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7212019 MATH_3075_3975_s12
httpslidepdfcomreaderfullmath30753975s12 14
MATH30753975
Financial Mathematics
Week 12 Solutions
Exercise 1 We consider the Black-Scholes model M = (B S ) with theinitial stock price S 0 = 9 the continuously compounded interest rate r = 001per annum and the stock price volatility σ = 01 per annum
1 Using the Black-Scholes call option pricing formula
C 0 = S 0N 1048616
d+(S 0 T )1048617minusKeminusrT N
1048616dminus(S 0 T )
1048617
we compute the price C 0 of the European call option with strike priceK = 10 and maturity T = 5 years We find that
d+(S 0 T ) = minus013578 dminus(S 0 T ) = minus035938
and thus C 0 = 059285
2 Using the Black-Scholes put option pricing formula
P 0 = K eminusrT N 1048616minus dminus(S 0 T )
1048617minus S 0N 1048616minus d+(S 0 T )
1048617we find that the price P 0 = 110514
3 The put-call parity relationship holds since
C 0minusP 0 = 059285minus 110514 = minus051229 = 9minus 10eminus005 = S 0minusKeminusrT
4 We now recompute the prices of call and put options for modified ma-turities T = 5 months and T = 5 days
bull We note that 5 months is equivalent to T = 0416667 and thus
d+(S 0 T ) = minus153541 dminus(S 0 T ) = minus159996
Hence C 0 = 0015315 and P 0 = 0973735
bull We note that 5 days is equivalent to T = 0013699 and thus
d+(S 0 T ) = minus898455 dminus(S 0 T ) = minus899615
Hence C 0 = 149E minus 21 and P 0 = 099863
bull The call option (respectively put option) price decreases to zero(respectively increases to K minus S 0 = 1) when the time to matu-rity tends to zero This is related to the fact that S 0 lt K andthus for short maturities it is unlikely (respectively very likely)that the call option (respectively put option) will be exercised atexpiration
1
7212019 MATH_3075_3975_s12
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Exercise 2 Assume that the stock price S is governed under the martingale
measure P by the Black-Scholes stochastic differential equation
dS t = S t1048616
r dt + σ dW t1048617
where σ gt 0 is a constant volatility and r is a constant short-term interestrate Let 0 lt L lt K be real numbers Consider the contingent claim withthe payoff X at maturity date T gt 0 given as
X = min1048616|S T minusK | L
1048617
1 It is easy to sketch the profile of the payoff X as the function of thestock price S T The decomposition of this payoff in terms of payoffs of
standard call and put options reads
X = L minus C T (K minus L) + 2C T (K )minus C T (K + L)
Note that other decompositions are possible
2 The arbitrage price πt(X ) satisfies for every t isin [0 T ]
πt(X ) = Leminusr(T minust) minus C t(K minus L) + 2C t(K )minus C t(K + L)
3 We will now find the limits of the arbitrage price limLrarr0 π0(X ) andlimL
rarrinfinπ0(X ) We observe the payoff X increases when L increases
Hence the price π0(X ) is also a increasing function of L Moreover
limLrarr0
π0(X ) = minusC 0(K ) + 2C 0(K )minus C 0(K ) = 0
By analysing the payoff X when L tends to infinity we obtain
limLrarrinfin
π0(X ) = P 0(K ) + C 0(K )
4 To find the limit lim σrarrinfin π0(X ) we observe that
limσ
rarrinfin
d+(S 0 T ) = infin limσ
rarrinfin
dminus(S 0 T ) = minusinfin
so that
limσrarrinfin
N 1048616
d+(S 0 T )1048617
= 1 limσrarrinfin
N 1048616
dminus(S 0 T )1048617
= 0
Hence the price of the call option satisfies for all strikes K isin R+
limσrarrinfin
C 0(K ) = S 0
This in turn implies that lim σrarrinfin π0(X ) = Leminusr(T minust)
2
7212019 MATH_3075_3975_s12
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Exercise 3 (MATH3975) We consider the stock price process S given by
the Black and Scholes model
1 We will first show that S t = eminusrtS t is a martingale with respect tothe filtration F = (F t)tge0 generated by the stock price process S Weobserve that this filtration is also generated by W Using the propertiesof the conditional expectation we obtain for all s le t
EP
1048616S t
F s1048617 = EP
983080S s eσ(W tminusW sminus 1
2σ2(tminuss)) F s
983081= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss) EP1048616eσ(W tminusW s)1048617
where in the last equality we used the independence of increments of the Wiener process Recall also that W t minus W s =
radic tminus s Z where Z sim
N (0 1) and thus
EP
1048616S t | F s
1048617 = S s eminus
1
2σ2(tminuss)
EP
1048616eσradic tminussZ 1048617
Let us finally recall that if Z sim N (0 1) then for any real number a
