Semester One Final Examinations, 2019
MATH3090 Financial Mathematics
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School of Mathematics & Physics EXAMINATION
Semester One Final Examinations, 2019
MATH3090 Financial Mathematics
This paper is for St Lucia Campus students.
Examination Duration:
Reading Time:
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This is a Closed Book Examination – no materials permitted During reading time – write only on the rough paper provided This examination paper will be released to the Library Materials Permitted In The Exam Venue:
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Calculators – Casio FX82 series or UQ approved (labelled)
Materials To Be Supplied To Students:
1 x 14-Page Answer Booklet
Instructions To Students:
• This exam paper contains five questions, each carries the number of marks shown.
• There are a total of 45 marks.
• Write your working and solution to each question in the answer
booklet provided.
• Additional exam materials (eg. answer booklets, rough paper) will be provided upon request.
120 minutes 10 minutes
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Question Mark
Page 1 of 7
Total ________
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Q 1. ABC Bank enters into a credit default swap as the protection buyer to insure the bank against default on a large fixed interest loan it has made with Queensland Mining Company. The loan ABC Bank has with Queens- land Mining Company is for $50 million over 3 years at a fixed rate of 8% with annual loan payments. ABC Bank is insured for 80% of the loss given default. The yearly probability of default is constant at 16% and the yearly recovery rate is constant at 90%. The notional principal is the loan amount.
Assume that the following yield spot curve is observed: y0,1 = 4%, y0,2 = 4.35%, and y0,3 = 4.68%, and the swap rate is 2%. Find the credit default swap value. [10 marks]
Solution. Fill in the table from the lecture notes. In each year the payout given default is
80% × (1 − R)C = $1.552 million.
Where C is obtained from the fixed interest loan formula using y = 0.08,
F = 50, T = 3 Event
Default t1 Default t
Swap Cashflows 1.552
−1, 1.552 −1, −1, 1.552 −1, −1, −1
PV Swap Cashflows
1.552 1.04
Probability 0.16 (0.84)0.16 (0.84)2 0.16 (0.84)3
−0.25 + 1.552 1.04 1.04352
Default t 3
No Default
− 0.25 − 0.25 1.04 1.04352
1.552 1.04683
−0.25 − 0.25 1.04 1.04352
0.25 1.04683
1.552 swap value = 0.16 1.04
+ (0.84)0.16 − 1.04 + 1.04352
0.25 0.25 1.552 + (0.84) 0.16 − 1.04 − 1.04352 + 1.04683
3 0.25 0.25 0.25 + (0.84) − 1.04 − 1.04352 − 1.04683
(The correct swap rate (which makes the swap value equal to zero) is about 9%.) There may be some rounding differences between this and your solution.
Page 2 of 12
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Q 2. The current price of a stock is $30. We assume that in 1 year, the stock price will be either $33 or $27 with physical probabilities 60% and 40%, respectively. The risk-free rate r is 5%. We want to price a European call written on this stock with strike K = 30 and maturity T = 1 year.
(a) Compute the call option price using the physical probabilities.
Solution: the calculated option price is e−0.05×1(3 × 0.6 + 0 × 0.4) = 1.722
(b) Compute the call option price using the risk-neutral probabilities.
Solution: The risk-neutral probability for the up-move is p= 30×e0.05×1 −27 =0.7564
e−0.05×1(3 × 0.7564 + 0 × 0.2436) = 2.1585
(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 the type-1 condition). [4 marks]
Solution: As an example, consider the case when the option is traded at a price C ̃0 > 2.1585. In this case, the option is priced more than the replicating portfolio, so we should short the option and long the replicating portfolio. More specifically, we form a portfolio Theta consisting of
(-1 option, α bank account, β stock, (C ̃0 − 2.1585) bank account)
replicating portfolio >0
where α and β are defined on L6.31. Page 3 of 12
The option price is
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Then, the time-0 value of this portfolio is
V0 =−C ̃0 +2.1585+C ̃0 −2.1585=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 −(33−30)+ = −3. The replicating portfolio gives
αerT +βSu =Cu =(33−30)+ =3
which offsets with -3.
But we have the positive amount from (C ̃0 − 2.1585) bank accounts. Thus VT > 0 with certainty.
