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Chapter 14
Solved Problems
14.1 Probability review
Problem 14.1. Let X and Y be two N0 -valued random variables such that X = Y + Z, where Z is a Bernoulli random variable with parameter p ∈ (0, 1), independent of Y . Only one of the following statements is true. Which one? (a) X + Z and Y + Z are independent (b) X has to be 2N0 = {0, 2, 4, 6, . . . }-valued (c) The support of Y is a subset of the support of X (d) E[(X + Y )Z] = E[(X + Y )]E[Z]. (e) none of the above Solution: The correct answer is (c). (a) False. Simply take Y = 0, so that Y + Z = Z and X + Z = 2Z. (b) False. Take Y = 0. (c) True. For m in the support of Y (so that P[Y = m] > 0), we have P[X = m] ≥ P[Y = m, Z = 0] = P[Y = m]P[Z = 0] = P[Y = m](1 − p) > 0. Therefore, m is in the support of X. (d) False. Take Y = 0. (e) False. Problem 14.2. A fair die is tossed and its outcome is denoted by X, i.e., X∼ 1 2 3 4 5 6 . 1/6 1/6 1/6 1/6 1/6 1/6 107
CHAPTER 14. SOLVED PROBLEMS After that, X independent fair coins are tossed and the number of heads obtained is denoted by Y. Compute: 1. P[Y = 4]. 2. P[X = 5|Y = 4]. 3. E[Y ]. 4. E[XY ]. Solution: 1. For k = 1, . . . , 6, conditionally on X = k, Y has the binomial distribution with parameters k and 1 . Therefore, 2 k −k i 2 , 0≤i≤k P[Y = i|X = k] = 0, i > k, and so, by the law of total probability.
6
P[Y = 4] =
k=1
P[Y = 4|X = k]P[X = k] (14.1) 5 −5 6 −6 + 2 + 2 ) 4 4
= 29 384
=
1 −4 6 (2
.
2. By the (idea behind the) Bayes formula P[X = 5|Y = 4] = = P[X = 5, Y = 4] P[Y = 4|X = 5]P[X = 5] = P[Y = 4] P[Y = 4]
5 4 1 6
2−5 ×
5 4
1 6 6 4
2−4
+
2−5
+
2−6
=
10 29
.
1 3. Since E[Y |X = k] = k (the expectation of a binomial with n = k and p = 2 ), the law of total 2 probability implies that 6 6
E[Y ] =
k=1
E[Y |X = k]P[X = k] =
1 6 k=1
k 2
=
7 4
.
4. By the same reasoning,
6 6
E[XY ] =
k=1 6
E[XY |X = k]P[X = k] = kE[Y |X = k]P[X = k] =
k=1
E[kY |X = k]P[X = k]
k=1...