MA3071 Financial Mathematics Year 2022-2023 Coursework 1
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MA3071 Financial Mathematics
Year 2022-2023
Coursework 1
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. If you are using MATLAB or Excel 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.
- If you use MATLAB, you will need to include your code in the appendix of your answers. Your code should be properly commented.
- If you use Excel, you should submit your answers along with a separate Excel file.
>> Computational mistakes will be penalized more in courseworks 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 discrete market, where the interest rate 3.96% p.a is fixed and the underlying asset price is defined by a 20_period (each period is one month) binomial tree model,
S0 = 4.35, St = St − 1 Yt
where Yt , t = 1, 2, . . . , 20 are independent random variables with
P (Yt = 1.01) + P (Yt = 0.99) = 1, t = 1, . . . , 4
P (Yt = 1.005) + P (Yt = 0.987) = 1, t = 5, . . . , 8
P (Yt = 1.015) + P (Yt = 0.97) = 1, t = 9, . . . , 12
P (Yt = 1.02) + P (Yt = 0.975) = 1, t = 13, . . . , 16
P (Yt = 1.01) + P (Yt = 0.99) = 1, t = 17, . . . , 20
For a European put option with the strike price K = 4.65 and time to maturity of 20 months, answer the following questions:
i) [10 marks] State the no-arbitrage condition.
ii) [40 marks] Find the no-arbitrage time 0 option price of the European put option.
iii) [20 marks] If the European put option can only be exercised when the underlying asset price goes below the level L = 4.2 during or at the end of the 20 months. Calculate the no-arbitrage option price at time 0 of this barrier put option.
An Asian option is a special type of option contract. For Asian options the payoff is determined by the average underlying asset price over some pre-set period of time. This is different from the case of the usual European option and American option, where the payoff of the option contract depends on the price of the underlying instrument at exercise. Asian options are thus one of the basic forms of exotic options.
Assume that an Asian call option can only be exercised at maturity, and the payoff is defined by
C = max(An _ K, 0), An = St
where An denotes the average price for the n periods and K is a fixed strike price.
iv) [30 marks] Let K = 4.65 and n = 20 months. Using the above binomial tree model, find the no-arbitrage option price at time 0 of this Asian call option.
Solutions
i) [10 marks] As the yearly interest rate is 3.96%, the monthly interest rate is ρ = = 0.33%. For each period in the 20_period tree, we have
ut = 1.01 > 1 + ρ = 1.0033 > dt = 0.99, t = 1, . . . , 4 ut = 1.005 > 1 + ρ = 1.0033 > dt = 0.987, t = 5, . . . , 8 ut = 1.015 > 1 + ρ = 1.0033 > dt = 0.97, t = 9, . . . , 12
ut = 1.02 > 1 + ρ = 1.0033 > dt = 0.975, t = 13, . . . , 16 ut = 1.01 > 1 + ρ = 1.0033 > dt = 0.99, t = 17, . . . , 20
Hence, the no arbitrage condition is satisfied for all periods in the binomial tree model.
ii) [40 marks] The binomial tree model is a time varying model with 20 periods. Thus, there will be 220 paths from time 0 to time 20.
Let S and Ci , i = 1, . . . , 220 denote the underlying asset price at time 20 and the payoff of the European put option corresponding to path i, respectively, e.g.,
C1 = max(K _ S 0) = max(K _ S0u1 . . . u20 , 0)
= max(4.65 _ 4.35 . 1.014 . 1.0054 . 1.0154 . 1.024 . 1.014 , 0) = 0 C22← = max(K _ S 0) = max(K _ S0 d1 . . . d20 , 0)
= max(4.65 _ 4.35 . 0.994 . 0.9874 . 0.974 . 0.9754 . 0.994 , 0) s 1.6025
Then, the arbitrage free q _probabilities are defined by
q) = = = 0.665, qt) = 0.335, t = 1, . . . , 4 q) = = = 0.9056, qt) = 0.0944, t = 5, . . . , 8
q) = = = 0.74, qt) = 0.26, t = 9, . . . , 12 q) = = = 0.6289, qt) = 0.3711, t = 13, . . . , 16
q) = = = 0.665, qt) = 0.335, t = 17, . . . , 20
Let Qi , i = 1, . . . , 220 denote the q _probability corresponding to path i, e.g.,
Q1 = qq . . . q
= 0.6654 . 0.90564 . 0.744 . 0.62894 . 0.6654 s 0.0012
Q22← = qq . . . q
= 0.3354 . 0.09444 . 0.264 . 0.37114 . 0.3354 s 1.0939 × 10 − 12
Hence, the option price at time 0 is
22←
V0 = (1 + ρ)−20 Ci Qi = 0.1157
i=1
which is calculated by MATLAB.
iii) [20 marks] For the barrier put option, it has the same setup as the above.
