Download Solutions to Problems - Probability with Engineering Applications | ECE 313 and more Assignments Statistics in PDF only on Docsity!
University of Illinois Fall 2008
ECE 313: Solutions to Problem Set 10
In Problems 1 and 2, φ(u) and Φ(u) denote the pdf and CDF of the standard Gaussian random variable.
- (a) E[X
2 k+ ] =
∞
−∞
u
2 k+ φ(u) du = 0 because the integrand is an odd function.
(b) Since dφ(u) = −uφ(u)du, we can integrate by parts to get
E[X
2 k ] =
∞
−∞
u
2 k φ(u) du =
∞
−∞
u
2 k− 1 · uφ(u) du = −
∞
−∞
u
2 k− 1
· dφ(u)
= −u
2 k− 1 φ(u)
∞
−∞
∞
−∞
(2k − 1)u
2 k− 2 φ(u) du = 0 + (2k − 1)
∞
−∞
u
2 k− 2 φ(u) du
= (2k − 1)E[X
2 k− 2 ] = (2k − 1)(2k − 3)E[X
2 k− 4 ] = · · · = (2k − 1)(2k − 3) · · · 3 · 1 · E[X
0 ]
= (2k − 1)(2k − 3) · · · 3 · 1 =
(2k)!
k k!
(c) Let Y =
X
2
R
. Then, Y takes on values in [0, ∞) and thus FY (v) = 0 for v < 0. For v ≥ 0, FY (v) =
P {Y ≤ v)} = P {
X 2
R ≤ v} = P {X
2 ≤ vR} = P {−
vR ≤ X ≤
vR} = Φ(
vR) − Φ(−
vR)
since X is a standard Gaussian random variable. Thus, fY (v) = 0 for v < 0 while for v ≥ 0,
fY (v) =
d dv FY (v) =
1 2
R y
[
φ(
vR) + φ(−
vR)
]
1 √ 2 π
R y exp
1 2 vR
∼ Gamma
1 2
R 2
pdf.
- (a) Since φ(·) is an even function, 1 − Φ(x) =
∫ (^) −x
−∞
t
tφ(t) dt =
∫ (^) x
∞
u
uφ(−u) du =
x
u
uφ(u) du.
Integrating by parts as in Problem 1(b) above, we have that for x > 0,
1 − Φ(x) = −
x
u
dφ(u) =
u
φ(u)
∞
x
x
u 2 · φ(u) du =
x
φ(x) −
x
u 2 · φ(u) du.
Since the integrand of the rightmost integral in the line above is positive, so is the integral.
Even though we do not know the exact value of the integral, we can nonetheless deduce that
1 − Φ(x) < x
− 1 φ(x) for x > 0.
(b) Next, writing the integrand above as
u 3
· (−uφ(u)) and repeating the integration by parts and
the argument about the value of an integral being positive, we get
1 − Φ(x) =
x
φ(x) −
x
u 3
· (−uφ(u)) du =
x
φ(x) −
u 3
φ(u)
∞
x
x
u 4
· φ(u) du
showing that 1 − Φ(x) >
x
− 1 − x
− 3
φ(x) for x > 0.
These results were proved in the note The Complementary Unit Gaussian Distribution Function Q(x)
available on the COMPASS web page for ECE 313.
- The probability that a device operates for at least 15 hours is P {X > 15 } = exp(−15). In general, it
is not possible to compute the probability that at least three of the six devices will operate for at least
15 hours from this information. However, if we assume that all six devices operate independently and
so their failures are independent of each other, then Y, the number of devices that are operational for
at least 15 hours is a binomial random variable with parameters (6, exp(−15)), and
P {Y ≥ 3 } =
6 ∑
i=
i
[exp(−15)]
i (1 − exp(−15))
6 −i ≈ 5. 7250 × 10
− 19 .
- (a) X is uniformly distributed on [0, 2 π). From the diagram below, it should be obvious that the
probability that the random chord is longer than the side of the inscribed equilateral triangle is
P { 2 π/ 3 < X < 4 π/ 3 } =
1 3
(b) Since the circle has radius 1, an arc of length X subtends angle X at the center of the circle.