For the case C ̃0 < 2.1585, 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.
Page 4 of 12
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Q 3. Consider a binomial model of the yield curve over 3 years where the spot yield for first year period y0,1 = 4%. The probability of an up movement in 1-year forward rates for year t = 2, 3 is pt = 0.5 + 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 = 4%. We get
y(u) = 0.04 × 1.6 = 0.064
y(d) = 0.04 × 0.9 = 0.036 y(ud) = 0.04 × 1.6 × 0.9 = 0.0576 y(uu) = 0.04 × 1.62 = 0.1024 y(dd) = 0.04 × 0.92 = 0.0324
[10 marks]
for the binomial model forward-rate lattice. (1.5 marks for the lat- tice)
Probability of an up movement in year 2 is p2 = 0.7 and in year 3 is p3 = 0.8.
To find y0,2 we assume the zero pays 1 at the end of year 2. We calculate that
The price now is
= 0.9398, = 0.9653.
(0.5 marks) (0.5 marks)
[p2P(u) + (1 − p2)P(d)] = 1 [0.7 × 0.9398 + 0.3 × 0.9653]
1+y0,1 1.04
= 0.9110 (0.5 marks)
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Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
y0,2 = 0.9110 − 1 = 4.7709%. (1 marks)
To calculate y0,3 we assume that the zero pays 1 at the end of year 3.
We calculate that
P (uu) = 1 = 1
= 0.9766, = 0.9455,
0.9119 (0.5 marks)
1 [p3P(ud) + (1 − p3)P(dd)] 1+y(d)
1 [0.8 × 0.9455 + 0.2 × 0.9686] 1.036
P (ud) = 1
P (dd) = 1
We then see that P(u) =
(0.5 marks) (0.5 marks) (0.5 marks)
= 0.8783. 1
1 3 0.8783
(0.5 marks)
− 1 = 4.4205%.
[p3P(uu) + (1 − p3)P(ud)] 1 [0.8 × 0.9765 + 0.2 × 0.9455]
= 0.9171. (0.5 marks) We then find that the price now is
P = 1 [p2P(u) + (1 − p2)P(d)] 1+y0,1
= 1 [0.7 × 0.9119 + 0.3 × 0.9171] 1.04
We already have y0,1 = 4%.
Page 6 of 12
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Using the formula from Part (a) of this question, we find that
y1,2 = 1+y and that
(1 + y0,2)2
1.04770922
1.04 −1=5.5475%.
(1 + y0,3)3
y2,3 = (1 + y0,2)2 − 1 = 1.0477092 − 1 = 3.7232%.
Page 7 of 12
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Q 4. Recall the Black–Scholes model discussed in class. Specifically, adopt the frictionless assumption and consider two basic assets, namely a bank account with dynamics
dBt = rBtdt, B0 = 1, and a non-dividend-paying stock with dynamics
dSt =μStdt+σStdWt, S0 >0.
Here, r > 0 is the constant risk-free rate, σ > 0 is the instantaneous volatility, μ is the drift, and W is Brownian motion under physical prob- abilities.
Let C(S,t) be the price of a T-expiry, K-strike European call option writ- ten at time t ∈ [0,T] on S. Recall the Black–Scholes Partial Differential Equation (PDE) that governs C(S, t):
∂C +rS∂C +1∂2Cσ2S2 =rC, (S,t)∈(0,∞)×[0,T), ∂t ∂S 2∂S2
C(S, T ) = (S − K)+.
(a) Let τ = T −t, z = ln(S)+(r−σ2/2)τ, and U(z,τ) = erτC(S,T −τ). Show that U(z,τ) satisfies the PDE
∂U−1σ2∂2U=0, (z,τ)∈R×(0,T], ∂τ 2 ∂z2
U(z,0)=f(z)=(ez −K)+ .
[3 marks] Solution: By the definition of U, and repeated applications of the
chain rule,
∂U 1 2∂2U ∂τ−2σ∂z2 =e
rτ 2 ∂C ∂C 1 2 2∂2C ∂C rC−(r−σ/2)S∂S−∂t −2σ S∂S2+S∂S
∂C ∂C 122∂2C rC − rS ∂S − ∂t − 2σ S ∂S2
which is identically 0 if and only if C satisfies the B-S PDE. Page 8 of 12
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics
Moreover, the initial/terminal data are equivalent: at τ = 0 we have U(z,0)=C(ez,T). SoU(z,0)=(ez−K)+ forallz∈R,iffC(S,T)= (S−K)+ forallS>0.
