Semester One Final Examinations, 2018
MATH3090 Financial Mathematics
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School of Mathematics & Physics EXAMINATION
Semester One Final Examinations, 2018
MATH3090 Financial Mathematics
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Question Mark
Page 1 of 2
Total ________
Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
Q 1. Consider a binomial model of the yield curve over 3 years where the spot yield for first year period y0,1 = 5%. The probability of an up movement in 1-year forward rates for year t = 2, 3 is pt = 0.4 + 0.1t, and 1-year forward rates can go up by a factor of u = 1.6 or down by a factor of d = 0.9. Calculate the zero-coupon bond yield curve and the implied 1-year forward rates embedded in this yield curve. Use discrete
compounding.
We already have y0,1 = 5%. We get
y(u) = 0.05 × 1.6 = 0.08 y(d) = 0.05 × 0.9 = 0.045 y(ud) = 0.05 × 1.6 × 0.9 = 0.072 y(uu) = 0.5 × 1.62 = 0.128 y(dd) = 0.5 × 0.92 = 0.0405
for the binomial model forward-rate lattice. tice)
Probability of an up movement in year 2 is p2 = 0.6 and in year 3 is p3 = 0.7.
To find y0,2 we assume the zero pays 1 at the end of year 2. We calculate that
The price now is
= 0.9259, = 0.9569.
(0.5 marks) (0.5 marks)
(1.5 marks for the lat-
[p2P(u) + (1 − p2)P(d)] = 1 [0.6 × 0.9259 + 0.4 × 0.9569]
1+y0,1 1.05
= 0.8936 (0.5 marks)
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Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
y0,2 = 0.8936 − 1 = 5.7849%. (0.5 marks)
To calculate y0,3 we assume that the zero pays 1 at the end of year 3.
We calculate that
P (uu) = P (ud) =
1 + y(uu) 1
= 1 1.0405
= 0.8865, = 0.9328,
(0.5 marks) (0.5 marks) (0.5 marks)
P (dd) = We then see that
P(u) = 1 1+y(u)
[p3P(uu) + (1 − p3)P(ud)] 1 [0.7 × 0.8865 + 0.3 × 0.9328]
0.8337 (0.5 marks)
1 [p3P(ud) + (1 − p3)P(dd)] 1+y(d)
1 [0.7 × 0.9328 + 0.3 × 0.9611] 1.045
= 0.9008. (0.5 marks) We then find that the price now is
P = 1 [p2P(u) + (1 − p2)P(d)] 1+y0,1
= 0.8196. (0.5 marks) 1
0.8196 − 1 = 6.8561%.
(0.5 marks)
= 1 [0.6 × 0.8337 + 0.4 × 0.9008] 1.05
We already have y0,1 = 5%.
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Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
Using the formula from Part (a) of this question, we find that (1 + y0,2)2 1.0578492
y1,2 = 1+y −1= 1.05 −1=6.5757%. (0.5marks) 0,1
(1 + y0,3)3 1.0685613
y2,3 = (1 + y0,2)2 − 1 = 1.0578492 − 1 = 9.0312%. (0.5 marks)
Q 2. In each of the following parts, you will be given two quantities to compare. Let a and b denote the left-hand-side and the right-hand-side, respectively. There are exactly seven possible answers. Under the assumptions given in each part,
• Answer: “=” if a = b must hold.
• Answer: “<” if a < b must hold.
• Answer: “>” if a > b must hold.
•Answer: “≤”ifa
impossible.
•Answer: “≥”ifa>banda=bareeachpossible;buta
Pick your answer and provide a brief justification.
Note: As usual, our standing assumptions of frictionless markets and no arbitrage apply to all parts. Aside from these general assumptions, each part stands on its own (even if they share the same notation), unless otherwise instructed.
Let P denote the physical probability measure, and let E denote expec- tation with respect to P.
Let P denote the risk-neutral probability measure, and let E denote ex- pectation with respect to P.
There exists a bank account Bt = ert, where r > 0 is constant.
There exists a stock S such that St > 0, ∀t > 0, with probability 1.
Let T and K denote positive constants. Page 4 of 10
Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
(a) Let a = S0 − K. Let b be the time-0 price of a T-expiry, K-strike European call option on S. Compare a and b. [2.5 marks] Solution: a < b
Justification:b≥S0−Ke−rT >S0−K=a.
(b) Let a and b respectively be the time-0 price of a T-expiry, K-strike European call and put option on the underlying asset S. Furthermore, let Z0 be the time-0 price of a T -maturity zero-coupon bond. Compare a and b when K = S0/Z0.
Solution: a = b by put call parity with K = S0/Z0.
(c) Asset S satisfies P(ST = 100) = 0.6andP(ST = 0) = 0.4. Compare S0 and 50. [2.5 marks] Solution: S0 <=> 50 since this is not a risk-neutral measure.
(d) Assets A and B are worth 100 and 110 today, respectively. Asset A will be worth 125 tomorrow with a physical probability of 0.2 and 85 otherwise. Asset B will be worth 125 tomorrow with physical probability 0.7 and 85 otherwise. Let a be today’s price of a European call on A with strike K = 100. Let b be today’s price of a European call on B with strike K = 100. Compare a and b.
[3 marks] Solution: a < b because the risk-neutral probability for the up-move
is higher for asset B.
