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of Illinois Page 1 of 4 Fall 2001
Assigned: Wednesday, October 31, 2001
Due: Wednesday, November 7, 2001
Reading: Ross, Chapter 5 and Chapter 6
Noncredit Exercises: Ross, Chapter 5: Problems 15-38; Chapter 6: Problems 1, 8-15, 20-
Problems:
1. [Read Example 3d on pp. 198-199 first.] Let the (straight) line segment ACB be a diameter
of a circle of unit radius and center C. Consider an arc AD of the circle where the length X
of the arc (measured clockwise around the circle) is a random variable uniformly
distributed on [0,2π). Now consider the “random chord” AD.
(a) Find the probability that the length L of the random chord is greater than the side of the
equilateral triangle inscribed in the circle.
(b) Express L as a function of the random variable X , and find the probability density function
for L.
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. The random variable X has probability density function f X (u) =
2(1 – u), 0 ≤ u ≤ 1 ,
0, elsewhere.
Let Y = (1 – X )^2.
(a) What is the CDF F Y (v) of the random variable Y? Be sure to specify the value of F Y (v)
for all v, –∞ < v < ∞.
(b) Show that the F Y (v) that you found in part (b) is a nondecreasing continuous function.
2.(a) Let 0 ≤ v ≤ 1. Then, F Y (v) = P{ Y ≤ v} = P{(1 – X )^2 ≤ v} = P{– v ≤ 1 – X ≤ v} = P{ X ≥ 1 – v}
= 1 – F X (1– v) = (1 – (1– v))^2 = v where we used the result that F X (u) = 1 – (1–u)^2.
Hence, F Y (v) =
0,^ u < 0,
v, 0 ≤ v ≤ 1, 1, v > 1.
(b) A sketch of the function F Y (v) reveals that it is a nondecreasing continuous function. It is not
differentiable at α = 0 or at β = 1.
3. The radius of a sphere is a random variable R with pdf f R (ρ) =
3 ρ^2 , 0 < ρ < 1 ,
0 elsewhere.
(a) Use LOTUS to find the average radius, average volume and average surface area of the
sphere. Does a sphere of average radius have average volume? Does a sphere of average
radius have average surface area?
(b) Find the CDF F V (α) and pdf f V (α) of V , the volume of the sphere.
(c) Find E[ V ] directly from this pdf. Do you get the same answer as in part (a)? Why not?
(d) If the sphere is made of metal and carries an electrical charge of Q coulombs, what is the
CDF F S (x) and the pdf f S (x) of the surface charge density S on the sphere?
3.(a) E[ R ] = ∫
3 ρ^3 dρ =
3 4 ;^ E[ V ] = E[4π R
3 /3] = ∫
4 πρ^5 dρ =
2 π 3 ;^ E[ A ] = E[4π R
2 ] = ∫
12 πρ^4 dρ =
12 π 5
of Illinois Page 2 of 4 Fall 2001
The average volume E[ V ] = E[4π R^3 /3] corresponds to a sphere of radius (1/2)1/3^ and the average surface area E[ A ] = E[4π R^2 ] to a sphere of radius (3/5)1/2. Note that E[ V ] = E[4π R^3 /3] ≠ 4 π(E[ R ])^3 /3, etc. This illustrates the general result that E[g( X )] hardly ever equals g(E[ X ]). You will save yourself a lot of grief if you keep this in mind: that E[g( X )] = g(E[ X ]) is a common misconception among the instochaste. Exercise: Find a function g(•) for which E[g( X )] does equal g(E[ X ]). (b) The volume V has values in the range (0, 4π/3). For any u, 0 < u < 4π/3, F V (u) = P{ V ≤ u}
= P{4π R^3 /3 ≤ u} = P{ R ≤
3 3u/4π } = F R (
3 3u/4π ) = 3u/4π since F R (ρ) = ρ^3 for 0 < ρ < 1. Hence, f V (u) is uniform on (0, 4π/3).
(c) Obviously, E[ V ] = midpoint of uniform pdf = 2π/3 as in part (a) (c) The electrical charge is uniformly distributed on the surface of the sphere. The surface charge density is
S = Q/4π R^2 > Q/4π. For x > Q/4π, F S (x) = P{ S ≤ x} = P{Q/4π R^2 ≤ x} = P{1 > R ≥ Q/4πx } = 1 – (Q/4πx)1.5. Hence, f S (x) = (3/2x)(Q/4πx)1.5^ for x > Q/4π, and 0 otherwise.
4. [“Give me an A! Give me a D! Give me a converter! What have you got? An A/D converter! Go Team!”]
A signal X is modeled as a unit Gaussian random variable. For some applications,
however, only the quantized value Y (where Y = α if X > 0 and Y = –α if X ≤ 0) is used.
Note that Y is a discrete random variable.
(a) What is the pmf of Y?
(b) Suppose that α = 1. If the signal X happens to have value 1.29, what is the error made in
representing X by Y? What is the squared-error? Repeat for the case when X happens to
have value π/4 and when X happens to have value –π/4.
