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University Problem Set #11: Solutions ECE 313
of Illinois Page 1 of 2 Spring 2002
1.(a) As is obvious from the figure, the chord is longer than the side of the inscribed equilateral triangle if 2π/3 < X < 4π/3. Hence, the desired probability is (4π/3–2π/3)/2π = 1/3 as in the second model in Ross. What is the geometrical relation between the two models? (b) Since the circle has radius 1, an arc of length θ subtends an angle θ at C. Also, the length of the chord joining the endpoints of the arc is 2 sin (θ/2). Hence, L = 2 sin( X /2). Note that as X increases from 0 to 2π, the chord length increases from 0 to 2 (at X = π), and then decreases to 0 (at X = 2π). For any x, 0 ≤ x ≤ 2, F L (x) = P{ L ≤ x} = P{2 sin( X /2) ≤ x} = 2P{0 ≤ X ≤ 2 arcsin(x/2)} (Why twice?) = 2(2 arcsin(x/2)/2π) = (2/π)arcsin(x/2).
Hence, f L (x) =
d dx
F L (x) =
^1
π 1–(x/2)^2
0 ≤ x ≤ 2,
0, otherwise.
2.(a) Obviously P{ Y = α} = P{ Y = –α} = 1/2.
(b) (1.29–1) = 0.29. (1.29 –1)^2 = 0.0841. (π/4–1) = –0.214…, (π/4–1)^2 = 0.046….
(–π/4–(–1)) = –0.214…, (–π/4–(–1))^2 = 0.046…. Note that the error for + X is the same as that for – X.
(c) E[ Z ] = ∫
0
∞
(u–α)^2 ƒ(u)du + ∫
0
(u+α)^2 ƒ(u)du = ∫
∞
(u^2 + α^2 )ƒ(u)du – 4α ∫
0
∞ uƒ(u)du = 1+α^2 –4α/ 2 π on
expanding out the quadratics, changing variables, and using the fact that E[ X^2 ] = σ^2 + μ 2 = 1. Note that uf(u) is a perfect integral. It is easy to show that E[ Z ] has minimum value 1–2/π at α = 2/π (d) From tables of Φ(•), we get P{ W = –3} = P{ W = +3} = Φ(–2.5) = 0.0062, P{ W = 0} = Φ(0.5) – Φ(–0.5) = 0.3830, P{ W = –1} = P{ W = +1} = Φ(1.5) – Φ(0.5) = 0.2417, and P{ W = –2} = P{ W = +2} = Φ(2.5) – Φ(1.5) = 0.0606. (e) P{ Z 2 = 1} = P{ W < 0} = 0.3085.
P{ Z 1 = 1} = P{ W = 2} + P{ W = 3} + P{ W = –1} + P{ W = –2} = 0. P{ Z 0 = 0} = P{ W = 2} + P{ W = 0} + P{ W = –2} = 0.5042 ≠ P{ Z 0 = 1}
d db(bt)exp(–bt
(^2) /2) = t•exp(–bt (^2) /2) – bt•exp(–bt (^2) /2)•t (^2) /2 is zero for b = 2/t. Thus, if we observe that X = t,
the maximum-likelihood estimate of b is 2/t. Reality check: If the observed value t is large, we estimate the value of b to be quite small. This makes sense. If the system lasted for a long time, its hazard rate can be expected to be small, and the hazard rate is proportional to b.
4.(a)
0.2•ln 2
f (u) 0
f (u) 1
(b) The maximum-likelihood decision chooses the hypothesis which has the larger pdf value at the observation. Since the two pdfs cross at θ = 0.2•ln 2, we see that the decision is to choose H 0 if X > θ and H 1 if X < θ.
(c) PFA= P{false alarm} = P{H 1 is chosen when in fact H 0 is the true hypothesis} = P{ X < θ when H 0 is true}
0
θ
f 0 (u)du = ∫
0
θ 5•exp(–5u)du = 1 – exp(–5θ) = 1/2.
PMD = P{missed detection} = P{ X > θ when H 1 is true} = ∫
θ
∞
f 1 (u)du = ∫
θ
∞ 10•exp(–10u)du = exp(–10θ) = 1/4.
