Math 697 Fall 2014: Week 4
Exercise 1 (More on the Markov property) The Markov property means that the
future depends on the present but not on the past, i.e.,
P cfw_Xn = in | Xn1 = in1 , X0 = i0 = P cfw_Xn = in | Xn1 = in1 .
1. Show that the Markov pro
Math 697 Fall 2014: Week 5
Exercise 1 (Card shuing) Suppose you have a deck of 52 cards. At each time you
shue the cards by picking a card at random and placing it on top of the deck. The state
space of this Markov chain consists of all permutation of the
Math 697 Fall 2014: Week 1
Exercise 1 For positive numbers a and b, the Pareto distribution with parameter a, b,
Pa,b has the p.d.f f (x) = aba xa1 for x b and f (x) = 0 for x < b. What is the inversion
method to generate Pa,b .
x
e
Exercise 2 The standar
Math 697 Fall 2014: Week 3
Exercise 1 (More algorithms to compute )
1. The estimator for constructed in class is based on the indicator RV
I = 1 if
V12 + V22 1
(1)
where V1 and V2 are uniform random variable on [1, 1]. As an alternative show that
you can
Math 697 Fall 2014: Week 2
Exercise 1 Consider a normal random variable Z with = 0 and = 1. Use Chernov
bounds to show that for any a > 0
2 /2
P (Z a) ea
Exercise 2 Consider a Poisson random variable N with mean . Use a Chernov bound
to show that for n >
Math 697 Fall 2014: Week 8
Exercise 1 Consider the Markov chain on S = cfw_0, 1, 2, 3, with transition probabilities
P (0, 0) = 1 p0 , P (0, 1) = p0
P (j, j 1) = 1 pj ,
P (j, j + 1) = pj .
(a) Under which conditions on pj is the Markov chain positive rec
Math 697 Fall 2014: Week 9
Exercise 1 Given a branching process with the following ospring distributions determine
the extinction probability a.
(a) p(0) = .25, p(1) = .4, p(2) = .35
(b) p(0) = 5, p(1) = .1, p(3) = .4
(c) p(0) = .62, p(1) = .30, p(2) = .0
Math 697 Fall 2014: Week 7
Exercise 1 On a chessboard compute the expected number of plays it takes a knight,
starting in one of the four corners, to return to its initial position if we assume that at
each each play it is equally likely to make any of it
Math 697 Fall 2014: Week 1-solutions
b
x
Exercise 1 The cdf of the Pareto is F (x) = 1
is X = bU 1/a)
a
for x b. Thus the inversion method
Exercise 2 The cdf of the standardized logistic distribution has the p.d.f F (x) =
U
Thus the inversion method is X
Math 697 Fall 2014: Week 3
Exercise 1 (More algorithms to compute )
1. Two estimators:
Sn =
1
n
n
2
2
Ik , with I = 1cfw_U1 +U2 1 and E[Sn ] =
4
and var(Sn ) =
(1
4
1
n
) =
4
k=1
1
0.1685
n
n
Sn =
J = (1 U 2 )1/2 with E[Sn ] =
Jk ,
4
and var(Sn ) =
1
n
Math 697 Fall 2014: Week 2-Solutions
2 /2
Exercise 1 The MGF of Z is M (t) = et
. So the Chernov bound is
2
et /2
2
P (Z > a) min ta = et /2ta .
t0 e
2 /2
The minimum of t2 /2 ta is t = a and so we have P (Z > a) ea
.
t
Exercise 2 The MGF of N is M (t) =
Math 697 Fall 2014: Week 4
Exercise 1 (More on the Markov property)
1. Using the Markov property and Bayes formula
P cfw_X0 = i0 | X1 = i1 , Xn = in
P cfw_X0 = i0 , X1 = i1 , Xn = in
=
P cfw_X1 = i1 , Xn = in
P cfw_Xn = in | Xn1 = in1 P cfw_X1 = i1 |