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Exercise 1 - solution
KTHICT/COS/RST: IK2507
Göran Andersson, goeran@kth.se
Problem 1
Consider the combined experiment consisting of four statistically independent flips of a coin. Suppose that on each flip a head
and a tail are equally likely outcomes.
a. Determine the probability that at most two heads occur in a single performance of the combined experiment, and
b. determine the probability that at least two heads occur in a single performance of the combined experiment.
Solution
a.
Let  be the number of flips that results in a head. Then  œ BinI4, 2 M.
1
4
PH § 2L = ‚PH = kL = ‚
k
2
k=0
2
k=0
1
2
k
1-
1
4-k
=
2
1
2
‚
4 2
k=0
11
4
=
k
16
b.
By symmetry the answer here is the same as in a.
PH ¥ 2L = ‚PH = kL =
2
k=0
1
2
‚
4 4
k=2
1
4
=
k
2
‚
4 4
k=2
1
4
=
4-k
2
‚
4 2
i=0
11
4
=
i
16
2
IK2507_Ex1_solution.nb
Problem 2
Suppose that a random variable X has a uniform probability density
f X HxL =
1
b-a
a§x§b
0
otherwise
a. Calculate the probability that
X§
2a+b
3
b. Let a = -1 and b = 2. In this case, calculate the probability that †X § § c for c = 1 ê 2 and c = 3 ê 2.
Solution
a.
f X HxL
1
b-a
a
x
b
Assume that a < b (the pdf would not make sense otherwise). Notice that
want to cover
1
3
of the distance to b. This means that
P X§
2a+b
=
3
1
3
2 a+b
3
We can also use an integral. Notice that a < b fl a <
P X§
=
2a+b
3
=‡
2 a+b
3
-¶
1
2a+b
b-a
3
f X HxL dx = ‡
-a =
a
1
3
2 a+b
3
< b.
1
b-a
dx
2 a+b
3
=a+
b-a
.
3
The density f X HxL is constant and we
IK2507_Ex1_solution.nb
b.
We have a = -1 Ï b = 2 fl b - a = 3 so
f X HxL =
With c = 1 ê 2:
P †X § §
1
1
3
-1 § x § 2
0
otherwise
=P -
1
§X§
1
2
2
2
3
3
3
= ‡ f X HxL dx = ‡
1ê2
1ê2
-1ê2
-1ê2
With c = 3 ê 2:
P †X § §
=P -
‡ 0 dx + ‡
-1
=
2
-3ê2
3ê2
-1
2
1
3
§X§
2
dx = 0 +
= ‡ f X HxL dx
3ê2
-3ê2
1 3
3 2
- H-1L =
5
6
1
3
dx =
1 1
3 2
- -
1
2
=
1
3
3
4
IK2507_Ex1_solution.nb
Problem 3
Let R be a random variable with the Rayleigh density
fR HrL =
r
b
-
e
r2
0§r
2b
0
otherwise
a. Determine the corresponding probability distribution function and plot your results.
b. Calculate (in terms of the parameter b) the radius R0 such that the probability that R > R0 is e-1 .
c. Calculate the probability density of R2 .
Solution
a.
The cdf is (assuming that r ¥ 0)
PHR § rL = ‡ fR HtL dt = ‡
r
r
-¶
0
t
-
e
b
u=
t2
2b
dt
du =
t2
2b
t
dt
b
r2
= ‡ e-u du = 1 - e
2b
-
r2
2b
0
The cdf is therefore
FR HrL =
-
1-e
0
FR HrL
r2
2b
0§r
otherwise
1
1-
1
‰
R0
r
IK2507_Ex1_solution.nb
b.
PHR > R0 L = e-1 ñ 1 - FR HR0 L = e-1
-
ñe
R20
= e-1 ñ -
2b
R20
= -1
R0 ¥0
2b
R0 =
2b
c.
Set A = R2 and assume that t ¥ 0
F A HtL = PHA § tL = PIR2 § tM = PIR §
tM
R¥0
-
= 1-e
t
2b
fl fA HtL =
d
dt
F A HtL =
d
dt
K1 - e
-
t
2b
This means that R2 œ expHlL with l = 1 ê H2 bL.
O=
1
2b
-
e
t
2b
5
6
IK2507_Ex1_solution.nb
Problem 4
Let X be a random variable with a probability density f X HxL which exists everywhere. Let Y be a new random variable defined as
Y =aX +b
where a and b are real constants and a is nonzero. Determine the probability density fY HyL of the random variable Y .
