Download Solutions to ECE 313 Final Exam, University of Illinois, Spring 2002 - Prof. Dilip Sarwate and more Exams Statistics in PDF only on Docsity!
University Solutions to Final Exam ECE 313
of Illinois Page 1 of 2 Spring 2002
1.(a) P{at least one correct} = P{A ∪ B ∪ C} = 1 – P{Ac^ ∩ Bc^ ∩ Cc} = 1 – P{none correct} = 1 – P{Ac)P(Bc)P(Cc} = 1 – 0.2×0.1×0.3 = 1 – 0.006 = 0. (b) P{only #2 correct} = P{Ac^ ∩ B ∩ Cc} =P{Ac)P(B)P(Cc} = 0.2×0.9×0.3 = 0.
(c) P{#3 correct | at least one correct} = P{C | A ∪ B ∪ C} = P{C ∩ [A ∪ B ∪ C]}/P{A ∪ B ∪ C}
= P{C}/P{A ∪ B ∪ C} = 0.7/0.994 = 50/
2.(a) P{ X > n} = P{mailman not bitten by a dog for the first n days} = P{not bitten on n-th day and not bitten on previous n–1 days} = P{not bitten on n-th | X > n–1}P{ X > n–1} = [1 –
n+ ] × P{ X > n–1} = n n+ × P{ X > n–1}
= n n+
×
n– n × P{ X > n–2} = n n+
×
n– n
×
n– n– × P{ X > n–3} = n n+
×
n– n
× … ×
×
× P{ X > 1}
n+ since P{mailman not bitten on first day} =
and all the intermediate factors cancel. Hence,
p X (n) = P{ X > n–1} – P{ X > n} =
n
n+
n(n+1) for n = 1 , 2, 3, …
(b) E[ X ] = ∑
n=
∞
n•p X (n) = ∑
n=
∞ n•
n(n+1)
n=
∞ 1 n+ = ∞ since the harmonic series diverges. Dog lovers
everywhere can breathe a sigh of relief! Note that we can also get this using E[ X ] = ∑
n=
∞ P{ X > n}…
3. Let q = 1–p. Then, P{ X is even} = pq + pq^3 + pq^5 + … = pq 1–q^2
q 1+q
which gives q =
and p =
Hence E[ X ] = 1/p = 5. 4. Which of the following statements are true for all random variables X and Y with identical finite variance σ^2? TRUE FALSE n n E[ X^2 ] = E[ Y^2 ] n n var( X + Y ) = 2σ^2 n n var( X – Y ) = 0 n n var( X + Y ) + var( X – Y ) = 4σ^2 n n var(2 X + 3 Y ) = var(3 X + 2 Y ) n n cov( X , Y ) ≤ σ^2 n n X + Y and X – Y are uncorrelated random variables n n X + Y and X – Y are independent random variables
5.(a) Y = exp( X ) takes on values in the range (1,∞) as X varies between (0,∞).
For v > 1, F Y (v) = P{ Y ≤ v} = P{exp( X ) ≤ v} = P{ X ≤ ln v} = 1 – exp(–λ(ln v)) = 1 – v–λ. Note λ > 0.
Hence, f Y (v) =
λv–λ–1 (^) , v > 1, 0, v ≤ 1. 6.(a) The pdfs are as shown in the sketch below.
–5 5
f (u) 0 f (u) 1
The maximum-likelihood decision chooses the hypothesis which has the larger pdf value at the observation. By inspection, we see that the decision is to choose H 1 if X > 0 and H 0 if X < 0. PFA= P{false alarm} = P{H 1 is chosen when in fact H 0 is the true hypothesis} = P{ X > 0 when H 0 is true}
0
∞
f 0 (u)du = ∫
0
∞
(1/2)•exp(–|u+5|)du = ∫
0
∞
(1/2)•exp(–u–5)du = (1/2)exp(–5) ∫
0
∞ exp(–u)du = (1/2)•exp(–5).
Similarly, PMD = P{missed detection} = (1/2)•exp(–5) also.
University Solutions to Final Exam ECE 313
of Illinois Page 2 of 2 Spring 2002
(b) The minimum-error-probability rule comparesΛ(u) =
f 1 (u) f 0 (u)
exp(–|u–5|) exp(–|u+5|)
exp(10),^ u > 5,
exp(2u), –5 ≤ u ≤ 5, exp(–10), u < –5,
to
π 0 π 1 which equals 1 2 when^ π^1 = 2π^0 =^
2 3 , and hence the minimum-error-probability rule chooses H^1 whenever exp(2 X ) > 1/2, i.e. X > –(1/2)•ln 2 = θ. Note that –5 < θ < 0 and that exp(–θ) = 2.
