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Exam: In-term 1 Page: 1 of 2 Date: Oct 07, 2008
University of Texas at Austin, Department of Mathematics M362M - Introduction to Stochastic processes I
In-term exam I
Note: you can leave your answers in terms of sums and/or expressions involving binomial coefficients.
Problem 1.1. (40pts) A fair die is tossed and its outcome is denoted by X, i.e.,
X ∼
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 12. Therefore,
P[Y = i|X = k] =
{(k i
2 −k, 0 ≤ i ≤ k 0 , i > k, and so, by the law of total probability.
P[Y = 4] =
∑^6
k=
P[Y = 4|X = k]P[X = k]
= 16 (2−^4 +
2 −^5 +
2 −^6 )
» = 38429
(2) By the (idea behind the) Bayes formula
P[X = 5|Y = 4] =
P[X = 5, Y = 4]
P[Y = 4]
P[Y = 4|X = 5]P[X = 5]
P[Y = 4]
4
2 −^5 × 16
1 6
2 −^4 +
4
2 −^5 +
4
2 −^6
» =^1029
(3) Since E[Y |X = k] = k 2 (the expectation of a binomial with n = k and p = 12 ), the law of total probability implies that
E[Y ] =
∑^6
k=
E[Y |X = k]P[X = k] = (^16)
∑^6
k=
k 2
» =^74
(4) By the same reasoning,
E[XY ] =
∑^6
k=
E[XY |X = k]P[X = k] =
∑^6
k=
E[kY |X = k]P[X = k]
∑^6
k=
kE[Y |X = k]P[X = k] = (^16)
∑^6
k=
1 2 k
2 » =^91 12
Course: M362M Instructor: Gordan ˇZitkovi´c Semester: Fall 2008
Exam: In-term 1 Page: 2 of 2 Date: Oct 07, 2008
Problem 1.2. (30pts) A fair coin is tossed repeatedly and the record of the outcomes is kept. Tossing stops the moment the total number of heads obtained so far exceeds the total number of tails by 3. For example, a possible sequence of tosses could look like HHTTTHTHHTHH. What is the probability that the length of such a sequence is at most 10?
Solution: Let Xn, n ∈ N 0 be the number of heads minus the number of tails obtained so far. Then, {Xn}n∈N 0 is a simple symmetric random walk, and we stop tossing the coin when X hits 3 for the first time. This will happen during the first 10 tosses, if and only if M 10 ≥ 3, where Mn denotes the (running) maximum of X. According to the reflection principle,
P[M 10 ≥ 3] = P[X 10 ≥ 3] + P[X 10 ≥ 4] = 2(P[X 10 = 4] + P[X 10 = 6] + P[X 10 = 8] + P[X 10 = 10])
= 2−^9
[(
)] »
=^1132
Problem 1.3. (30pts) The generating function of the N 0 -valued random variable X is given by
PX (s) =
s 1 +
1 − s^2
(1) Compute p 0 = P[X = 0]. (2) Compute p 1 = P[X = 1]. (3) Does E[X] exist? If so, find its value; if not, explain why not.
Solution: (1) p 0 = PX (0), so p 0 = 0, (2) p 1 = P (^) X′ (0), and P (^) X′ (s) =
1 − s^2 (1 +
1 − s^2 )
so p 1 = 12. (3) If E[X] existed, it would be equal to P (^) X′ (1). However, lim s↗ 1
P (^) X′ (s) = +∞,
so E[X] (and, equivalently, P (^) X′ (1)) does not exist.
Formulas.
(1) If X has the binomial distrubution with parameters n and p, then P[X = k] =
(n k
pk(1−p)n−k, for k = 0,... , n, E[X] = np, Var[X] = np(1 − p). (2) If {Xn}n∈N 0 is a simple random walk, then
P[Xn = k] =
( (^) n (n+k)/ 2
p
n+k (^2) q
n−k (^2) , k ∈ {−n, −n + 2,... , n − 2 , n} , 0 , otherwise. (3) If Mn = max(X 0 , X 1 ,... , Xn) is the running maximum of the symmetric random walk, then P[Mn = k] = P[Xn = k] + P[Xn = k + 1], for k = 0,... , n.
Course: M362M Instructor: Gordan ˇZitkovi´c Semester: Fall 2008
