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UNIVERSITY OF ILLINOIS, URBANA-CHAMPAIGN
Department of Electrical and Computer Engineering
ECE 313: Probability with Engineering Applications-Fall 2003
Problem Set 5 Solution
Problems to be turned in:
- (a) When f (x) = a + bx, x ∈ R, we may apply linearity of the expectation as follows: E[a + bX] = a + bE[X].
That is, E[f (X)] = f (E[X]), which is a version of Jensen’s inequality. (b) When f (x) = x^2 , x ∈ R, we can appeal to the variance formula,
E[X^2 ]^ −^ (E[X])^2 =^ V ar(X)^ ≥^0. That is, E[f (X)] ≥ f (E[X]). (c) Figure 1 shows the plot of f (x) = x−^1 and fa(x) with a = 1. Since f (x) ≥ fa(x) for all x > 0, we conclude that
E
[
X
]
= E[f (X)] ≥ E[fa(X)] = E[a−^1 − a−^2 (X − a)] = a−^1 − a−^2 (μ − a) ,
where E[X] = μ. The best bound is obtained on setting a = μ, giving
E
[
X
]
≥ a−^1 − a−^2 (a − a) = μ−^1 =
E[X]
- Let E be the event that a randomly chosen pregnant women has an ectopic pregnancy and S the event that the chosen person is a smoker. Then, the problem states that P (E|S) = 2P (E|SC^ ) and P (S) = 0.32. Hence,
P (S|E) =
P (SE)
P (E)
P (E|S)P (S)
P (E|S)P (S) + P (E|SC^ )P (SC^ )
2 P (S)
2 P (S) + P (SC^ )
- First, note that
P (A | B) = P (B)−^1 P (A ∩ B), P (B | A) = P (A)−^1 P (A ∩ B).
Hence, the first equality follows from the assumption that P (A) = P (B). For the second identity, use the formula based on Axiom III as follows:
P (Ω) = P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
Thus, 1 = 2P (A) − P (A ∩ B), and we obtain, P (A | B) = P (B)−^1 P (A ∩ B) = P (B)−^1 [−1 + 2P (A)].
Using P (A) = P (B) once again, this identity establishes the result.
−4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−
0
2
4
6
8
10 f(x)f a(x), a=
Figure 1: The plot of f (x) and fa(x) with a = 1.
- (a) Let A be the event that an examinee has correctly answered the question and B be the event that an examinee knew the answer. Then,
P (A) = p 0 = p + 1 − p 6 P (B) = p P (A ∩ B) = P (B) = p
P (B|A) =
P (B ∩ A)
P (A)
p p + 1 − 6 p
6 p 5 p − 1
(b) The probabilities of a correct answer are independent, and identically distributed, with proba- bility p 0. Consequently, the number of incorrect answers W is binomial (N, q) with q = 1 − p 0. (Here, p 0 is the probability that any given answer is correct.) (c) Since P (K = k) =
(N
k
pk(1 − p)N^ −k^ and P (W = w) =
(N
w
qw(1 − q)N^ −w,
P (W = w|K = k) =
N − k w
)w( 1 6
)N −k−w
P (K = k|W = w) =
P (K = k) P (W = w) · P (W = w|K = k)
(N
k
pk(1 − p)N^ −k (N w
qw(1 − q)N^ −w^
N − k w
)w( 1 6
)N −k−w ,
where q = 1 − p 0 = 1 − (p + 1 − 6 p) = 56 (1 − p).
- (a) The likelihood matrix for observation XY given the hypothesis is the following:
XY 00 01 02 10 11 12 20 21 22 H 0 0.56 0.16 0.08 0.07 0.02 0.01 0.07 0.02 0. H 1 0.01 0.01 0.08 0.03 0.03 0.24 0.06 0.06 0.
