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University of Illinois ECE 313 FALL 2007 Professors Tamer Ba¸sar & Ada Poon October 8, 2007
TEST I
7:00 p.m. - 8:30 p.m.
NOTE: This is a closed-book closed-notes (and closed-neighbor) exam, with only one sheet of notes (8 12 ′′^ × 11 ′′) allowed. Also, no calculators, laptops, palm pilots, and the like are allowed.
Name (Last, First) : ................................................................
Section : © Section C, 10 MWF © Section D, 11 MWF
Problem 1 : ........................ (2 × 10 = 20 points)
Problem 2 : ........................ (4 × 10 = 40 points)
Problem 3 : ........................ (14 + 6 = 20 points)
Problem 4 : ........................ (6 + 7 + 7 = 20 points)
TOTAL : ........................ (100 points)
Problem 1 (20 points)
Let E, F , and G be three events defined on a common sample space, with 0 < P (E) < 1, 0 < P (F ) < 1, and 0 < P (G) < 1. (Note the strict inequality in each case.) Read each of the following statements carefully, and check the corresponding True box if the statement is always true (that is, it holds for all E, F , G); otherwise check the corresponding False box. Each correct choice counts +2 points, whereas an incorrect choice counts −1 point; so guess at your own risk. You can use the space provided at the bottom as well as back of the previous page for scratch work.
TRUE FALSE
t t P (EF ) + P (Ec^ ∪ F c) = 1
t t P (E ∪ F ) = 1 − P (Ec|F c)P (F c)
t t P (E|F c) + P (E|F ) = 1
t t P (EF |E) = P (EF |F )
t t P (E|F ) = P (E|F G)P (G|F ) + P (E|F Gc)P (Gc|F )
t t P (EF G) ≤ min
P (E), P (F ), P (G)
t t P (E|F )P (F ) + P (Ec|F )P (F ) = P (F )
t t If P (EF G) = P (E)P (F )P (G), then E, F , G are independent
t t If E and F are mutually exclusive, P (E|F ) = P (E)
t t If the occurrence of event F makes event E more likely, then the occurrence of E necessarily makes F also more likely.
END OF PROBLEM 1
(vi) Let X be a discrete random variable with probability mass function (pmf)
P (X = i) =
{ (^0). 4 for i = − 1
- 2 for i = 1
- 4 for i = 2
The expected value of X, E[X], is a. 0. 6 b. 1 c. 1. 6 d. 2. 2
(vii) Let X be as defined in (vii) above. E[(X^2 − 1)^2 ] is a. 1. 2 b. 3. 6 c. 2 d. none of these
(viii) Y is a discrete random variable with mean 3 and variance 4. Use Chebyshev’s inequality to obtain the value of K such that P (|Y − 3 | > 4) ≤ K. a. K = 1 b. K = 0. 5 c. K = 2 d. K = 0. 25
(ix) You have two biased coins. The probability of heads for the first one is equal to 0.6 and for the second one equal to p, where p is unknown. Consider the experiment where you flip the coins together a total of 500 times (independent trials), and record the number of times both show heads. Let that number be 120. What is the maximum likelihood estimate, ˆp, for p? a. 0. 36 b. 0. 4 c. 0. 5 d. none of these
(x) In the experiment in (ix) above, you now replace the second coin with an unbiased one. Let X be the number of times both coins show heads when flipped together 500 times. What is the most likely value for X? That is, find the integer k that maximizes the quantity P (X = k). a. 150 b. 151 c. 120 d. none of these
END OF PROBLEM 2
Problem 3 (14+6 = 20 points)
The probability of a bit error in a communication line is 10−^3. We want to compute the probability that a block of 2, 000 bits has 5 or more errors.
(i) Compute the desired error probability when bit error is modeled as a Poisson random variable. [You can leave the final answer in terms of exponentials.]
Probability when modeled as Poisson =
(ii) Now consider modeling the bit error as a Binomial random variable. (a) Write down the expression for the desired error probability under this modeling assumption. (b) Can you use the result in part (i) above to compute this quantity? Why?
(a) Expression for probability when modeled as Binomial =
(b)
END OF PROBLEM 3
(iii) Lastly, we introduce a cost function C 00 = − 1 , C 01 = 3, C 10 = 6, C 11 = −3, where Cij is the cost of ruling that X = i when in fact it is X = j, i, j = 0, 1. Let α and β be as in part (ii) above, and p = 0.25. Find the Bayes’ minimum average cost decision rule.
Bayes’ minimum average cost DR =
END OF PROBLEM 4 / END OF TEST
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