University Problem Set #11: Solutions ECE 313
of Illinois Page 2 of 2 Spring 2002
(d) Λ(u) =
10•exp(-10u) 5•exp(-5u)
= 2•exp(–5u) decreases from 2 at u = 0 to 1 at u = 0.2•ln 2 to 0 at u = ∞. We compare
Λ(u) to π 0 /π 1 to get that Λ(u) > π 0 /π 1 if u < –0.2•ln(π 0 /2π 1 ) = ξ, that is, the Bayesian rule chooses H 0 if X > ξ and H 1 if X < ξ. Note that ξ < 0 if π 0 > 2π 1.
(e) Now ξ = –0.2•ln(1/4) = 0.2•ln(4). Proceeding as in part (c), we readily get that PFA = 1 – exp(–5ξ) = 3/
while PMD = exp(–10ξ) = 1/16. The average error probability is thus (3/4)•(1/3) + (1/16)•(2/3) = 7/24. Notice that since π 0 < π 1 , the Bayesian decision rule accepts a larger PFA in return for a smaller PMD because PFA has a smaller weight than PMD in the expression for the average error probability.
(f) The decision rule that always chooses H 0 makes an error precisely in those instances when H 1 is the true
hypothesis. Hence its average error probability is just π 1 , the probability that H 1 is the true hypothesis.
(h) The Bayesian decision rule compares the likelihood ratio Λ(u) = 2•exp(–5u) to π 0 /π 1. But, since Λ(u) has
maximum value 2, it is always smaller than π 0 /π 1 if π 0 > 2π 1 , that is, if π 0 > 2/3. It thus achieves average error probability π 1 < 1/3. In contrast, for these prior probabilities, the ML rule achieves an average error probability (1/2)•π 0 + (1/4)•π 1 > (1/2)•(2π 1 ) + (1/4)•π 1 = 1.25π 1.
5.(a) No, the unconditional pdf of X is given by [(a 2 π)–1^ exp(–u^2 /2a^2 ) + (b 2 π)–1^ exp(–u^2 /2b^2 )]/2, which is not a Gaussian pdf.
(b) Λ(u) =
f 1 (u) f 0 (u)
(a 2 π)exp(–u^2 /2b^2 ) (b 2 π)exp(–u^2 /2a^2 )
a b
•exp
–u^2
b^2
a^2
(c) Suppose that the observation X has value u. The maximum-likelihood decision rule says that H 1 is chosen
as the true hypothesis if Λ(u) > 1 and H 0 is chosen if Λ(u) < 1. Thus, H 1 is chosen if ln(a/b) – (u^2 /2)(b–2^ – a–2^ ) > 0. This is equivalent to the statement that the rule chooses H 1 whenever the observation X is such that
| X | > ab
ln b^2 – l n a 2 b^2 – a 2
= c.
Note that f 0 (0) = 1/(a 2 π) > 1/(b 2 π) = f 1 (0) and the two pdf curves cross each other at ±c. P(false alarm) = P{| X | > c |H 0 is true) = 2Q(c/a). P(missed detection) = P{| X | < c |H 1 is true) = 1 – 2Q(c/b).
6.(a) In the figure below, I use • to denote a probability mass of 1/12 and • to denote a probability mass of 1/6.
2
1
3
1 2 3 4 5
4
(a) p X (0) = 1/6 + 1/12 = 1/4, p X (1) = 1/3, p X (3) = 1/4, p X (5) = 1/6.
p Y (–1) = 2×1/12 + 1/6 = 1/3, p Y (3) = 1/3, p Y (4) = 1/3.
(b) Since, for example, p X (3)p Y (3) = (1/3)×(1/4) =1/12 ≠ p X , Y (3,3) = 0, the random variables are not
independent. We can also see this via the eyeball test: there are empty spots in the grid. (c) P{ X ≤ Y } = p X , Y (0,3) + p X , Y (1,3) + p X , Y (1,4) + p X , Y (3,4) = 1/2.
P{ X + Y ≤ 8} = 1 – p X , Y (5,4) = 11/12.
(d) Given Y = 3, X takes on values 0, 1 and 5 with conditional probabilities (1/6)/(1/3), (1/12)/(1/3), and (1/12)/(1/3), that is, 1/2, 1/4 and 1/4 respectively. Hence, E[ X | Y =3] = 0•(1/2) + 1•(1/4) + 5•(1/4) = 3/2, and E[ X^2 | Y =3] = 0^2 •(1/2) + 1^2 •(1/4) + 5^2 •(1/4) = 26/4, giving var( X | Y =3) = 26/4 – 9/4 = 17/4.