Solution
The cdf FY HyL is assuming that a > 0
FY HyL = PHY § yL = PHa X + b § yL = P X §
fY HyL =
d
dy
FY HyL =
y-b
d
dy
FX
a
=
a
= FX
y-b
a
y-b
1
a
y-b
fX
a
a < 0:
FY HyL = PHY § yL = PHa X + b § yL = P X ¥
y-b
fY HyL =
1
d
dy
FY HyL =
The pdf fY HyL is therefore
fY HyL =
dy
y-b
1
†a§
d
fX
a
1 - FX
y-b
a
=-
a
a
= 1 - FX
y-b
fX
a
y-b
a
IK2507_Ex1_solution.nb
Problem 5
Consider a binomial random variable Yn , n = 1, 2, 3, …, which has the probability density
PHYn = kL =
Determine EHYn L
n!
k ! Hn - kL !
pk H1 - pL
n-k
Solution
The expectation value is according to the definition with q = 1 - p
EHYn L = ‚k PHYn = kL = ‚k
The expression k
k
n
n
k=0
k=0
n
k
n
n k n-k
n k n-k
p q = ‚k
p q
k
k
k=1
can be rewritten in a smart manner:
n!
n ÿHn - 1L ÿ… ÿHn - k + 1L
n
=k
=k
k
k ! Hn - kL !
k!
= n
Hn - 1L ÿ … ÿ HHn - 1L - Hk - 1L + 1L
\ EHYn L = ‚n
n
k=1
= n p‚
n-1
i=0
Hk - 1L !
=n
n-1
k-1
n
n - 1 k n-k
n - 1 k-1 n-1-Hk-1L
p q =n p‚
p q
k-1
k-1
k=1
n - 1 i n-1-i
p q
= n p H p + qLn-1 = n p ÿ1n-1 = n p
i
Alternative
The partial derivative is powerful tool in this case.
n
∑ pk
∑ n n
n k n-k
n
qn-k = p
p q =‚
p
pk qn-k
‚
‚k
k
k
k
∑
p
∑
p
k=0
k=0
k=0
n
= p
∑
∑p
H p + qLn = p n H p + qLn-1 = n p
7
8
IK2507_Ex1_solution.nb
Problem 6
Consider a sine wave process 8Xt , 0 § t<, where
Xt = V cosHw tL
Show whether this is a stationary process in the strict sense.
Solution
We assume that w is a constant. The first moment or expectation value is
E@Xt D = E@V cosHw tLD = E@V D cosHw tL
Since E@Xt D is time dependent (if E@V D ∫ 0) this is not a strict stationary process.
Another way to see this is to look at the distribution:
F X Hx, tL = PHXt § xL = PHV cosHw tL § xL
F X Hx, t + tL = PHXt+t § xL = PHV cosHw Ht + tLL § xL ∫ PHV cosHw tL § xL = F X Hx, tL
Since cos Hw Ht + tLL ∫ cosHw tL this is not a strict stationary process. In other words: time shift influence the cdf.
IK2507_Ex1_solution.nb
Problem 7
Let 8Xt , -¶ < t < ¶< be a strictly stationary real random process. Show that
a. R X H-tL = R X HtL, i.e. the autocorrelation function of a stationary real random process is an even function of its argument.
b. †R X HtL§ § EIXt2 M = R X H0L, i.e. the autocorrelation function of stationary, real random, process has its maximum magnitude at the origin in t.
Solution
a.
We use the fact that Xt is (wide-sense) stationary then R X Ht, t + tL = R X HtL.
R X HtL = E@Xt Xt+t D = E@XHt+tL-t Xt+t D
h E@XT-t XT D = E@XT XT+H-tL D = R X H-tL
b.
2
HXt ≤ Xt+t L2 ¥ 0 ñ Xt2 + Xt+t
¥ ¡ 2 Xt Xt+t
Using the fact that Xt is (wide-sense) stationary and forming the expectation value gives
2
EAXt2 + Xt+t
E = 2 EAXt2 E ¥ ≤2 E@Xt Xt+t D
This means that
R X H0L ¥ †R X HtL§
QED
9
10
IK2507_Ex1_solution.nb
Problem 8
Wide-sense stationary. Let the random process 8Zt , -¶ < t < ¶< be such that
ó
Zt = Y cosHtL + X sinHtL
for all values of t, where X and Y are independent binary random variables each of which assumes the values -1 and +2 with
probabilities
2
3
and
1
3
respectively.
1. Calculate E@Zt D and RZ Ht1 , t2 L.
2. Show that this process is stationary in the wide sense but not in the strict sense. (Hint: Consider EAZt3 E.)
Solution
1.
E@X D = E@Y D = H-1L ÿ
2
3
+ 2ÿ
1
3
= 0, EAX 2 E = EAY 2 E = H-1L2 ÿ
2
3
+ 22 ÿ
1
3
=2
Using the independence we have:
E@X Y D = E@X D E@Y D = 0
Expansion of Zt1 Zt2 gives
Zt1 Zt2 = HY cosHt1 L + X sinHt1 LL HY cosHt2 L + X sinHt2 LL
h X 2 sinHt1 L sinHt2 L + X Y HsinHt1 L cosHt2 L + cosHt1 L sinHt2 LL + Y 2 cosHt1 L cosHt2 L
Now forming the expectation values gives
E@Zt D = E@X cosHtL + Y sinHtLD = E@X D cosHtL + E@Y D sinHtL = 0
RZ Ht1 , t2 L = E@Zt1 Zt2 D = E@HX cosHt1 L + Y sinHt1 LL HX cosHt2 L + Y sinHt2 LLD
0
0
h EAX 2 E sinHt1 L sinHt2 L + E@X Y D H…L + EAY 2 E cosHt1 L cosHt2 L
0
h 2 sinHt1 L sinHt2 L + 2 cosHt1 L cosHt2 L = HcosHt1 - t2 L - cosHt1 + t2 LL + HcosHt1 - t2 L + cosHt1 + t2 LL
2
h 2 cosHt1 - t2 L
2
IK2507_Ex1_solution.nb
2.
