Download Math489/889 Stochastic Processes and Advanced Mathematical

Math489/889
Stochastic Processes and
Advanced Mathematical Finance
Solutions for Homework 7
Steve Dunbar
Due Mon, November 2 , 2009
1. Time to review all of the information we have about coin-tossing fortunes and random walks. You have a record of 100 tosses of a coin
made in class on September 21, 2009. The essence of all the following
questions in Problem 1 is to estimate in various ways an answer to the
following question: How ”typical” was your coin flip sequence? All 2100
coin flip sequences are equally likely of course, so yours is neither more
nor less typical than any other in that way. However, some sets or
events of coin flip sequences are more or less “typical” as measured by
the probability of a corresponding event. How “typical” as measured
by the probability of events containing your sequence is your sequence
in one or more of these categories?
(a) What is your value of T100 , and the number of heads and the
number of tails in your record? Let ² = |T100 | and use Chebyshev’s
inequality as in the proof of the Weak Law of Large Numbers to
provide an upper bound on the probability that for all possible
records |T100 | > ².
(b) For your record let a be the absolute value of the difference of the
number of Heads you obtained and the expected number of heads.
Use the Central Limit Theorem to estimate the probability
P [|Tn − nµ| ≤ a]
1
(c) Use a full blank 8.5 × 11 piece of paper and draw the record of the
random walk you recorded as a graph of Tn versus n. Choose the
time ( or step scale) so that the full time axis for your graph is 10
inches. What is the appropriate scale for the space (or Tn ) axis?
(d) What is the excess of Heads over Tails in your record? What is the
probability that among all possible records, one would observe an
excess of heads over tails at least as large as what you observed?
Use the half-integer correction to provide a good estimate.
(e) What is the amount of time your random walk was on the positive
side? What is the probability of spending at least this much time
on the positive side, among all possible walks?
2. If you buy a lottery ticket in 50 independent lotteries, and in each
lottery your chance of winning a prize is 1/100, write down and evaluate
the probability of winning and also approximate the probability using
the Central Limit Theorem.
(a) exactly one prize,
(b) at least one prize,
(c) at least two prizes.
Explain with a reason whether or not you expect the approximation to
be a good approximation.
Solution: The exact probabilities are easy:
(a)
(b)
µ ¶µ
¶1 µ
¶49
50
1
99
≈ 0.3055586198
1
100
100
¶0 µ
¶50
µ ¶µ
1
99
50
≈ 0.3949939329
1−
0
100
100
(c)
µ ¶µ
¶0 µ
¶50 µ ¶ µ
¶2 µ
¶49
50
1
50
1
99
99
−
≈ 0.08943531310
1−
1
0
100
100
100
100
2
The normal approximations (from the Central Limit Theorem using
the half-integer, or histogram area corrections) are
h
i
3/2−50·(1/100)
√
√
(a) P 1/2−50·(1/100)
<
Z
<
≈ 0.4223907551,
99·50/100
99·50/100
h
i
√
(b) P 1/2−50·(1/100)
< Z] = 1/2 = 0.5,
99·50/100
h
i
√
(c) P 3/2−50·(1/100)
< Z ≈ 0.0776092449.
99·50/100
The normal approximations are not good since the rule of thumb npq >
18 is not satisfied, in fact npq ≈ 1/2.
3. Find a number k such that the probability is about 0.6 that the number
of heads obtained in 1000 tossings of a fair coin will be between 440
and k.
Solution: Since we wish to count the number of heads, let Xi = 1
with probability 1 if the fair coin comes up heads on the ith flip, and
Xi = 0 with probability
P 1/2 if it comes up tails. Then µ = 1/2 and
σ 2 = 1/4, The Sn = ni=1 Xi counts the number of heads. We seek k