It can be shown that the PDE in part (a) has solution
∞ U(z,τ)=
1 −(z−ζ)2/(2σ2τ)
where f(ζ) = (eζ − K)+, as defined in part (a).
(b) Using the above result, show that
z+σ2τ/2 z −lnK +σ2τ z −lnK
e f(ζ)dζ ,
U(z,τ) = e N σ√τ − KN σ√τ ,
where N(x) is the cumulative distribution function of the standard
normal distribution, i.e. N(x) = √
U(z,τ)= √2exp−2σ2τ(e−K)dζ −∞ 2πσ τ
∞ 1 (z−ζ)2ζ +
e−v /2dv .
∞ 1 (z−ζ)2ζ
= √2exp−2σ2τ(e−K)dζ
2 exp − 2σ2τ 2πσ τ
logK e dx−K
2 exp − 2σ2τ dζ
(log K−z−σ2τ)/σ
edζ−K z+xσ√τ+σ2τ
1 (z−ζ)2ζ ∞
z+σ2τ/2∞ =e
(log K−z−σ2τ)/σ
z+σ2τ/2 z −logK +σ2τ z −logK
1 −(x+σ√τ)2
√ √2πe 2 dx−K τ
√ √2πe dx τ
(log K−z)/σ
(log K−z)/σ =e N σ√τ −KN σ√τ
(c) Carry out the steps transforming the solution U(z,τ) in part (b) to C(S, t) to show that C(S, t) satisfies the Black–Scholes formula
C(S,t)=SN(d1)−Ke−r(T−t)N(d2), t∈[0,T), Page 9 of 12
Semester One Final Examinations, 2019
MATH3090 Financial Mathematics
ln(Ser(T −t)/K) d1,2 = σ√T − t
σ√T − t ± 2 .
[3 marks] Substituting z = log(S) + (r − σ2/2)τ and U = erτ C into (b),
rτ rτ log(Serτ/K)+σ2τ/2 log(Serτ/K)−σ2τ/2
e C=Se N σ√τ −KN σ√τ Simplifying gives the BS formula.
Page 10 of 12
Semester One Final Examinations, 2019 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) State the mean–variance optimisation problem for a portfolio com-
posed of risky assets and the risk-free asset.
Solution: See slide 27 of notes:
1T w=argmin 2wΣw
suchthat Rf +w·(μ−Rf1)=μP.
(b) Solve the mean–variance optimisation problem in part (a) using the method of Lagrange multipliers. In particular, derive an expression
L ( w , λ ) = 12 w T Σ w + λ μ P − R f − w · ( μ − R f 1 ) Page 11 of 12
for the optimal portfolio’s variance, denoted by σP . Then express μP in terms of σP . The Lagrangian is:
(0.5marks) (0.5marks)
Semester One Final Examinations, 2019 MATH3090 Financial Mathematics The partial derivatives of the Lagrangian are:
∇L= ∂ L=Σw−λ(μ−Rf1), (i) ∂w
∂ L=μP −Rf −w·(μ−Rf1) (ii) ∂λ
Solution: From (i):
Σw − λ(μ − Rf 1) = 0 ⇒ w = λΣ−1(μ − Rf 1). (1 mark)
From (ii):
μP −Rf =(μ−Rf1)·w =λ(μ−Rf1)TΣ−1(μ−Rf1)
⇒ λ = μP − Rf . (1 mark) (μ−Rf1)TΣ−1(μ−Rf1)
Finding the optimal variance involves just writing out
σP = w Σw (0.5 marks)
and some simple arithmetic (0.5 marks) to find that 2 T ( μ P − R f ) 2
σ P = w Σ w = ( μ − R f 1 ) T Σ − 1 ( μ − R f 1 ) . WehaveμP =Rf +(μ−Rf1)TΣ−1(μ−Rf1)σP
END OF EXAMINATION
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