ACTSC 446/846 Winter 2013 Mathematical Models for Finance
Assignment 2 Department of Statistics and Actuarial Science University of Waterloo, Canada
Due: Thursday Feb 26th, 2013 in class. Hard copy please. No electronic version. To earn the full credit of
Mathematical Models in Finance
ACTSC/STAT 446/846, Fall 2014
This is an early version of the outline as of June 2014
Instructor:
Ruodu Wang, M3 3122, ext.31569, [email protected]
Lectures:
1:30-3:00 Tuesdays and Thursdays, RCH 301.
Tutorials:
Only used fo
Department of Statistics and Actuarial Science
University of Waterloo
ACTSC/STAT 446/846 - Midterm Exam
February 23, 2010
Last name:
First name:
Course: 446 or 846 (circle one)
Section: 1 or 2 (circle one)
(Write your UWID number on the back of this page
STAT/ACTSC 446/846
Assignment #4 (due November 23, 2009)
1. Let cfw_Wt be dened by the SDE dWt = dt + dBt , where Bt is a standard Brownian motion. Use Itos formula to write the following processes Yt in the form of a stochastic integral (dYt = u(Yt , t)
QUESTION 1:
If F2 > F1e r ( t2 t1 )
an investor could make a riskless profit by
(a) taking a long position in a futures contract which matures at time t1
(b) taking a short position in a futures contract which matures at time t 2
When the first futures co
STAT/ACTSC 446/846
Assignment #1 (due September 28, 2007) Introduction to derivatives
Note: When handing in your assignment, please use a cover page showing only your UWID number and section (lecture) number. Please write your name on the rst actual page
Problem 1. The process
Zt =
t
0
es dBs
has an integral representation with u = 0 and v(s) = es . The process cfw_Xt can be
obtained from cfw_Zt through the transformation Xt = g(t, Zt ) with
g(t, z) = et [x + z],
for which we have
gt = et [x + z], gz =
STAT/ACTSC 446/846
Assignment #3 (due November 9, 2007)
Note:When handing in your assignment, please use a cover page showing only your UWID number and section (lecture)
number. Please write your name on the rst actual page of your assignment. ACTSC/STAT
ACTSC 446/846 Winter 2013
Mathematical Models for Finance
Assignment 2
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Due: Thursday Feb 26th, 2013 in class. Hard copy please. No electronic version.
To earn the full credit of
1. Consider a single—period market model with N securities, of which the prices at time t are
denoted by a vector S(t) = (5105), . . . , SN(t) for t = O and l.
(a) Give a formal deﬁnition of an arbitrage opportunity. You must state your answer accurately
1. Describe the prot from the following portfolio: a long forward contract on an underlying
asset, and a long European put option on the asset with the same maturity as the forward
contract. The strike price of the option is equal to the forward price of
Example 5.1. Suppose S(0) = (100, 400, 500) and
50, 200, 300
S(1, ) = 100, 400, 500 .
50, 200, 250
Consider a trading strategy = (1, 1, +1)T , and respectively
calculate the portfolio value at time 0 and time 1 for the trading
strategy.
6
Example 5.2. Con
ACTSC 446 Fall 2014
Mathematical Models for Finance
Lecture 4
Properties of Option Prices
Ruodu Wang
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Email: [email protected]
Ruodu Wang
ACTSC446 Fall 14 L4
Previously on 446
Le
Example 3.2. Calculate the price of a 3-month European call
option by a 3-step binomial model. Assume K = 100, S0 = 100,
u = 1.0594, d = 0.9439, and r = 0.05.
18
Question Set 05 Solution
1.
ACTSC 446/846, Fall 2016
(a) The result is obviously true for l = 0 since X is adapted to P. Assume it is true for some
integer l > 0. Then,
E [X (k + l + 1|Pk )]
= E [E [X (k + l + 1|Pk+l )] |Pk ] by properties of conditional
Question Set 05 Solution
ACTSC 446/846, Fall 2016
1. We are given S (0) = (10.8, 10) and S (1, ) =
12, 13
8, 3
.
(a) The state price vector = (1 , 2 ) should be the strictly positive solution to equation
S (0) = S (1, ) , i.e.,
10.8 = 121 + 82
10 = 131 +
Example 4.3. (Digital option) Consider the elementary single-period binomial
market model with parameters: r = 0.03 (continuously compounded interest
rate), S0 = 1, u = 2, d = 1/2 . Compute the price of a digital call option with
strike price K which is c
Example 3.2 (Revisit). Calculate the price of a 3-month European call option by a 3-step binomial model, using the formula
developed in the last slide. Assume K = 100, S0 = 100, u = 1.0594,
d = 0.9439, and r = 0.05.
22
ACTSC 446 Fall 2014
Mathematical Models for Finance
Lecture 3
Basics of Financial Markets
Ruodu Wang
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Email: [email protected]
Ruodu Wang
ACTSC446 Fall 14 L2
Previously on 446
Le
ACTSC 446 Fall 2014
Mathematical Models for Finance
Lecture 5
Properties of Option Prices (2)
Ruodu Wang
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Email: [email protected]
Ruodu Wang
ACTSC446 Fall 14 L5
Previously on 44
ACTSC 446 Winter 2013
Mathematical Models for Finance
Lecture 19
Review of B-S model and Greeks
Ruodu Wang
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Email: [email protected]
Ruodu Wang
ACTSC446 Winter13 L18
Review of Le
Actsc 446/846 (Winter 2016)
Solution to Assignment 1
1. (a) The payo to a long forward at expiration is equal to:
Payo to long forward = ST K
Therefore, we can construct the following table:
Price of asset in 6 months
40
45
50
55
60
Agreed forward price
5
ACTSC 446 Fall 2014
Mathematical Models for Finance
Lecture 1
Review of Financial Markets
Ruodu Wang
Department of Statistics and Actuarial Science
University of Waterloo, Canada
Email: [email protected]
Ruodu Wang
ACTSC446 Fall 14 L1
Lecture 1
Financial
Fact 2.1. We have
S0 ! CA(S0, K, T ) ! CE (S0, K, T )
K ! PA(S0, K, T ) ! PE (S0, K, T )
Proof.
29
Fact 2.2. We have
CA(S0, K, T ) ! CE (S0, K, T ) ! (PV(ST ) PV (K)+
PA(S0, K, T ) ! PE (S0, K, T ) ! (PV (K) PV(ST )+
Proof.
30
Early exercise for American
Question Set 09 Solution
1.
ACTSC 446, Winter 2016
(a) We first compute E[eBt ]. Let Zt = eBt and apply Itos formula, we get
1
dZt = Zt dBt + 2 Zt dt
Z t2
Z
1 2 t
Zt = Z0 +
Zs dBs +
Zs ds
2
0
0
Z t
Z t
1 2
Zs ds
E[Zt ] = E[Z0 ] + E
Zs dBs + E
2
0
0
Z