so that
P [440 ≤ S1000 ≤ k] = 0.6
This is equivalent to
·
¸
440 − (1/2)1000
Sn − (1/2)1000
k − (1/2)1000
√
√
√
P
≤
≤
= 0.6
(1/2) 1000
(1/2) 1000
(1/2) 1000
By the CLT, this can approximated with
√
Φ((k − 500)/(5 2)) − Φ(−3.795) = 0.6
where Φ(x) is the c.d.f. function for the N (0, 1) random variable. Evaluating with a table
√
Φ((k − 500)/(5 2)) − 0.000075 = 0.6
so
and
√
Φ((k − 500)/(5 2)) = 0.600075
√
(k − 500)/(5 10) = 0.25354
so k = 504.
3
4. Suppose you bought a stock at a price b + c, where c > 0 and the
present price is b. (Too bad!) You have decided to sell the stock after
30 more trading days have passed. Assume that the daily change of the
company’s stock on the stock market is a random variable with mean
0 and variance σ 2 . That is, if Sn represents the price of the stock on
day n with S0 given, then
Sn = Sn−1 + Xn , n ≥ 1
where X1 , X2 , . . . are independent, identically distributed continuous
random variables with mean 0 and variance σ 2 . Write an expression
for the probability that you do not recover your purchase price. Justify with reference to a mathematical theorem why you can write this
expression.
Solution:
Since we explicitly assume that the daily changes are
independent, identically distributed continuous random variables with
mean 0 and variance σ 2 , the hypotheses of the Central Limit Theorem
are satisfied. Then
"
#
30
X
P S0 +
Xi < b + c = P [S3 0 < c]
i=0
·
¸
c
≈P Z< √
σ 30
µ
¶
c
√
=Φ
σ 30
5. A bank has $1,000,000 available to make for car loans. The loans are in
random amounts uniformly distributed from $5,000 to $20,000. Make
a model for the total amount that the bank loans out. How many loans
can the bank make with 99% confidence that it will have enough money
available?
Let X1 , X2 , X3 . . . be a sequence of random variables representing the
individual loan amounts. These random variables may reasonably be
assumed to be independent, and of course are identically distributed
random variables on the interval [5000, 20000]. Then E [Xi] = 12500
and Var [Xi ] = 18,750,000 so σ = 4330.127020. Then the total loan
amount is Sn = X1 + · · · + Xn . We seek P [Sn > 1000000] ≤ 0.01.
4
h
This is approximately the probability P Z >
example from tables) this requires
1000000−12500n
√
4330.127020 n
i
. Note (for
1000000 − 12500n
√
> 2.326347874
4330.127020 n
or n < 73.10947801 so the bank can expect to make about 73 loans.
6. Evaluate p2k,2n and graph the probability mass function for 2n = 30
and all admissible values. Use this to show that p2k,2n is a probability
mass function for 2n = 30. (Some computer software will make this
easy and pleasant, but is not necessary.)
Solution:
2k
p2k,30−2k
0 0.1444644481
2 0.0747229903
4 0.0581178814
6 0.0503688305
8 0.0459889322
10 0.0433609932
12 0.0418395549
14 0.0411363691
16 0.0411363691
18 0.0418395549
20 0.0433609932
22 0.0459889322
24 0.0503688305
26 0.0581178814
28 0.0747229903
30 0.1444644481
7. Required for Mathematics Graduate Students, Extra Credit
for anyone else By actually evaluating the integral, show that
Z
√
1
2
1 α
p
= arcsin( α)
π 0
π
x(1 − x)
5
p
Use this to show that 1/(π x(1 − x)) is a probability density function
for 0 < x < 1.
√
√
Solution: Let u = x, so that du = 1/(2 √x) dx. Changing the
limits of integration u = 0 when x = 0 and u = α when x = α. After
the substitution, the integral becomes
1
π
Define
Z
√
α
√
0
√
2
2
du = arcsin( α).
π
1 − u2


0
α≤0
F (α) =
arcsin( α) 0 < α < 1

1
1≤α
√
2
π
Then F (α) ≥ 0, limα→−∞ F (α) = 0 and limα→+∞ F (α) = 1 and F (α)
is non-decreasing. Hence F (α) is a valid c.d.f. and the given function
is a valid p.d.f.
6