the system.
a) Calculate the probability of system failure.
b) If the system is working properly, what is the probability that the component C is working
properly?
Solution:
a) If F denotes the event: “the system is working properly”, we have to calculate: P (F c ) = 1 − P (F ).
Therefore:
P (F )
=
P ((A ∩ B) ∪ (A ∩ C) ∪ (D ∩ E))
1
Using the notation: AB = A ∩ B, AC = A ∩ C and DE = D ∩ E, then
P (F )
= P (AB ∪ AC ∪ DE)
= P (AB) + P (AC) + P (DE) − P (AB ∩ AC) −
P (AB ∩ DE) − P (AC ∩ DE) + P (AB ∩ AC ∩ DE)
= P (A)P (B) + P (A)P (C) + P (D)P (E) − P (A)P (B)P (C) −
P (A)P (B)P (D)P (E) − P (A)P (C)P (D)P (E) + P (A)P (B)P (C)P (D)P (E)
= (0.8)2 + (0.8)(0.9) + (0.8)2 − (0.8)2 (0.9) − (0.8)4 − (0.8)3 (0.9) + (0.8)4 (0.9)
=
0.92224.
(1)
Finally P (F c ) = 1 − P (F ) = 1 − 0.92224 = 0.07776.
b) Using Bayes theorem:
P (C|F )
=
P (F |C)P (C)
P (F )
Since: P (F |C) = P (A ∪ (D ∩ E)) = P (A) + P (D ∩ E) − P (A ∩ D ∩ E) = 0.8 + 0.82 − 0.83 = 0.928,
it follows that
P (C|F )
Q2.
=
P (A ∪ (D ∩ E))P (C)
(0.928)(0.9)
=
= 0.90562
P (F )
(0.92224)
(1.5p) A discrete random variable has the following probability function:
xi
P (X = xi )
1
0.15
2
a
3
b
4
c
5
0.1
and we know that P (X < 4) = 0.65 and P (X > 2) = 0.6.
a) Calculate E(X) and V ar(X).
b) Write and draw the distribution function of the random variable X.
c) Complete the following MATLAB code to approximate P (X ≤ 3).
u=rand(1000,1);
x=1*(0<u & u<=0.15)+ 2*(0.15<u & u<=0.40)+3*(0.40<u & u<=____________)
+4*(0.65<u & u<=0.9)+5*(0.9<u & u<=1);
w=(____________);
p=sum(w)/1000;
Solution:
a) First we calculate the values of a, b and c. Solving the following system:
0.15 + a + b = 0.65
b + c + 0.1 = 0.6
0.15 + a + b + c + 0.1 = 1
2
we get that a = b = c = 0.25, then
E(X)
=
1(0.15) + 2(0.25) + 3(0.25) + 4(.25) + 5(0.1) = 2.9.
To calculate the variance we have to do:
E(X 2 )
=
12 (0.15) + 22 (0.25) + 32 (0.25) + 42 (.25) + 52 (0.1) = 9.90,
and then,
V ar(X)
=
2
E(X 2 ) − (E(X)) = 9.90 − 2.92 = 1.49 .
b)

