Solved Problems

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...