From 1. we have that E@Zt D = 0, independent of t. Setting t1 = t and t2 = t + t gives
RZ Ht, t + tL = 2 cosHt - Ht + tLL = 2 cosHtL
independent of t. This shows that Zt is wide sense stationary. Using the hint we have:
EAX 3 E = EAY 3 E = H-1L3 ÿ
2
3
+ 23 ÿ
1
3
= 2, EAX 2 Y E = E@X D2 E@Y D = 0, EA X Y 2 E = 0
EAZt3 E = EAHY cosHtL + X sinHtLL3 E
h EAX 3 E sin3 HtL + 3 EAX 2 Y E sin2 HtL cosHtL + 3 EAX Y 2 E sinHtL cos2 HtL + EAY 3 E cos3 HtL
h 2 Isin3 HtL + cos3 HtLM
Since the 3rd moment EAZt3 E clearly is time-dependent the process cannot be strict sense stationary.
11
12
IK2507_Ex1_solution.nb
Problem 9
Consider a wide sense stationary process 8Xt , -¶ < t < ¶<, with E@Xt D = 0, " t, and a periodic function f HtL = f Ht + TL. Prove
that Y HtL = f HtLÿ Xt is cyclostationary with period T.
Solution
The autocorrelation is
RY Ht, t + tL = E@Y HtL Y Ht + tLD = E@ f HtL ÿ Xt ÿ f Ht + tL ÿ Xt+t D
h f HtL f Ht + tL E@Xt Xt+t D = f HtL f Ht + tL Rx HtL
Since f HtL is periodic we have
RY Ht + T, t + T + tL = f Ht + TL f Ht + T + tL Rx HtL = f HtL f Ht + tL Rx HtL = RY Ht, t + tL
This means that RY Ht, t + tL is periodic in t with period T, i.e. Y HtL is wide sense cyclostationary.
IK2507_Ex1_solution.nb
13
Problem 10
Suppose that 8Xt , -¶ < t < ¶< is a zero mean, wide-sense stationary random process whose spectral density
S X H f L = 0, " f > f0 . The process Xt is input in a modulator characterized by the sinusoidal random process 8Wt , -¶ < t < ¶<,
which
Wt = cosH p t + ZL
ó
where f p =
p
2p
p f0 , Z œ U@0, 2 pD and Xt and Z are independent random variables for all t. The modulator output is such that
Yt = Xt Wt = Xt cosH p t + ZL.
1. Show that Yt is wide sense stationary and that its auto correlation function is given by
RY HtL =
1
R X HtL cosH p t L
2
2. Show that the spectral density of the output process is
SY H f L =
1
SX I f - f pM +
SX I f + f pM
4
4
3. Assuming that S X is rectangular, sketch SY as function of the frequency.
1
Solution
1.
E@Yt D = E@Xt Wt D = E@Xt D E@Wt D = E@Xt Dÿ 0 = 0
indep
RW HtL = E@Wt Wt+t D = E@cosH p t + ZL cosH p Ht + tL + ZLD
h
1
2
E@cosH p tL + cosH p H2 t + tL + 2 ZLD =
1
cosH p tL
2
RY HtL = E@Yt Yt+t D = E@Xt Wt Xt+t Wt+t D = E@Xt Xt+t D E@Wt Wt+t D
indep
h R X HtL RW HtL =
1
2
R X HtL cosH p tL =
This shows that Yt is wide sense stationary.
1
2
R X HtL cosI2 p f p tM
14
IK2507_Ex1_solution.nb
2.
SY H f L = 't @RY HtLD H f L = ‡
¶
-¶
h
1
h
1
h
1
h
1
2
4
4
4
‡
¶
-¶
K‡
RY HtL ‰-i 2 p f t dt
R X HtL cosI2 p f p tM ‰-i 2 p f t dt =
¶
-¶
R X HtL ‰-i 2 p I f - f p M t dt + ‡
¶
-¶
1
4
‡
¶
-¶
R X HtL I‰i 2 p f p t + ‰-i 2 p f p t M ‰-i 2 p f t dt
R X HtL ‰-i 2 p I f + f p M t dtO
I't @R X HtLD I f - f p M + 't @R X HtLD I f + f p MM
IS X I f - f p M + S X I f + f p MM
3.
SX H f L
- f0
- fp
f
SY H f L
f0
f p - f0
f p + f0
f