0





0.15



 0.4
F (x) =

0.65





0.9



1,
if
if
if
if
if
if
x<1
1≤x<2
2≤x<3
3≤x<4
4≤x<5
x ≥ 5.
Figure 2: Distribution function.
c) Complete the following code in MATLAB to approximate P (X ≤ 3).
3
(2)
u=rand(1000,1);
x=1*(0<u & u<=0.15)+ 2*(0.15<u & u<=0.40)+3*(0.40<u & u<=0.65)
+4*(0.65<u & u<=0.9)+5*(0.9<u & u<=1);
w=(x<=3);
p=sum(w)/1000;
Q3.
(1.5p) In an industrial manufacturing process, the failures that may occur are classified into 3
types:
- Failures caused by breakdowns produced in the machinery (B).
- Failures caused by the presence of pollutants in the raw materials used (P).
- Failure due to other causes (O).
If such failures appear independenlty of each other according to Poisson processes with parameters
λB = 0.25 failures per week, λP = 0.50 failures per week, and λO = 0.125 failures per week, respectively,
answer the following questions:
a) The probability that in a given week no failures has occurred due to breakdowns produced in the
machinery.
b) The probability that in a given week at least a failure has occurred.
c) The probability that the time between two consecutive failures exceeds 4 weeks.
Solution:
Let XB ∼ P oisson(0.25), XP ∼ P oisson(0.50), and XO ∼ P oisson(0.125), three indepen-
dent r.v.
a)
Pr(XB = 0) =
e−λB λ0B
= e−0.25 = 0.7788.
0!
b) Let XT = XB + XP + XO , due to the reproductive property of the Poisson distribution, it follows
that XT ∼ P oisson(λT = 0.25 + 0.50 + 0.125 = 0.875). Therefore,
Pr(XT ≥ 1) = 1 − Pr(XT = 0) = 1 −
e−λT λ0T
= 1 − e−0.875 = 0.5831.
0!
c)
Pr(T > 4) = 1 − FT (4) = 1 − (1 − e−0.875·4 ) = e−3.5 = 0.0302,
where T ∼ Exponential(λT = 0.875).
4
Statistics Exam
Telecommunication Engineering
June 24th , 2013
Problems
1h 30m
P1.
(2.5p)We select at random a point from the interval (0, 0.5) following an U (0, 0.5) distribution,
and another point, independently, also from the the interval (0, 0.5) following an U (0, 0.5). Let X1 be a
random variable that represents the position of the first point and let X2 be another random variable
that represents the position of the second point. Answer the following questions:
a) Write down the joint density function of the vector (X1 , X2 ), f(X1 ,X2 ) (x1 , x2 ) and its support.
b) Determine the density function of
X1
X2 .
c) Show that the mean (expectation) of
d) Calculate P X1 < X22 .
X1
X2
is infinite.
Solution:
a) Using the independence between X1 and X2 :
f(X1 ,X2 ) (x1 , x2 ) = fX1 (x1 ) · fX2 (x2 ) =
1
1
·
=4
0.5 − 0 0 − (−0.5)
for (x1 , x2 ) ∈ (0, 0.5) × (0, 0.5), that is, 0 < x1 < 0.5 and 0 < x2 < 0.5.
b) The transformations x1 /x2 and x2 are 1-1 and derivable hence:
f(Y1 ,Y2 ) (y1 , y2 ) = f(X1 ,X2 ) (x1 (y1 , y2 ), x2 (y1 , y2 )) · |J|
where J is the Jacobian. The inverse transforms are:
The Jacobian is:
x1
= y1 y2
x2
= y2
y
2
J =
0
y1
1
= y2
thus
f(Y1 ,Y2 ) (y1 , y2 ) = 4|y2 | = 4y2
for 0 < y2 < 0.5 and 0 < y1 y2 < 0.5 (see the diagram in Figure 3) and 0 otherwise.
The (marginal) density of Y1 = X1 /X2 is:
( R 0.5
0.5
4y2 dy2 = 21 = 2y22 0 = 0.5
0
1/2y1
R 1/2y1
=
fY (y1 )
| 1{z }
4y2 dy2 = 2y22 0
= 2y12
0
R
∞
−∞
1
f (y1 ,y2 )dy2
(
=
0.5
1
2y12
for 0 < y1 < 1
for 1 < y1 < ∞
5
for 0 < y1 < 1
for 1 < y1 < ∞
3
2
1
Y2=1/(2y1)
-3
-1
-2
1
2
3
-1
-2
-3
Figure 3: Support of (Y1 , Y2 )
c) The expectation of Y1 is:
Z ∞
1
1
y1 dy1 +
y1 2 dy1
2
2y
0
1
1
1 1
∞
+ [ln(y1 )]1 = ∞
4 2
Z
E[Y1 ]
=
=
1
d) The probability is 0.25:
Y1
Z
z }| {
P (X1 /X2 < 0.5) =
0.5
Z
fY1 (y1 )dy1 =
−∞
0
0.5
1
dy1 = 0.25
2
P2.
(2.5p) In a communication system the following random signal is used, Y (t) = A · sin(ωt + φ),
where A is a random variable that follows a Bernoulli distribution with parameter p > 0, φ is another
random variable, independent of A, that follows a uniform distribution on the interval (−π, π) and ω is
a known constant.
Useful relationships:
cos(a + b)
=
cos(a) cos(b) − sin(a) sin(b)
cos(a − b)
=
cos(a) cos(b) + sin(a) sin(b)
sin(a + b)
=
sin(a) cos(b) + cos(a) sin(b)
sin(a − b)
=
sin(a) cos(b) − cos(a) sin(b)
1. Calculate the (statistical) mean of the process Y (t).
2. Calculate the (statistical) autocorrelation function of Y (t).
3. Is the process Y (t) weakly stationary?.
6
4. Check if the process Y (t) is ergodic for the mean and autocorrelation.
Solution:
a) The process mean is:
= E[A · sin(ωt + φ)]
E[Y (t)]
· E[sin(ωt + φ)]
1
sin(ωt + φ) dφ
p·
2π
−π
p
π
( − cos(ωt + φ))|−π
2π
p
×0=0
2π
=
indep. E[A]
Z π
=
=
=
b) The autocorrelation function of the process is:
RY (t, t + τ )
=
E[Y (t)Y (t + τ )]
=
E[A · sin(ωt + φ)A · sin(ω(t + τ ) + φ)]
=indep.
z }| {
E[A2 ] ·E[sin(ωt + φ) sin(ω(t + τ ) + φ)]
p
=
=
=
=
using b = ωt + φ, a = (ω(t + τ ) + φ)


