UNIVERSITY OF CALIFORNIA, SAN DIEGO
Electrical & Computer Engineering Department
ECE 250 - Winter Quarter 2018
Random Processes
Solutions to P.S.
... [Show More] #2
1. Probabilities from a cdf. Let X be a random variable with the cdf shown below.
F(x)
x
1
2/3
1/3
1 2 3 4
1/3x
Find the probabilities of the following events.
(a) fX = 2g.
(b) fX < 2g.
(c) fX = 2g [ f0:5 ≤ X ≤ 1:5g.
(d) fX = 2g [ f0:5 ≤ X ≤ 3g.
Solution:
(a) There is a jump at X = 2, so we have
PfX = 2g = PfX ≤ 2g - PfX < 2g
= F (2) - F (2-)
=
2 3
-
1 3
=
1 3
:
1
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(b) PfX < 2g = F (2-) = 1 3.
(c) since fX = 2g and f0:5 ≤ X ≤ 1:5g are two disjoint events,
P(fX = 2g [ f0:5 ≤ X ≤ 1:5g) = PfX = 2g + Pf0:5 ≤ X ≤ 1:5g
=
1 3
+ F (1:5) - F (0:5-)
=
1 3
+
1 3
-
1 3
× 0:52
=
7
12
:
(d) We have
P(fX = 2g [ f0:5 ≤ X ≤ 3g) = Pf0:5 ≤ X ≤ 3g
= F (3) - F (0:5-)
=
5 6
-
1 3
× 0:52
=
3 4
:
2. Gaussian probabilities. Let X ∼ N(1000; 400). Express the following in terms of the Q
function.
(a) Pf0 < X < 1020g.
(b) PfX < 1020jX > 960g.
Solution: Using the fact that Xσ-µ ∼ N(0; 1), thus F (x) = Φ(x-σµ) = 1 - Q(x-σµ).
(a) We have
Pf0 < X < 1020g = Q 0 -20 1000 - Q 102020 - 1000 = Q(-50) - Q(1):
(b) We have
PfX < 1020jX > 960g = Pf960 < X < 1020g
PfX > 960g
=
Q(960-201000) - Q(102020 -1000)
Q(960-201000)
=
Q(-2) - Q(1)
Q(-2) :
2
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3. Laplacian. Let X ∼ f(x) = 1 2e-jxj.
(a) Sketch the cdf of X.
(b) Find PfjXj ≤ 2 or X ≥ 0g .
(c) Find PfjXj + jX - 3j ≤ 3g .
(d) Find PfX ≥ 0jX ≤ 1g .
Solution:
(a) We have
FX(x) = Z-1 x 1 2e-jujdu = 11 2e-x;12e-x; if if x < x ≥ 00:
-6 -4 -2 0 2 4 6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
F
X(x)
Figure 1: cdf of X
(b) We have
PfjXj ≤ 2 or X ≥ 0g = PfX ≥ -2g
= 1 - PfX < -2g
= 1 - Z-1 -2 1 2e-jxjdx
= 1 -
1 2
e-2:
(c) We have
PfjXj + jX - 3j ≤ 3g = Pf0 ≤ X ≤ 3g
= Z03 1 2e-jxjdx
=
1 2
-
1 2
e-3:
3
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(d) We have
PfX ≥ 0 j X ≤ 1g = Pf0 ≤ X ≤ 1g
PfX ≤ 1g =
FX(1) - FX(0-)
FX(1) =
1=2 - 1=2e-1
1 - 1=2e-1 =
1 - e-1
2 - e-1 :
4. Distance to the nearest star. Let the random variable N be the number of stars in a region
of space of volume V . Assume that N is a Poisson r.v. with pmf
pN(n) = e-ρV (ρV )n
n! ; for n = 0; 1; 2; : : : ;
where ρ is the "density" of stars in space. We choose an arbitrary point in space and define
the random variable X to be the distance from the chosen point to the nearest star. Find the
pdf of X (in terms of ρ).
Solution: The trick in this problem, as in many others, is to find a way to connect events
regarding X with events regarding N. In our case, for x ≥ 0:
FX(x) = PfX ≤ xg
= 1 - PfX > xg
= 1 - PfNo stars within distance xg
= 1 - PfN = 0 in sphere centered at origin of radius xg
= 1 - e-ρ 43 πx3:
Now differentiating, we get
fX(x) = 4πρx2e-ρ 43 πx3:
For x < 0, both the cdf and the pdf are zero everywhere.
5. Uniform arrival. The arrival time of a professor to his office is uniformly distributed in the
interval between 8 and 9 am. Find the probability that the professor will arrive during the
next minute given that he has not arrived by 8:30. Repeat for 8:50.
Solution: For convenience, let us denote the length of one hour by a.
Then, without loss of generality, we can consider the random variable T , denoting the arrival
time of the professor in his office, to be distributed uniformly in [0; a].
Let 0 < η < 1.
Then we have to calculate the probability that T will lie between ηa and ηa + a=60, given
that T does not lie in [0; ηa], for η = 1=2 and η = 5=6.
4
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