MA3071 Financial Mathematics Year 2022-2023 Coursework 2
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MA3071 Financial Mathematics
Year 2022-2023
Coursework 2
MARKING CRITERIA:
>> Courseworks are marked out of 100 points. The number of marks of each main question is indicated at the beginning of each.
>> Clearly justify and explain your answers. You are expected to use MATLAB to do the
calculations. A printout of your answers without the full explanation of the formulas you are using and of your reasoning will not score full marks.
- You are required to submit a single PDF file (containing justifications, explanations, and codes for each question). You can copy or screenshot your codes into the PDF file without providing the code files in any format.
- You can submit your answers up to 3 attempts when submitting via Blackboard. Only the last attempt of your submission will be assessed. Email submissions won’t be accepted.
- For late submissions, your coursework will be penalized by 30% marks.
>> Computational mistakes will be penalized more in coursework than in exam mark-ing, since you have plenty of time and tools to check your calculations when doing coursework. Please note: Any numerical result should be rounded to four decimal places.
Question [100 marks]
Consider a continuous time market where the interest rate is fixed and 3.96% p.a. A com- pany’s stock price in this market is currently 4.25 and the future price is modelled by
dSt = 0.0396St dt + 0.2St dBt
where {Bt ; t ≥ 0} is a standard Brownian motion.
a) For a European call option on this stock with a strike price 4.65 and time to maturity of
7 months, answer the following questions:
i) [10 marks] Calculate the current price of the call option using the Black-Scholes model.
ii) [20 marks] Using the Monte-Carlo method with M = 10k , k = 3, 4, 5, 6, 7, 8, calcu- late the current price of the call option.
iii) [10 marks] Compare the results you obtained in ii) with the true price. Comment on errors and complexities of Monte-Carlo methods.
b) [20 marks] For an exotic option payoff C = max(4.65 − sinh(ST ), 0) and time to maturity of 7 months, estimate the current option price using the Monte-Carlo method.
Rainbow options are usually calls or puts on the best or worst of n underlying assets. Con- sider a Rainbow call on max option, giving the holder the right to purchase the maximum asset at the strike price at expiry, whose payoff is defined by
max ←max ╱S, S , · · · , Sn)、− K, 0]
where S) is the price of ith underlying asset at time T and K is a fixed strike price.
Assume that a Rainbow call on max option can only be exercised at the maturity and the prices of all underlying assets are independent. Let choose ρ = 0.0396, K = 4.65, n = 5 and time to maturity of 7 months. S i = 1, · · · , 5 are defined by
dSi) = 0.0396Si)dt + σi Si)dBt ,
i = 1, · · · , 5
where σ 1 = 0.2, σ2 = 0.25, σ3 = 0.3, σ4 = 0.35, σ5 = 0.4 and S = 3.5, S = 4, S = 4.5, S = 5, S = 5.5.
c) [40 marks] Using the Monte-Carlo method and choose M = 108 , find the time 0 option price of the Rainbow option.
Solutions
a) i) [10 marks] We know S0 = 4.25, ρ = 0.0396, K = 4.65, σ = 0.2, T = . Thus, the current price of the call option using the Black-Scholes model is
g(0, S0 ) = S0 Φ(d1 ) − Ke-ρT Φ(d2 )
= 4.25Φ(d1 ) − 4.65e-0.0396 × Φ(d2 )
where
d1 = = = −0.3612 d2 = d1 − 0.2 ′ = −0.5140
Hence,
Φ(d1 ) = 0.3590
Φ(d2 ) = 0.3036
g(0, S0 ) = 4.25Φ(d1 ) − 4.65e-0.0396 × Φ(d2 ) = 0.1459
which are calculated by MATLAB.
1 S0=4 .25; 2 K=4 . 65; 3 T=7/12; 4 sigma=0 .2; 5 rho=0 . 0396; 6 7 d1=1/(sigma *sqrt (T)) * (log (S0/K)+(rho+sigmaˆ2/2) *T) 8 d2=d1-sigma *sqrt (T) 9 normcdf(d1) 10 normcdf(d2)
11 BS |
ii) [20 marks] As we know S0 = 4.25, ρ = 0.0396, K = 4.65, σ = 0.2, T = . Thus, the current price of the call option can be calculated by the Monte-Carlo method as following,
g(0, S0 ) ≈ e-ρT i max ╱S0 e /ρ - 、T+σzi^T − K, 0、
= e-0.0396 × i max ╱4.25e/0.0396- 、× +0.2zi^ − 4.65, 0、
where zi , i = 1, · · · , M are random observations from a N(0, 1) distribution.
Then, we use the Monte-Carlo method with M = 10k , k = 3, 4, 5, 6, 7, 8, calculate the current price of the call option by MATLAB. And the results are listed below,
Options |
g(0, S0 ) |
M |
Estimates |
Relative errors |
CPU time (s) |
Call |
0.1459 |
103 104 105 106 107 108 |
0.1533 0.1466 0.1464 0.1457 0.1458 0.1459 |
0.0501 0.0043 0.0030 0.0014 7.0325×10-4 2.2434×10-4 |
0.000262 0.002176 0.005344 0.059244 0.501530 4.696607 |
where the Relative error = │ │ .
1 S0=4 .25; 2 K=4 . 65; 3 T=7/12; 4 sigma=0 .2; 5 rho=0 . 0396; 6 7 d1=1/(sigma *sqrt (T)) * (log (S0/K)+(rho+sigmaˆ2/2) *T); 8 d2=d1-sigma *sqrt (T);
9 BS 10 11 tic; 12 k=8; %k=3,4,5,6,7,8 13 M=10ˆk; 14 C=0; 15 a=rho-sigmaˆ2/2; 16 for i=1:M 17 z=randn; 18 claim=max (S0 *exp (a *T+z *sigma *sqrt (T))-K,0); 19 C=C+claim; 20 end
21 MC 22 toc; 23 relative error=abs ((MC price -BS price)/BS price) |
iii) [10 marks] From the above Table 1, we can see the larger the value of M, the closer the estimated value is to the true value. When M increases exponentially, the relative error decreases. Of course, the code running consumes more time.
b) [20 marks] As we know S0 = 4.25, ρ = 0.0396, σ = 0.2, T = . For an exotic option payoff C = max(4.65 − sinh(ST ), 0), the current option price can be estimated using the Monte-Carlo method,
g(0, S0 ) ≈ e-ρT i max ╱4.65 − sinh ╱S0 e /ρ - 、T+σzi^T 、, 0、 = e-0.0396 × i max ╱4.65 − sinh ╱4.25e/0.0396- 、× +0.2zi^ 、, 0、
where zi , i = 1, · · · , M are random observations from a N (0, 1) distribution.
Then, we estimate the current option price using the Monte-Carlo method with M = 108 by MATLAB, and the result is g(0, S0 ) ≈ 3.3984 × 10-6 ≈ 0.
1 S0=4 .25; 2 K=4 . 65; 3 T=7/12; 4 sigma=0 .2; 5 rho=0 . 0396; 6 7 M=10ˆ8; 8 C=0; 9 a=rho-sigmaˆ2/2; 10 for i=1:M 11 z=randn; 12 ST=S0 *exp (a *T+z *sigma *sqrt (T)); 13 claim=max (K-sinh (ST),0); 14 C=C+claim; 15 end 16 price=exp (-rho *T)*C/M |
c) [40 marks] For a Rainbow call on max option, the payoff is
max ←max ╱S, S , · · · , Sn)、− K, 0] where S) is the price of ith underlying asset
at time T and K is a fixed strike price.
2023-01-08