The University of Sydney
Discipline of Finance
FINC6000 – Quantitative Finance and Derivatives
Assignment – 2018
– Due date and time: October 15, 2017 by 17:00.
– Please include a cover page containing only the student numbers of each member in your group.
Cover page does not contribute to the total number of pages for the assignment.
– Assignments must be typed and submitted via Turnitin through Canvas.
– Penalty of 10% per calendar day, or part thereof, will apply to late submissions.
– An Excel spreadsheet, normal_random.xlsx, has be uploaded to Canvas and contains 8,191 × 8
array of standard normal random variates required for the assignment.
– Do not attach large printouts of your Excel spreadsheet to the assignment, and only include
values specifically requested.
This assignment is concerned with the pricing of a spread range accrual on two correlated assets
Xt and Yt under the Black-Scholes model.
Assume that the prices, Xt and Yt, of a risky assets at time t satisfy the equations dXt = rXtdt + XXtdwX
t , (1) dYt = rYtdt + Y YtdwY t , (2) under the risk-neutral measure Q, where wX t and wY
t are standard Q-Wiener process with cor(dwX t , dwY t ) = dt, and r, X, , , X0, and Y0 are constants. Moreover, assume that the market is complete.
Let 0 = t0 < t1 < t2 < · · · < tn be a sequence of times and consider a European derivative with payoff at time T of
hT = R Xn i=1 i1 L<Xti −Yti<U , (3) where 0 < R 2 R, tn T, L < U 2 R, and i is the year fraction for the interval [ti−1, ti]. It is known that Xt and Yt are lognormal random variables, and that the difference of lognormal random variables is not lognormal. In this assignment we will value the derivative using normal approximation for Xti − Yti , and alternatively by using Monte Carlo simulation.
1. This question guides you to compute the distributional properties of Xt − Yt. [20 marks]
(a) By letting xt = lnXt and yt = ln Yt, and applying Itô’s lemma, show that [3 marks]
dxt = r − 12 2X dt + X dwX t , dyt = r − 12 2Y dt + Y dwY t .
(b) By integrating the two expressions in part (a) and using wX0 = wY0 = 0, show that Xt = X0e r−12
2X t+XwX t , Yt = Y0e r−12 2Y t+Y wY t . 2
Hence conclude that the means and the variances of Xt and Yt are given by EQ[Xt] = X0ert, varQ[Xt] = X2
0 e2rt e2X t − 1, EQ[Yt] = Y0ert, varQ[Yt] = Y 2 0 e2rt e2Y t − 1.
You may use the fact that EQ h e−12 2t+wt i = 1 for any 2 R+, where wt is a standard
Q-Wiener process. [8 marks]
(c) Show that you may write XwX t +Y wY t = p 2X + 2Y + 2XY wt, where wt is another standard Q-Wiener process. Hence, or otherwise, show that [6 marks]
covQ [Xt, Yt] = X0Y0e2rt eXY t − 1.
(d) Conclude that EQ[Xt − Yt] = (X0 − Y0)ert and [3 marks] varQ [Xt − Yt] = X2 0 e2rt e2X
t − 1
+ Y 2
0 e2rt
e2Y
t − 1
− 2X0Y0e2rt
eXY t − 1
.
2. In this question, assume that X0 = 10.25 Y0 = 10, X = 16%, Y = 18%, = 0.6, r = 4%,
R = 6%, L = −0.25, U = 0.25, ti = 0.25i and
i = 0.25 for 1 i 4, and T = 1. [16 marks]
(a) For each i 2 {1, 2, 3, 4}, compute the mean and the variance of Xti−Yti using the expressions
obtained in Question 1, and fill in the second and third columns of the following table,
rounding each value to 4 decimal places. [8 marks]
i EQ [Xti − Yti ] varQ [Xti − Yti ] Q[Xti − Yti < L] Q[Xti − Yti < U]
1 – – – –
2 – – – –
3 – – – –
4 – – – –
(b) Using the values of EQ [Xti − Yti ] and varQ [Xti − Yti ] computed in part (a), and making
the assumption that Xti − Yti is normally distributed, compute Q[Xti − Yti < L] and
Q[Xti − Yti < U], and fill in the fourth and fifth columns in the table given in part (a).
Explain how you computed the values, and round the values to 4 decimal places. [6 marks]
(c) Hence, or otherwise, compute the price of the derivative with payoff given by (3) under
the assumption that Xti − Yti is normally distributed for 1 i 4. Round the price to 6
decimal places. [2 marks]
3. Using the parameters given in Question 2, and the standard normal random numbers contained
in the file normal_random.xlsx, simulate the prices of the two assets Xt and Yt, and compute the
price of the derivative with payoff given in (3) by following the steps below. In this question,
you should use simulation step size of t = 0.25, and use the following equations to update the
log asset prices
xti+1 = xti +
r −
12
2X
t + X
p
t 1,
yti+1 = yti +
r −
12
2Y
t + Y
p
t 1 + Y
p 1 − 2 p t 2, where 1 and 2 are pair of independent standard normal random numbers used to simulate
over a given step and path. [12 marks]
(a) Simulate the asset price paths as described above, and fill in the following table with the values you obtained.
3 path lnX0.25 ln Y0.25 lnX0.5 ln Y0.5 lnX0.75 ln Y0.75 lnX1 ln Y1
2 – – – – – – – –
3 – – – – – – – –
4 – – – – – – – –
5 – – – – – – – –
Note that the required values are the asset prices and not the log-asset prices that you simulate. For each path, use the first two normal random variates for the first step, next two normal random variates for the next step, and so on, and round the values to 4 decimal places. [6 marks]
(b) Using the simulated paths, compute the Monte Carlo price of the derivative with the payoff given in (3). Give a very brief explanation of how you computed the price and round the value to 6 decimal places. [3 marks]
(c) Provide two reasons for the difference you observe with the price obtained in Question 1, and explain how you would be able to obtain a more accurate approximation to the correct price. [3 marks]
4. Let qi = Q[L < Xti − Yti < U] for 1 i n, so that qi are the probabilities of observing L < Xti − Yti < U under Q. Suppose that rather than being a payoff on a derivative, the amount given in (3) is the interest paid per $1 invested. [7 marks]
(a) What should be the value of R so that a risk-neutral investor would be indifferent to investing in this instrument and the risk free asset? Derive an expression for R in terms of other variables. [3 marks]
(b) Using the parameter values given in Question 2, and using the normal approximation, compute the corresponding value of R and round to 4 decimal places. [2 marks]
(c) Compute the simple annual interest rate equivalent to the continuously compounded rate of r = 4%, and explain the difference between this rate and the value of R obtained in part (b). In particular, explain why one is much larger than the other. [2 marks]