PFA =
θ
∞ 1
2 •exp(–|u+5|)du =^ ∫
θ
∞ 1 2 •exp(–u–5)du =^
1
2 exp(–5)^ ∫
θ
∞ exp(–u)du = 1 2 •exp(–5–θ) =
exp(–5) 2
. Similarly,
PMD =
θ 1
2 •exp(–|u–5|)du =^ ∫
θ 1 2 •exp(u–5)du =^
1
2 exp(–5)^ ∫
θ exp(u)du = 1 2 •exp(–5+θ) =
exp(–5) 2 2
1 2 PFA.
Finally, the average error probability is π 0 PFA + π 1 PMD = 2•exp(–5)/3. More generally, the threshold θ equals (1/2)•ln( π 0 /π 1 and the average error probability is π 0 π 1 exp(–5) which has maximum value (1/2))exp(–5) if π 0 = π 1 = 1/2. Of course, all the above applies only if exp(–10) < (π 0 /π 1 ) < exp(10).
7. The joint pdf has value 2 on the triangular region shown in the left-hand figure below. For any a, 0 ≤ a < ∞, P{ Z ≤ a} = P{ Y / X ≤ a} = P{ Y ≤ a X } = P{( X , Y ) lies in the shaded region shown} = 2×((1/2)× 1 ×a/(a+1)) = a/(a+1). Hence, f Z (a) = 1/(1+a)^2 for a ≥ 0 and 0 for a < 0.
(1/(a+1),a/(a+1))
v
u
v
Figure for Problem 7
u
Problem 8: joint pdf
1 u
Problem 8: pdf of X
1 X
MMSE estimator
Y ^
8. The joint pdf has value 2/3 on the region shown in the second figure above. (a) f X (u), the value of the marginal pdf of X at u, equals the cross-sectional area of the pdf surface at u. Hence, we get that f X (u) has constant value 2/3 for 0 < u < 1, and decreases to 0 as u increases from 1 to 2, as showb in the third figure above. More formally,
f X (u) =
2/3^ for 0 < u < 1,
(2/3)•(2–u) for 1 ≤ u < 2, 0 otherwise.
It is easily verified that the area under the pdf is 1.
(b) The MMSE estimator for Y given X is the mean of the conditional pdf of Y given the value of X. Also, the conditional pdf is simply the cross-section of the joint pdf surface “normalized” to have area 1. It is easily seen that if X = a where 0 < a < 1, then the conditional pdf of Y is uniform on (1–a, 2–a) and hence has mean (3/2) – a which varies from 3/2 at a = 0 to 1/2 at a = 1, while if 1 ≤ a < 2, then the conditional pdf of Y is uniform on (0, 2–a) and thus has mean 1 – a/2 which varies from 1/2 at a = 1 to 0 at a = 2. Thus, the MMSE estimator Y ^is (3/2)– X if 0 < X < 1, and 1 – X /2 if 1 ≤ X < 2, as illustrated in the right-hand figure above. Note that the function is piecewise linear, and thus is different from the linear MMSE estimator. 9.(a) Z = 5 X + Y is a Gaussian random variable with mean E[ Z ] = E[5 X + Y ] = 5×E[ X ] + E[ Y ] = 5×0 + 7 = 7
and variance var( Z ) = 5^2 var( X ) + 1^2 var( Y ) + 2× 5 × 1 ×cov( X , Y ) = 25×4 + 16 + 10ρ var( X )var( Y ) = 100 + 16 + 10×(1/16)× 2 ×4 = 121 = 11^2. Hence, f Z (w) = 1/(11 2 π)•exp(–(w–7)^2 /242), –∞ < w < ∞. (b) P{ Y > 3 X } = P{3 X – Y < 0}. But 3 X – Y is a Gaussian random variable with mean
E[3 X – Y ] = 3×E[ X ] – E[ Y ] = –7 and variance 3^2 var( X ) + (–1)^2 var( Y ) + 2× 3 ×(–1)×cov( X , Y ) = 9×4 + 16 –6ρ var( X )var( Y ) = 36 + 16 – 6×(1/16)× 2 ×4 = 49 = 7^2. Hence, 3 X – Y is N (–7,7^2 ) and
thus P{3 X – Y < 0} = Φ( )