Furthermore, the length L of the chord is 2 sin(X /2), increasing from 0 when X = 0 to 2 when
X = π and decreasing back to 0 at X = 2π. For any x, 0 < x < 2,
FL(x) = P {L ≤ x} = P {2 sin(X /2) ≤ x} = 2 · P { 0 ≤ X ≤ 2 arcsin(x/2)} =
π
arcsin
x
Hence,
fL(x) =
d
dx
FL(x) =
π
1 − (x/2) 2
, 0 ≤ x ≤ 2 ,
0 , otherwise.
- (a) The pdfs are as shown below. Note that it is immediately obvious that small values of X result
in decisions in favor of H 1 and large values of X result in decisions in favor of H 0.
(b) Λ(u) =
f 1 (u)
f 0 (u)
10 · exp(− 10 u)
5 · exp(− 5 u)
= 2 · exp(− 5 u) which has value 2 at u = 0 and decays away to 0
as u → ∞. Note that Λ(u) > 1 for u < 0 .2 ln 2. Thus, the likelihood ratio test is equivalent to de-
ciding in favor of H 1 if the observed value of X is smaller than the threshold 0.2 ln 2. Equivalently,
Γ 1 = (0, 0 .2 ln 2), Γ 0 = (0.2 ln 2, ∞).
(c) PFA =
Γ 1
f 0 (u) du =
0 .2 ln 2
0
5 · exp(− 5 u) du = − exp(− 5 u)
0 .2 ln 2
0
PMD =
Γ 0
f 1 (u) du =
∞
0 .2 ln 2
10 · exp(− 10 u) du = − exp(− 10 u)
∞
0 .2 ln 2
= 0−(− exp(−2 ln 2)) =
ML Decision Rule: Compare Λ(u) to 1, or equivalently, compare ln Λ(u) to 0:
ln Λ(u) ≷ 0
ln
σ 0
σ 1
exp
u
2
[
σ 2 0
σ 2 1
]})
ln
σ 0
σ 1
u 2
[
σ
2 0
σ
2 1
]
u
2 ≷
2 ln
σ 1 σ 0
1 σ^20
1 σ^21
|u| ≷
2 ln
σ 1 σ 0
1 σ^20
1 σ^21
|u| ≷
4 ln
2 ln 2 = 1.1774 = ξ
So the ML decision rule becomes: Choose H 1 if |X | ≥ ξ, otherwise choose H 0.
Bayes Decision Rule: Compare Λ(u) to
π 0
π 1
, or equivalently, compare ln Λ(u) = ln
σ 0
σ 1
u
2
[
σ 2 0
σ 2 1
]
to ln
π 0
π 1
. The statement of the general rule is complicated because of the need for
allowing for both possibilities: σ
2 0 > σ
2 1 and^ σ
2 0 < σ
2
- For our specific problem with^ σ
2 0 = 1, σ
2 1 = 2,
ln
u 2
≷ ln
π 0
π 1
⇒ u
2 ≷ 4
ln
2 + ln
π 0
π 1
Note that the right hand side is negative for π 0 <
and in this case, the decision is in favor
of H 1 for all observed values of X. For π 0 ≥
, the decision is in favor of H 1 if |u| exceeds
2 ln 2 + 4 ln(π 0 /π 1 ) =
ξ + 4 ln(π 0 /π 1 ). Define γ =
2 ln 2 + 4 ln(π 0 /π 1 ), π 0 ≥ 1 1+
√ 2
0 , π 0 <
1 1+
√ 2
So the Bayes decision rule becomes: Choose H 1 if |X | ≥ γ, otherwise choose H 0.
Notice that γ > ξ if π 0 > π 1 ; that is, because H 0 is the more likely hypothesis, the Bayes decision
rule is playing the odds and making decisions in favor of H 0 even when |X | exceeds the ML
threshold ξ by a little bit (but not by a lot). On the other hand, when π 1 /π 0 >
2, the Bayes
decision rule just ignores X and always chooses H 1 , the more likely hypothesis.
(b) Under H 0 , X ∼ N (0, 1).
PFA = P {|X 0 | ≥ 1. 1774 } = 1 − P {|X 0 | ≤ 1. 1774 } = 2 (1 − Φ(1.1774)) ≈ 0. 2390
Under H 1 , X ∼ N (0, 2).
PMD = P {|X 1 | ≤ 1. 1774 } = P
- 1774 √ 2
X √ 2
- 1774 √ 2
