University Solutions to Final Exam ECE 313
of Illinois Page 1 of 2 Spring 2000
1.(a) P(A|B) = P(B|A) ⇒ P(AB)/P(B) = P(AB)/P(A) ⇒ P(A) = P(B) or P(AB) = 0 or both. Note that P(AB) = 0
does not necessarily mean that A and B are disjoint events. Thus, ➂ is a true statement, but ➁ is not
necessarily true. But ➂ implies that P(Ac) = P(Bc), so ➃ is a true statement too. Thus, none of the five
choices given correctly describes which are the true statements.
(b) The value of a pdf can exceed 1 (e.g. uniform distribution on (a,b) with b–a < 1. Also, if ➂ were true, the
area under the pdf would be ∞ rather than 1. Thus,Only ➁ and ➃ are true statements.
(c) None of the five choices correctly describes which are the true statements.
2. NO. FX(u) = 1 – FX(–u) for all u, –∞ < u < ∞.
YES.P{X > α} = 1 – F X(α) = FX(–α) for all α, – ∞ < α < ∞.
NO. FY(v) = P{Y ≤ v} = FX(v) – FX(–v) for v ≥ 0, and FY(v) = 0 for v < 0.
YES. fY(v), the derivative of FY(v), is fX(v) + fX(–v) = 2fX(v) for v ≥ 0, and 0 for v < 0.
NO. FX(–w) decreases from 1 to 0 as w increases from –∞ to +∞, and thus cannot be a CDF.
YES. FZ(w) = 1 – FX(–w) has derivative fX(–w).
NO. A pdf cannot be negative…
YES. E[X] = 0 by the symmetry of the pdf (Note: the mean exists because the variance is finite).
NO. var(Y) = E[Y2] – (E[Y])2. But, LOTUS gives E[Y2] = E[X2] = var(X) = 9 since E[X] = 0.
If E[Y] = 3, then var(Y) would be 0… which it is not.
YES. LOTUS gives E[Y2] = E[X2] = var(X) = 9 since E[X] = 0.
MAYBE. var(Y) = E[Y2] – (E[Y])2. Since E[Y] > 0, var(Y) can be less than 8. For example, if X is
uniformly distributed on [–3 3,+3 3] with variance (6 3)2/12 = 9, then Y is uniformly
distributed on [0,3 3] and thus var(Y) = (3 3)2/12 = 2.25.
YES. var(Y) = E[Y2] – (E[Y])2 = 9 – (E[Y])2 < 9 since E[Y] > 0.
YES. Using LOTUS, E[XY] = ∫
0
∞
u2fX(u) du + ∫
–∞
0
–u2fX(u) du = 0.
YES. cov(X,Y) = E[XY] – E[X]E[Y] = 0.
NO. Knowing the value of X tells you the value of Y exactly!
YES. By LOTUS, E[XZ] = E[–X2] = –E[X2] = –var(X) = –9.
YES. X + Y = X + |X| = {0, if X ≤ 0,
2X > 0, if X > 0. Thus, P{X + Y ≤ 0} = P{X ≤ 0} = 1/2.
YES. By Chebyshev’s inequality, FY(6) = FX(6) – FX(–6) = P{–6 ≤ X ≤ 6} ≥ 1 – (σ/6)2 = 3/4.
MAYBE. If X is a Gaussian random variable, then P{X ≤ 6) = Φ(2) = 0.9772 according to the table of
values of Φ(x).
YES. P{X2 + 4X + 3 < 0} = P{(X + 1)(X + 3) < 0} = P{–3 < X < –1} = FX(–1) – FX(–3).
YES. P{X2 – 4X + 3 > 0} = P{(X – 1)(X – 3) > 0} = P{X > 3} + P{X < 1} = 1 – FX(3) + FX(1)
= FX(–3) + 1 – FX(–1) = P{X ≤ –3} + P{X > –1} = P{X2 + 4X + 3 > 0}.
3. The Karnaugh map on the left is marked with letters a-h denoting the probabilities of the eight sets.
a b c d
e f gh
A
B
C
0.2
A
B
C
0.2
A
B
C
00.2
0.20.1 0.1 0.1
0.1
0.10.10.1 0.2
0
0.20.1
Since (b+d+e) + (c+f+h) + g = 0.8, and (c+f+h)+g = 0.6,we readily obtain that a = 0.2 and b+d+e = 0.2.
Next, adding together the three equations c+d+g+h = 0.5, b+c+f+g = 0.5, e+f+g+h = 0.5, we get
(b+d+e) + 2(c+f+h) + 3g = 1.5, i.e. 2(c+f+h) + 3g = 1.3 which combined with (c+f+h)+g = 0.6 gives
(c+f+h) = 0.5 and g = 0.1. Hence, P{all three events occurred} = g = 0.1, and P{all three | at least two}
= g/(c+f+h+g) = 0.1/0.6 = 1/6. On the other hand, P{at least two | A} = (c+g+h)/(1/2) = 2(c+g+h) cannot
be determined from the given information. For example, the two Karnaugh maps shown on the right are
consistent with the given data, and give different values for the desired probability.
