University Final Exam Solutions ECE 313
of Illinois Page 1 of 3 Fall 1999
1.(a) P(A|B) > P(A) is FALSE. Conditional probabilities can be smaller than unconditional probabilities.
P(A|B) + P(A|Bc) = 1 is FALSE. If A ⊂ B, then P(A|Bc) = 0 while P(A|B) = P(A)/P(B) ≠ 1 in all cases.
P(A|B) + P(Ac|Bc) = 1 is FALSE. If A ∩ B = ∅, then P(A|B) = 0 and P(Ac|Bc) ≠ 1 in all cases.
P(A|B) + P(Ac|B) = 1 is TRUE. Conditional probabilities are a probability measure.
If P(A) = P(B), then P(A|B) = P(B|A) is TRUE. P(A|B) = P(A ∩ B)/P(B) = P(A ∩ B)/P(A) = P(B|A).
If P(A|B) = P(B|A), then P(A) = P(B) is FALSE! If A ∩ B = ∅, then P(A|B) = P(B|A) = 0; P(A) ≠ P(B).
If P(A|B) = P(B|A), then A and B are independent is FALSE. Independence requires P(A ∩ B) = P(A)P(B)
P(A ∩ B) ≤ P(A)P(B) with equality if and only if A and B are independent is FALSE. Note that the
alleged result is equivalent to P(A|B) ≤ P(A) which is just the complement of the first question.
P(B|A)P(A) + P(A|Bc)P(Bc) = P(A) is TRUE. The first term equals P(A|B)P(B).
P(A|B) = P(B|A)P(B)/P(A) is FALSE. If If A ∩ B ≠ ∅, it holds only if P(A) = P(B).
(b) ①P(A ∩ B) ≤ min{P(A), P(B)} ➁P(A ∪ B) ≥ [P(A) + P(B)]/2
➂P(A ∩ B) ≥ P(A) + P(B) – 1. ➃P(A ∩ B) ≤ P(A |B)
Since A ∩ B ⊂ A, P(A ∩ B) ≤ P(A). Similarly, P(A ∩ B) ≤ P(B) ⇒ P(A ∩ B) ≤ min{P(A), P(B)}.
Since A ⊂ A ∪ B, P(A) ≤ P(A ∪ B). Similarly, P(B) ≤ P(A ∪ B); hence,P(A ∪ B) ≥ [P(A) + P(B)]/2.
Since 1 ≥ P(A ∪ B) = P(A) + P(B) – P(A ∩ B), it follows that P(A ∩ B) ≥ P(A) + P(B) – 1.
Since P(B) < 1, it follows that P(A ∩ B) = P(A|B)P(B) ≤ P(A|B). Thus, all four statements are true.
(c) ① P{X > b} = 1 – FX(b). ➁If FX(a) < FX(b), then a < b.
➂If a < b, then FX(a) < FX(b). ➃FX(u) = 1/2 for some u, –∞ < u < ∞.
The first two properties hold for all CDFs since FX(u) is a right-continuous non-decreasing function. But, if
a < b, it is possible that FX(a) = FX(b) because the CDF did not increase between those points (it cannot
decrease, of course.) As a counterexample for ➃, note that for a Bernoulli random variable with parameter
p ≠ 1/2, FX(u) takes on values 0, p, and 1 only. Thus, ➂ and ➃ are not properties of all CDFS.
(d) X and Y have identical finite variance σ2
① E[X2] = E[Y2]➁var(4X – 5Y) = var(4Y – 5X)
➂|cov(X, Y)| ≤ σ2➃X + Y and X – Y are uncorrelated RVs
E[X2] = σ2 + {E[X]}2 ≠ E[Y2] = σ2 + {E[Y]}2 unless it so happens that E[X] = E[Y].
var(4X – 5Y) = 16•var(X) + 25•var(Y) –2•4•5•cov(X,Y) = 41σ2 – 40•cov(X,Y).
var(4Y – 5X) = 16•var(Y) + 25•var(X) –2•4•5•cov(Y,X) = 41σ2 – 40•cov(X,Y). Hence, equality holds.
|cov(X,Y)| = |ρvar(X)var(Y) | = |ρ|•σ2 ≤ σ2 since |ρ| ≤ 1.
cov(X + Y, X – Y) = var(X) – var(Y) – cov(X,Y) + cov(Y,X) = σ2 – σ2 – cov(X,Y) + cov(X,Y) = 0.
Thus, ➁, ➂, and ➃ are true statements.
2. 5 arrivals occur in (0,4], 2 in (3,4] and 4 in (3,6]. These events are not independent, but all three occur if and
only if there are 3 arrivals in (0,3], 2 arrivals in (3,4], and 2 arrivals in (4,6]. The latter three events are
independent since they correspond to disjoint time intervals. Hence,
P{X = 5, Z = 4 | Y = 2} = P{X = 5, Z = 4, Y = 2}/P{Y = 2}
= P{3 arrivals in (0,3], 2 arrivals in (3,4], and 2 arrivals in (4,6]}/P{2 arrivals in (3,4]}
= P{3 arrivals in (0,3], 2 arrivals in (4,6]} = P{3 arrivals in (0,3]}P{2 arrivals in (4,6]}
= exp(–3λ)•(3λ)3
3! exp(–2λ)•(2λ)2
2! = 9•exp(–5λ)•(λ)5.
