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University of Illinois Spring 2008
ECE 313: Solutions to Problem Set 10
- (a) I can take on values in the range (−I 0 , ∞). (b) FI (v) = 0 for v < −I 0. For any v > −I 0 , FI (v) = P {I ≤ v} = P {I 0 (exp(V) − 1) ≤ v} = P {V ≤ ln(1 + v/I 0 )} = FV (ln(1 + v/I 0 )). (c) For v > −I 0 ,
fI (v) = fV (ln(1+v/I 0 )) (^) 1 + 1 v/I 0
× I^1
0
= fV^ (ln(1 + v + I^ v/I^0 )) 0
{ I 0 / 2
(v 1 +I 0 )^2 ,^ v^ ≥^0 , 2 I 0 ,^ −I^0 < v <^0 ,^
Note that the pdf has constant value 1/(2I 0 ) from v = −I 0 to v = 0.
- (a) Y takes on values in [0, 1] and hence FY = 0 for v < 0, and FY (v) = 1 for v > 1. For 0 ≤ v ≤ 1, FY (v) = P {Y ≤ v} = P {X 2 ≤ v} = P {−
v ≤ X ≤
v} =
v. Hence fY (v) = 2 √^1 v if 0 ≤ v ≤ 1, and fY (v) = 0, otherwise. (b) Z takes on values in [− 1 , 1] and hence, FZ = 0 for v < −1, and FZ (v) = 1 for v > 1. For 0 ≤ v ≤ 1, FZ (v) = P {Z ≤ v} = P {g(X ) ≤ v} = P {X ≤
v} = 12 [1 +
v]. For − 1 ≤ v ≤ 0, FZ (v) = P {Z ≤ v} = P {g(X ) ≤ v} = P {X ≤
−v} = 12 [1 −
−v]. Hence, fZ (v) = 4 √^1 |v| if 0 ≤ |v| ≤ 1, and fZ (v) = 0, otherwise. Note that the pdf is an even function, and approaches +∞ as v approaches 0 from either side.
- (a) The pmf of Y is pY (α) = pY (−α) =^12.
(b) E[Z] =
0
(u−α)^2 φ(u) du+
−∞
(u+α)^2 φ(u) du =
−∞
(u^2 +α^2 )φ(u) du− 4
0
αuφ(u) du
= 1 + α^2 − 2
π
α where we have used the facts that the standard Gaussian random variable has variance 1, the area under the pdf φ(u) is 1, and
0 u^ exp(−u
(^2) /2) du = 1 (cf. Problem 5(b) of Problem Set 1) in arriving at the result. E[Z] has minimum value 1 − (^) π^2 at α =
2 /π. (c) pW (3) = pW (−3) = Φ(− 2 .5) = 0.0062. pW (2) = pW (−2) = Φ(2.5) − Φ(1.5) = 0.0606. pW (1) = pW (−1) = Φ(1.5) − Φ(0.5) = 0.2417. pW (0) = Φ(0.5) − Φ(− 0 .5) = 0.3830. (d) Z 2 , Z 1 , Z 0 are Bernoulli random variables with parameters p 2 = P {W < 0 } = 0.3085, p 1 = P {W ∈ {− 2 , − 1 , 2 , 3 }} = 0.3691, and p 0 = P {W ∈ {− 3 , − 1 , 1 , 3 }} = 0. 4958 respectively.
- (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 } = 13.
A DB A x B
D
C
(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 2
. Hence,
fL(x) = d dx
FL(x) =
π
1 − (x/2)^2
, 0 ≤ x ≤ 2 , 0 , otherwise. .
- (a) The pdfs are as shown below.
(^00) 0.2 0.4 0.6 0.8 1 1.
1
2
3
4
5
6
7
8
9
10
u
f^ (
u)
(^) f f^0 (u) 1 (u)
(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 deciding 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) .5 ln 2
0
5 · exp(− 5 u) du = − exp(− 5 u)
0 .2 ln 2
0
2 −^ (−1) =
PMD =
Γ 0
f 1 (u) du =
0 .5 ln 2
10 · exp(− 10 u) du = − exp(− 10 u)
∞
0 .2 ln 2 = 0 − (− exp(−2 ln 2)) =^1 4
(d) Λ(u) = 2 · exp(− 5 u) > π^0 π 1
for u < 0 .2 ln
2 π 1 π 0
= 0.2 ln 2 + 0.2 ln
π 1 π 0
= ξ. Thus, the minimum-error-probability decision rule is equivalent to deciding in favor of H 1 if the observed value of X is smaller than ξ. Note that ξ < 0 if π 0 > 2 π 1 , that is, if π 0 > 2 /3. (e) If π 0 = 1/3, then ξ = 0.2 ln 4. Hence,
PFA =
∫ (^0) .2 ln 4
0
5 · exp(− 5 u) du = − exp(− 5 u)
0 .2 ln 4
0
4 −^ (−1) =
PMD =
0 .2 ln 4
10 · exp(− 10 u) du = − exp(− 10 u)
∞
0 .2 ln 4
= 0 − (− exp(−2 ln 4)) =
The average error probability thus is P¯e =
3 PFA^ +
3 PMD^ =^
- Note that since^ π^0 < π^1 ,
Bayes Decision Rule: Compare Λ(u) to π π^01 , or equivalently, compare log Λ(u) to log π π^01 :
log Λ(u) ≷ log π π^0 1 log
σ 0 σ 1 exp
u^2 2
[
σ^20 −^
σ 12
]})
≷ log
π 0 π 1
log
σ 0 σ 1 +^
u^2 2
[
σ 02 −^
σ 12
]
≷ log
π 0 π 1
u^2 ≷
2 log
σ 1 σ 0 ·^
π 0 π 1
1 σ^20 −^
1 σ^21
|u| ≷
2 log
σ 1 σ 0 ·^
π 0 π 1
1 σ^20 −^
1 σ^21
= γ
So the Bayes decision rule becomes: Choose H 1 if |u| ≥ γ, otherwise choose H 0.
(c) σ^20 = 1, σ^21 = 4 ⇒ ξ =
2 log( 21 ) (^11) − 14 = 1.^3596. Under H 0 , X ∼ N (0, 1). PFA = P {|X 0 | ≥ 1. 3596 } = 1 − P {|X 0 | ≤ 1. 3596 } = 2 (1 − Φ(1.3596)) ≈ 0. 177
Under H 1 , X ∼ N (0, 4). PMD = P {|X 1 | ≤ 1. 3596 } = P
− 1.^35962 ≤ X 2 ≤ 1.^35962
