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Main Examination period 2021 – May/June – Semester B
Online Alternative Assessments
MTH6155: Financial Mathematics II, sample exam
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Examiners: I. Goldsheid
c Queen Mary University of London (2021) Continue to next page
MTH6155 (2021) Page 2
1. The following convention is used in this paper. If Y(t) is a random process then Yt
may be used to describe the same process; a similar convention applies to any other
random process. In particular, both W(t) and Wt denote the standard Wiener process.
2. E denotes the expectation over a risk-neutral probability.
3. Time involved in calculations should be expressed in years. E. g., 3 months should
be converted into 0.25 years.
4. The precision of calculations should be to 3 decimal places.
5. You may use without proof the equality E(ebWt) = eb
2
2
t, where b is any real number.
Question 1 [20 marks].
Let Wt be the standard Wiener process. Define the process Xt by
Xt = e
θtWe2θt , for some constant θ > 0.
(a) Compute μm := E[(Xt)m] for all integer m > 0. [4]
(b) Compute Cov {Xt, Xs}. [4]
(c) Is Xt a Wiener process [2]
(d) Does this process have independent increments [5]
(e) What is the distribution of the increment Xt Xs, where t > s [5]
Question 2 [8 marks].
Suppose that the odds of m possible outcomes of an experiment are oi > 0, where
i = 1, . . . ,m. In other words, the return function is given by
ri(j) =
{
oi if j = i;
1 if j 6= i.
Use the Arbitrage Theorem to show that either
m∑
i=1
(1+ oi)
1 = 1 ,
or there is an arbitrage opportunity. [8]
Question 3 [7 marks]. The price S(u), 0 6 u 6 t, of the share is driven by a
geometric Brownian motion: S(u) = Seμu+σW(u). A proportional dividend on this share
is paid continuously at rate q > 0 and is reinvested in the share. The continuously
compounded interest rate is r. Compute the no-arbitrage price of a derivative with
expiration time t and payoff function
R(t) = [S(t/3)S(2t/3)S(t)]
1
3 . [7]
c Queen Mary University of London (2021) Continue to next page
MTH6155 (2021) Page 3
Question 4 [26 marks]. The price S(t) of a share follows the GBM with
parameters S = ￡40, μ = 0.02, σ = 0.18. the continuously compounded interest rate is
r = 6%.
Consider the option whose expiration time T is 15 months and whose payoff function is
R(S(T)) =
{
￡35 if S(T) 6 ￡35,
0 if S(T) > ￡35.
(a) Compute the no-arbitrage price of this option. [5]
(b) What is the probability that this option will be exercised [4]
(c) If you are the seller of this option, what should be your hedging strategy
Namely, how many shares must be in your portfolio and how much money should
be deposited in the bank at any time t, 0 6 t 6 T , in order for you to be able to
meet your obligation at time T [10]
(d) In one year the price of the share has dropped by ￡2. How many shares should be
in your hedging portfolio and how much money should be deposited in the bank [7]
Question 5 [8 marks]. The price S(t) of a share follows the geometric Brownian
motion S(t) = Seμt+σW(t). The parameters S and μ are given, but the volatility sigma
is not known.
You are asked to compute the implied volatility as an estimate for the parameter σ
using the data provided by the market.
What data you may want to use What is the volatility smile [8]
c Queen Mary University of London (2021) Continue to next page
MTH6155 (2021) Page 4
Question 6 [16 marks]. The price S(t), t > 0, of a share is described by the
following stochastic differential equation:
dSt = aStdt+ σ(t)StdWt, with initial value S(0) = S0,
where a is a constant and σ(t), t > 0, is a positive function
(a) Solve this equation. [8]
(b) Write down S(t) with S0 = 28, a = 0.5, and σ(t) = [0.1+ 0.05pi sin(pit)]
1
2 and
compute the probability that in a years time the price of the share will be less
than it is now. [8]
Question 7 [13 marks]. A company has just issued zero-coupon bonds with
expiration time of 2 years and the total nominal value of ￡3 million. The total value of
the company now stands at ￡4 million. A continuously compounded interest rate is 5%
per annum. The total value of the company follows the Geometric Brownian motion
with parameters μ = 0.4 and σ = 0.2.
(a) Under the Merton model, find the current value of the shareholders’ equity. [7]
(b) In 1 year time, the company’s value drops by 20%. What is the probability of the
company’s default on its obligation to bondholders. [6]
(c) This problem may be used as a replacement for (a) and (b).
After the value drops, the company decides to issue new bonds with the same
maturity date. It is known that this move was the cause of the shareholders’
equity drop by 70%. What is the debt of the company to bondholders at the end
of the 2 years’ period
End of Paper – An appendix of 1 page follows.
c Queen Mary University of London (2021) Continue to next page
MTH6155 (2021) Page 5
Table of the cumulative standard normal distribution
Φ (x) =
1√
2pi
x∫
∞
e t
2/2dt, Φ ( x) = 1 Φ (x)
x 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.7580 0.7611 0.7642 0.7673 0.7703 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916
2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936
2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990
3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993
3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995
3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998
3.5 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998
3.6 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.7 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.8 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.9 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
End of Appendix.
c Queen Mary University of London (2021)