EP
1048616eaZ
1048617 = e
1
2a2
By setting a = σradic
t
minuss we obtain
EP
1048616S t | S u u le s
1048617 = S s eminus
1
2σ2(tminuss) e
1
2σ2(tminuss) = S s
which shows that S is a martingale under P
2 To compute the expectation EP(S t) we observe that
EP(S t) = ert EP
(S t) = ert EP(S 0) = ertS 0 = ertS 0
To compute the variance VarP(S t) we recall that
Var P(S t) = EP(S 2
t)minus 1048667
EP(S t)
10486692
where in turnEP
(S 2t
) = S 20e2rt EP
983131e2σW tminusσ2t983133
= S 20e2rteσ2tEP
983131e2σW tminus 1
2(2σ)2t
983133
= S 20e2rteσ2tEP
983131eaZ minus
1
2a2
983133= S 20e2rteσ
2t
where we denote a = σradic
t Hence Var P(S t) = S 20e2rt1048616
eσ2t minus 1
1048617
3
7212019 MATH_3075_3975_s12
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Exercise 4 (MATH3975) We denote by v the Black-Scholes call option pri-
cing that is the function v R+ times [0 T ] rarr R such that C t = v(S t t) for allt isin [0 T ]
1 We wish to check that the pricing function of the European call optionsatisfies the Black-Scholes PDE
partv
partt +
1
2 σ2s2
part 2v
parts2 + rs
partv
parts minus rv = 0 forall (s t) isin (0infin)times (0 T )
with the terminal condition v(s T ) = (sminusK )+
To this end we use the partial derivatives from Section 55 in the course
notes (note that v(s t) = c(s T minus t))
vs(s t) = N (d+(s T minus t))
vss(s t) = n(d+(s T minus t))
sσradic
T minus t
vt(s t) = minus sσ
2radic
T minus tn(d+(s T minus t))minusKreminusr(T minust)N (dminus(s T minus t))
where n is the density function of the standard normal distributionHence
minus sσ
2radic
T minus tn(d+(s T minus t)) minusKreminusr(T minust)N (dminus(s T minus t))
+ 1
2 σ2s2
n(d+(s T minus t))
sσradic
T minus t+ rsN (d+(s T minus t))minus rv(s t) = 0
where we have also used the equality
v(s t) = sN (d+(s T minus t))minusKeminusr(T minust)N (dminus(s T minus t))
2 We need to show that for every s isin R+
limtrarrT
v(s t) = (sminusK )+
For this purpose we observe that d+(s K ) and dminus(s K ) tend to infin(respectively minusinfin) when t rarr T and s gt K (respectively s lt K )Consequently N (d+(s K )) and N (dminus(s K )) tend to 1 (respectively 0)when t rarr T and s gt K (respectively s lt K ) This in turn impliesthat v(s T ) tends to either sminusK or 0 depending on whether s gt K ors lt K The case when s = K is also easy to analyse
4
7212019 MATH_3075_3975_s12
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Exercise 2 Assume that the stock price S is governed under the martingale
measure P by the Black-Scholes stochastic differential equation
dS t = S t1048616
r dt + σ dW t1048617
where σ gt 0 is a constant volatility and r is a constant short-term interestrate Let 0 lt L lt K be real numbers Consider the contingent claim withthe payoff X at maturity date T gt 0 given as
X = min1048616|S T minusK | L
1048617
1 It is easy to sketch the profile of the payoff X as the function of thestock price S T The decomposition of this payoff in terms of payoffs of
standard call and put options reads
X = L minus C T (K minus L) + 2C T (K )minus C T (K + L)
Note that other decompositions are possible
2 The arbitrage price πt(X ) satisfies for every t isin [0 T ]
πt(X ) = Leminusr(T minust) minus C t(K minus L) + 2C t(K )minus C t(K + L)
3 We will now find the limits of the arbitrage price limLrarr0 π0(X ) andlimL
rarrinfinπ0(X ) We observe the payoff X increases when L increases
Hence the price π0(X ) is also a increasing function of L Moreover
limLrarr0
π0(X ) = minusC 0(K ) + 2C 0(K )minus C 0(K ) = 0
By analysing the payoff X when L tends to infinity we obtain
limLrarrinfin
π0(X ) = P 0(K ) + C 0(K )
4 To find the limit lim σrarrinfin π0(X ) we observe that
limσ
rarrinfin
d+(S 0 T ) = infin limσ
rarrinfin
dminus(S 0 T ) = minusinfin
so that
limσrarrinfin
N 1048616
d+(S 0 T )1048617
= 1 limσrarrinfin
N 1048616
dminus(S 0 T )1048617
= 0
Hence the price of the call option satisfies for all strikes K isin R+
limσrarrinfin
C 0(K ) = S 0
This in turn implies that lim σrarrinfin π0(X ) = Leminusr(T minust)
2
7212019 MATH_3075_3975_s12
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Exercise 3 (MATH3975) We consider the stock price process S given by
the Black and Scholes model
1 We will first show that S t = eminusrtS t is a martingale with respect tothe filtration F = (F t)tge0 generated by the stock price process S Weobserve that this filtration is also generated by W Using the propertiesof the conditional expectation we obtain for all s le t