a+b
a−b
}|
{
z
z}|{
1 

p · E cos( ωτ ) − cos(ω(2t + τ ) + 2φ)
2
π
1
cos(ωτ ) −
cos(ω(2t + τ ) + 2φ) dφ
2π
−π
(
π )
p
1 cos(ωτ ) (− sin(ω(2t + τ ) + 2φ)
2
4π −π
p
2
Z
p cos(ωτ )
2
c) Since the process mean is a constant that does not depend on t and the autocorrelation function only
depends on τ , the process is weakly stationary.
d) The temporal process mean is:
µT
z
MY
µT
MY
}|
{
T
1
Y (t)dt
= lı́m
T →∞ 2T −T
Z T
1
=
A · sin(ωt + φ)dt
2T −T
T !
A
cos(ωt + φ) =
−
2T
w
−T
=
Z
lı́m µT = 0 = E[Y (t)]
T →∞
Hence the process is ergodic for the mean.
The temporal autocorrelation is:
7
∗
{
T
1
Y (t)Y (t + τ )dt
AY = lı́m
T →∞ 2T −T
Z T
1
∗ =
A · sin(ωt + φ)A · sin(ω(t + τ ) + φ)dt
2T −T
Z
A2 T 1
=
{cos(ωτ ) − cos(ω(2t + τ ) + 2φ)} dt
2T −T 2
(
T )
sin(ω(2t + τ ) + 2φ)
A2
cos(ωτ )2T −
=
4T
2ω
−T
z
}|
Z
0
2
AY
=
=
z
2
}|
{
sin(ω(2T + τ ) + 2φ) sin(ω(−2T + τ ) + 2φ)
−
2ω
2ω
A cos(ωτ )
A
− lı́m
T
→∞
2
4T
A2 cos(ωτ )
6= RY (t, t + τ ).
2
Hence the process is not ergodic for the autocorrelation.
8