However, only if S20 < 4.2, then the option holder can exercise the option right. Under this barrier, if S > 4.2, then Ci = 0. If S20i < 4.2, then Ci = max(K _ S 0) = 4.65 _ S
After the adjustment of Ci , the barrier option price at time 0 is
22←
V0 = (1 + ρ)−20 Ci Qi = 0.0406
i=1
which is calculated by MATLAB.
iv) [30 marks] Let S t = 1, . . . , 20, i = 1, . . . , 220 denote the underlying asset price at time t corresponding to path i, e.g.,
S = S0u1 , S = S0u1u2 , . . . , S = S0u1u2 . . . u20
Then, the payoff corresponding to path i is
Ci = max ╱ Si) _ 4.65, 0\ , i = 1, . . . , 220
The Asian call option price at time 0 is
22←
V0 = (1 + ρ)−20 z Ci Qi = 0.0103
i=1
which is calculated by MATLAB.
MATLAB Codes
〇 S0=4 .35; 2 rho=(3 . 96/100)/12;%monthly interest rate 3 n=20; 4 K=4 . 65; 5 u=[1 . 01 *ones(1,4) 1 . 005 *ones(1,4) 1 . 015 *ones(1,4) 1 . 02 *ones(1,4) ... 1.01 *ones(1,4)]; 6 d=[0 . 99 *ones(1,4) 0 . 987 *ones(1,4) 0 . 97 *ones(1,4) 0 . 975 *ones(1,4) ... 0.99 *ones(1,4)]; f A %%%%%%%%%%%compute qu for each period d qu=zeros (1,n); 〇← for i=1:n 〇〇 qu(i)=(1+rho-d(i))/(u(i)-d(i)); 〇2 end 〇3 qd=1-qu; 〇4 〇5 %%%%%%%%%%%define the tree and q-probability for each path 〇6 St=[[S0; zeros (2ˆn- 1,1)] zeros (2ˆn,n)]; 〇f Q=[[1; zeros (2ˆn- 1,1)] zeros (2ˆn,n)]; 〇A for j=2:n+1 〇d for i=1:2ˆ(j- 1) 2← if rem (i,2)==0 2〇 St((i- 1) *2ˆ(n- j+1)+1,j)=St((i-2) *2ˆ(n- j+1)+1,j- 1) *d(j- 1); 22 Q((i- 1) *2ˆ(n- j+1)+1,j)=Q((i-2) *2ˆ(n- j+1)+1,j- 1) *qd(j- 1); 23 else 24 St((i- 1) *2ˆ(n- j+1)+1,j)=St((i-rem (i,2)) *2ˆ(n- j+1)+1,j- 1) *u(j- 1); 25 Q((i- 1) *2ˆ(n- j+1)+1,j)=Q((i-rem (i,2)) *2ˆ(n- j+1)+1,j- 1) *qu(j- 1); 26 end 2f end 2A end 2d 3← %%%%%%%%%%%%ii) European put 3〇 payoff=zeros (2ˆn,1); 32 for i=1:2ˆn 33 payoff(i)=max (K-St(i,n+1),0); |
34 end 35 V0=payoff' *Q(:,n+1)* (1+rho)ˆ(-n) 36 3f %%%%%%%%%%%%iii) Barrier European put 3A L=4 .2; 3d payoff=zeros (2ˆn,1); 4← for i=1:2ˆn 4〇 if St(i,n+1)<L 42 payoff(i)=max (K-St(i,n+1),0); 43 else 44 payoff(i)=0; 45 end 46 end 4f V0=payoff' *Q(:,n+1)* (1+rho)ˆ(-n) 4A 4d %%%%%%%%%%%%iv) Asian call 5← for j=1:n 5〇 for i=1:2ˆn 52 if St(i,j)==0 |
2023-01-08