Q 3. The current price of the stock is $50. We assume that in 1 year, the stock price will be either $55 or $45 with physical probabilities 70% and 30%, respectively. The risk-free rate r is 3%. We want to price a European call written on this asset with the strike K = 50 and the maturity T = 6 months.
(a) Compute the call option price using the physical probabilities.
Solution: the calculated option price is
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Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
e−0.03×0.5((55 − 50)+ × 0.7 + (45 − 50)+ × 0.3) = 3.448
(b) Compute the call option price using the risk-neutral probabilities.
Solution: The risk-neutral probability for the up-move is p = 50 × e0.03×0.5 − 45 = 0.5756
e−0.03×0.5((55 − 50)+ × 0.5756 + (45 − 50)+ × 0.4244) = 2.8351
(c) Show that if the option is traded at a price different than the price in (b), then there is an arbitrage. In particular, specify a static portfolio that meets the definition of arbitrage (which you are to verify, using either the type-1 or type-2 condition). [4 marks]
Solution: As an example, consider the case when the option is traded at a price C ̃0 > 2.8351. In this case, the option is priced more than the replicating portfolio, so we should short the option and long the repli- cating portfolio. More specifically, we form a portfolio Θ consisting of
(-1 option, α bank account, β stock, (C ̃0 − 2.8351) bank account)
replicating portfolio >0 where α and β are defined on L6.31.
Then, the time-0 value of this portfolio is
V0 =−C ̃0 +2.8351+C ̃0 −2.8351=0.
At time T, no matter which state, up or down, the stock ends up in, the long position of the replicating portfolio will offset with the short position in the option. For example, if the stock ends up in the up-state, then the payoff for the (-1 option) is −(55−50)+ = −5. The replicating portfolio gives
αerT +βSu =Cu =(55−50)+ =5 Page 6 of 10
The option price is
Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
which offsets with -5.
But we have the positive amount from (C ̃0 − 2.8351) bank accounts. Thus VT > 0 with certainty.
For the case C ̃0 < 2.8351, we can construct arbitrage similarly. To get full marks, the students must present details for this case too. If not, deduct 1 mark.
(d) Does the above result mean that the physical probabilities are irrele- vant to option pricing? Justify your answer.
No. The physical probabilities indirectly relevant because they affect the price of the underlying asset S. For example, if the physical probabilities change from (0.6, 0.4) for (up-move, down-move) to, for instance, (0.9,0.1) for (up-move, down-move), this will increase S0, which in turn will increase the risk-neutral probability of the up-move
pu = S0erT −Sd. Su − Sd
Through this mechanism, the option price will also increase.
Q 4. We denote by r > 0 the risk-free interest rate. Recall the Black–Scholes model and the Black–Scholes formula for a T-expiry, K-strike European call option written on S having a positive constant instantaneous volatil- ity σ:
CBS(S,t,K,T,r,σ)=SN(d1)−Ke−r(T−t)N(d2), t∈[0,T).
Here, N(x) is the cumulative distribution function of the standard normal
distribution, i.e.
d1,2 = You are to verify that
e−v /2dv ,
2π ln(Ser(T −t)/K)
σ√T − t 2 .
C(S,t) := CBS(S,t,K,T,r,σ) Page 7 of 10
Semester One Final Examinations, 2018 MATH3090 Financial Mathematics satisfies the Black-Scholes PDE with the appropriate boundary condition.
(a) Compute ∂C/∂t, ∂C/∂S, and ∂2C/∂S2. [3 marks] (b) Verify that C(S, t) satisfies
∂C+rS∂C+1∂2Cσ2S2=rC, (S,t)∈(0,∞)×[0,T). ∂t t ∂S 2 ∂S2 t
[3 marks] (c) Verify that for each fixed S, we have the limit C(S, t) → (S − K)+ as
Page 8 of 10
Semester One Final Examinations, 2018 MATH3090 Financial Mathematics
Q 5. Suppose that investors care only about the mean μP and the variance σP2 ofportfolioreturns,andthattheywanttominimiseσP2 subjecttoatarget portfolio return μP . Let r be the risk-free interest rate, and suppose that there are N risky assets. Also let
ω = (ω1,…,ωN) ∈ RN be the risky asset portfolio weights vector,
μ = (μ1,…,μN) ∈ RN the risky asset mean returns vector,
1 = (1,…,1) ∈ RN an N-dimensional vector of ones, and
σ11 … σ1N Σ= . .
. . σN1 … σNN
the N × N risky asset variance–covariance matrix.
(a) What percent of the portfolio is invested in the risk-free asset?
1 − ω · 1. (Exercise own judgment when allocating marks.)
(b) What are the expressions for μP and σP2 for a portfolio composed of risky assets and the risk-free asset? [2 marks] Solution:
See L10.30: μP = (1−ω·1)r+ω·μ = r+ω·(μ−r1) (1 mark) and σP2 = ωTΣω (1 mark).
(c) State the mean–variance optimisation problem for a portfolio com- posed of risky assets and the risk-free asset.
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Semester One Final Examinations, 2018
Solution: See L10.31:
w=argmin 2wΣw
such that r + w · (μ − r1) = μP .
MATH3090 Financial Mathematics
(1.5marks) (1.5 marks)
Page 10 of 10
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