(c) We wish to design the quantizer so as to minimize the squared-error. However, since X
(and Y ) are random, we can only minimize the squared-error in the probabilistic (that is,
average) sense. Now, part (b) shows that the squared-error depends on the value of X ,
and can be expressed as Z = ( X – Y )^2 = g( X ) =
( X – α)^2 if X > 0
( X + α)^2 if X ≤ 0.
So we want to choose α so that E[ Z ] is as small as possible. Use LOTUS to e-zily find
E[ Z ] as a function of α, and then find the value of α that minimizes E[ Z ].
(d) We now get more ambitious and use a 3-bit A/D converter which first quantizes X to the
nearest integer W in the range –3 to +3. Thus, W = 3 if X ≥ 2.5, W = 2 if 1.5 ≤ X < 2.5,
etc. Note that W is a discrete random variable. Find the pmf of W.
(e) The output of the A/D converter is a 3-bit 2's complement representation of W. Suppose
that the output is ( Z 2 , Z 1 , Z 0 ). What is the pmf of Z 2? of Z 1? of Z 0?
(f) Noncredit exercise (but a real-life engineering problem!): Suppose that W
takes on values –3α, –2α, –α, 0, +α, +2α, +3α and quantization is as before: X is
mapped to the nearest W value. What value of α minimizes E[( X – W )^2 ]?
4.(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.
of Illinois Page 4 of 4 Fall 2001
(d) Λ(u) = exp(–|u–1|)/exp(–|u+1|) =
exp(2),^ u > 1,
exp(2u), –1 ≤ u ≤ 1, exp(–2), u < –1.
I told you those absolute-value signs were
tricky! Λ(u) = exp(±2) if u = ±1.2, ±1; Λ(u) increases from exp(–2) to exp(2) as u increases from –1 to 1.
(e) Assuming that exp(–2) < (π 0 /π 1 ) < exp(2), Λ(u) > π 0 /π 1 for u > (1/2)ln(π 0 /π 1 ) = ln( π 0 /π 1 ). Hence,
the Bayesian decision rule is to choose H 1 if X > ln( π 0 /π 1 )and H 0 if X < ln( π 0 /π 1 ). On the other
hand, if (π 0 /π 1 ) > exp(2), then Λ(u), which has maximum value exp(2), can never exceed (π 0 /π 1 ) and the Bayesian decision is to always decide that H 0 is the true hypothesis. Similarly, if (π 0 /π 1 ) < exp(2), then Λ(u), which has minimum value exp(–2), can never be smaller than (π 0 /π 1 ) and the Bayesian decision is to always decide that H 1 is the true hypothesis.
(f) If π 0 = 2π 1 = 2/3, the Bayesian decision chooses H 1 whenever X > ln( π 0 /π 1 )= ln( 2) = θ > 0.
PFA = ∫
θ
∞
(1/2)•exp(–|u+1|)du = ∫
θ
∞
(1/2)•exp(–u–1)du = (1/2)exp(–1) ∫
θ
∞ exp(–u)du = (1/2)•exp(–1–θ) = 1/(2 2e).
Similarly, PMD = ∫
θ
(1/2)•exp(–|u–1|)du = ∫
θ
(1/2)•exp(u–1)du = (1/2)exp(–1) ∫
θ exp(u)du = (1/2)•exp(–1+θ)
= 1/( 2e) = 2PFA. Finally, the average error probability is π 0 PFA + π 1 PMD = 2/3e. More generally, the average error probability is π 0 π 1 exp(–1) which has maximum value (1/2))exp(–1) if π 0 = π 1 = 1/2. Of course, all the above applies only if exp(–2) < (π 0 /π 1 ) < exp(2).
(g) 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) If π 0 > exp(2)/[exp(2)+1], then π 1 = 1 – π 0 < 1/[exp(2)+1], and thus (π 0 /π 1 ) > exp(2). It follows that the
likelihood ratio, which has maximum value exp(2) can never exceed π 0 /π 1 and hence the Bayesian decision rule is to always decide that H 0 is the true hypothesis. The error probability is thus π 1 < 1/[exp(2)+1].
7. The random variable X models a physical parameter. If hypothesis H 0 is true, then, f 0 (u),
the pdf of X , is Gaussian with mean 0 and variance a^2. On the other hand, if hypothesis
H 1 is true, then f 1 (u), the pdf of X , is Gaussian with mean 0 and variance b^2 > a^2.
(a) Suppose that H 0 and H 1 have equal probability. Thus, for i = 0, 1, the pdf of X when
hypothesis Hi is true can be thought of as the conditional pdf of X given that Hi occurred,
i.e. f X |H
i
(u|Hi). Write an expression for the unconditional pdf of X. Is the unconditional
pdf of X a Gaussian pdf?
(b) What is the likelihood ratio? Simplify your answer.
(c) What is the maximum-likelihood decision rule, and what are the false alarm probability and
the missed detection probability of this rule?
7.(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).
