ECE 341: Probability and Random Processes for Engineers, Spring 2012
Homework 6
Name:
Assigned: 02.15.2012
Due: 02.22.2012
Problem 1. A certain deep space transmitter uses on-off modulation of a laser to send a bit, with
value either zero or one. If the bit is zero, the number of photons, X, arriving at the receiver has
the Poisson distribution with mean λ0 = 2; and if the bits is one, X has the Poisson distribution
with mean λ1 = 6. A decision rule is needed to decide, based on observation of X, whether the bit
was a zero or a one.
1. Suppose we decide that a 1 is sent if P [X = k|one is sent] > P [X = k|zero is sent]. What
does this decision rule simplify to in terms of X?
2. Suppose we know that the probability of sending a zero is π0 and that of sending a one is
π1 . Suppose now that we decide a 1 is sent if P [X = k, one is sent] > P [X = k, zero is sent].
What does this decision rule simplify to now in terms of X if π0 = 5π1 ?
Solution 1:
1. Decision Rule is in favor of 1 if
P [X = k|one is sent] ≥ P [X = k|zero is sent]
Thus
P [X = k|one is sent]
≥1
P [X = k|zero is sent]
(1)
(2)
(3)
But it is given that the number of photons X arriving at the receiver has Poisson distribution, then
λk
e−λ1 k!1
P [X = k|one is sent]
λ1
3k
= ( )k e−(λ1 −λ0 ) = 3k e−4 ≈
=
k
λ
P [X = k|zero is sent]
λ0
54.6
e−λ0 k!0
Thus based on this result, we decide that a bit 1 is sent if
1
3k
54.6
≥ 1, that is if X = k ≥ 4.
(4)
3.1. CUMULATIVE DISTRIBUTION FUNCTIONS
71
The sum of a series is, by definition, the limit of the sum of the first n terms as n → ∞. Therefore,
n
�
P {X < c} = lim
P (Gk ) = lim P {G1 ∪ G2 ∪ · · · ∪ Gn } = lim P {X ≤ cn } = lim FX (cn ).
n→∞
n→∞
n→∞
n→∞
2. Now Decision Rulek=1
is in favor of 0 if
That is, P {X
<=
c} k,
=one
FX (c−).
The
conclusion
of the proposition follows directly from (5)
the
P [X
is sent]
≥ Psecond
[X = k,
zero is sent]
first: P {X = c} = P {X ≤ c} − P {X < c} = FX (c) − FX (c−) = �FX (c).
(6)
P [X = k|one is sent]P [one is sent] ≥ P [X = k|zero is sent]P [zero is sent]
(7)
(8)
Example 3.1.3 Let X have the CDF shown in Figure 3.2.
P [X = k|one is sent]π1 ≥ P [X = k|zero is sent]π0
(9)
(a) Determine all values of u such
that P {X = u} > 0. (b) Find P {X ≤ 0}. (c) Find P {X < 0}.
(10)
P [X = k|one is sent]
π0
!
≥
=5
(11)
P [X = k|zero is sent]
π1
"#$
(12)
P [X = k|one is sent]
3k
=
≥5
P [X = k|zero is sent]
54.6ï!
(13)
0
Thus the condition above is satisfied when X ≥6
Figure 3.2: An example of a CDF.
Problem 2. Let X have CDF shown in Fig. . Find the numerical values of the following
quantities:
Solution: (a) The CDF has only one jump, namely a jump of size 0.5 at u = 0. Thus, P {X =
[X and
≤ 1]there are no values of u with u �= 0 such that P {X = u} > 0.
0}1.=P0.5,
(b) P {X ≤ 0} = FX (0) = 1. (c) P {X < 0} = FX (−0) = 0.5.
2. P [X ≤ 10]
3. P [X ≥ 10]
Example
4. P [X =3.1.4
10] Let X have the CDF shown in Figure 3.3. Find the numerical values of the
following quantities:
[|X≤−1},
5| ≤
(a)5. PP{X
(b)0.1]
P {X ≤ 10}, (c) P {X ≥ 10}, (d) P{X=10}, (e) P {|X − 5| ≤ 0.1}.
F (u)
X
1.0
0.75
0.5
.
0.25
0
Solution 2:
u
10
5
15
Figure 3.3: An example of a CDF.
• P [X ≤ 1] = FX [1] = 0.05 where FX (1) is calculated from the equation of the line FX [x] =
0.05x when x < 5
• P [X ≤ 10] = FX [10] = 0.75 from the graph.
• P [X ≥ 10] = 1 - P (X < 10) = 1 - FX (10− ) = 0.5.
2
• P [X = 10] = FX (10+) − FX (10−) = 0.75 - 0.5 = 0.25.
• P [|X − 5| ≤ 0.1] = P [4.9 ≤ X ≤ 5.1] = P (X ≤ 5.1) − P (X < 4.9) = 0.5 - 0.245 = 0.255.
where FX (4.9) is calculated from the equation of the line FX [x] = 0.05x when x < 5
Problem 3. Exponential random variables have a nice property termed the “memoryless property”, which means that P [T > s + t|T > s] = P [T > t], where T is exponentially distributed.
Show this property.
Solution 3: The aim is to prove that exponential random variables have property of being memoryless i.e P [T > s + t|T > s] = P [T > t].
Using Bayes Rule
P [T > s + t|T > s] =
P [T > s + t, T > s]
P [T > s]
P [T > s + t]
=
P [T > s]
e−λ(s+t)
e−λs
−λt
=e
= P [T > t]
=
(14)
(15)
(16)
(17)
Problem 4. Suppose Y = X 2 , where X ∼ N (µ, σ 2 ) with µ = 2, σ 2 = 3. Find the pdf of Y .
d
(HINT: dx
Φ(x) = √12π exp(−x2 /2).)
Solution 4: The RV Y = g(X). The RV X can take on all real values and Y take on only values
over R+
• if y < 0 then FY (y) = 0 since P (Y ≥ 0) = 1
• if y ≥ 0 then
3
√
√
FY (y) = P (X 2 ≤ y) =P (− y ≤ X ≤ y)
(18)
√
√
√
√
− y−2
− y−2
y−2
y−2
X −2
= P( √
≤ √
≤ √ ) = Φ( √ ) − Φ( √
) (19)
3
3
3
3
3
Taking the derivative with respect to y, using the chain rule and the hint that
fY (y) =
(
√
( y−2)2
√ 1 [exp(−
)
6
24πy
0
4
+ exp(−
√
( y+2)2
)]
6
d
dx Φ(x)
y≥0
y<0
=
√1
2π
exp(−x2 /2)