EP
1048616S t
F s1048617 = EP
983080S s eσ(W tminusW sminus 1
2σ2(tminuss)) F s
983081= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss) EP1048616eσ(W tminusW s)1048617
where in the last equality we used the independence of increments of the Wiener process Recall also that W t minus W s =
radic tminus s Z where Z sim
N (0 1) and thus
EP
1048616S t | F s
1048617 = S s eminus
1
2σ2(tminuss)
EP
1048616eσradic tminussZ 1048617
Let us finally recall that if Z sim N (0 1) then for any real number a
EP
1048616eaZ
1048617 = e
1
2a2
By setting a = σradic
t
minuss we obtain
EP
1048616S t | S u u le s
1048617 = S s eminus
1
2σ2(tminuss) e
1
2σ2(tminuss) = S s
which shows that S is a martingale under P
2 To compute the expectation EP(S t) we observe that
EP(S t) = ert EP
(S t) = ert EP(S 0) = ertS 0 = ertS 0
To compute the variance VarP(S t) we recall that
Var P(S t) = EP(S 2
t)minus 1048667
EP(S t)
10486692
where in turnEP
(S 2t
) = S 20e2rt EP
983131e2σW tminusσ2t983133
= S 20e2rteσ2tEP
983131e2σW tminus 1
2(2σ)2t
983133
= S 20e2rteσ2tEP
983131eaZ minus
1
2a2
983133= S 20e2rteσ
2t
where we denote a = σradic
t Hence Var P(S t) = S 20e2rt1048616
eσ2t minus 1
1048617
3
7212019 MATH_3075_3975_s12
httpslidepdfcomreaderfullmath30753975s12 44
Exercise 4 (MATH3975) We denote by v the Black-Scholes call option pri-
cing that is the function v R+ times [0 T ] rarr R such that C t = v(S t t) for allt isin [0 T ]
1 We wish to check that the pricing function of the European call optionsatisfies the Black-Scholes PDE
partv
partt +
1
2 σ2s2
part 2v
parts2 + rs
partv
parts minus rv = 0 forall (s t) isin (0infin)times (0 T )
with the terminal condition v(s T ) = (sminusK )+
To this end we use the partial derivatives from Section 55 in the course
notes (note that v(s t) = c(s T minus t))
vs(s t) = N (d+(s T minus t))
vss(s t) = n(d+(s T minus t))
sσradic
T minus t
vt(s t) = minus sσ
2radic
T minus tn(d+(s T minus t))minusKreminusr(T minust)N (dminus(s T minus t))
where n is the density function of the standard normal distributionHence
minus sσ
2radic
T minus tn(d+(s T minus t)) minusKreminusr(T minust)N (dminus(s T minus t))
+ 1
2 σ2s2
n(d+(s T minus t))
sσradic
T minus t+ rsN (d+(s T minus t))minus rv(s t) = 0
where we have also used the equality
v(s t) = sN (d+(s T minus t))minusKeminusr(T minust)N (dminus(s T minus t))
2 We need to show that for every s isin R+
limtrarrT
v(s t) = (sminusK )+
For this purpose we observe that d+(s K ) and dminus(s K ) tend to infin(respectively minusinfin) when t rarr T and s gt K (respectively s lt K )Consequently N (d+(s K )) and N (dminus(s K )) tend to 1 (respectively 0)when t rarr T and s gt K (respectively s lt K ) This in turn impliesthat v(s T ) tends to either sminusK or 0 depending on whether s gt K ors lt K The case when s = K is also easy to analyse
4
7212019 MATH_3075_3975_s12
httpslidepdfcomreaderfullmath30753975s12 34
Exercise 3 (MATH3975) We consider the stock price process S given by
the Black and Scholes model
1 We will first show that S t = eminusrtS t is a martingale with respect tothe filtration F = (F t)tge0 generated by the stock price process S Weobserve that this filtration is also generated by W Using the propertiesof the conditional expectation we obtain for all s le t
EP
1048616S t
F s1048617 = EP
983080S s eσ(W tminusW sminus 1
2σ2(tminuss)) F s
983081= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss)
EP
1048616eσ(W tminusW s) | F s
1048617= S s eminus
1
2σ2(tminuss) EP1048616eσ(W tminusW s)1048617
where in the last equality we used the independence of increments of the Wiener process Recall also that W t minus W s =
radic tminus s Z where Z sim
N (0 1) and thus
EP
1048616S t | F s
1048617 = S s eminus
1
2σ2(tminuss)
EP
1048616eσradic tminussZ 1048617
Let us finally recall that if Z sim N (0 1) then for any real number a
EP
1048616eaZ
1048617 = e
1
2a2
By setting a = σradic
t
minuss we obtain
EP
1048616S t | S u u le s
1048617 = S s eminus
1
2σ2(tminuss) e
1
2σ2(tminuss) = S s
which shows that S is a martingale under P
2 To compute the expectation EP(S t) we observe that
EP(S t) = ert EP
(S t) = ert EP(S 0) = ertS 0 = ertS 0
To compute the variance VarP(S t) we recall that
Var P(S t) = EP(S 2
t)minus 1048667
EP(S t)
10486692
where in turnEP
(S 2t
) = S 20e2rt EP
983131e2σW tminusσ2t983133
= S 20e2rteσ2tEP
983131e2σW tminus 1
2(2σ)2t
983133
= S 20e2rteσ2tEP
983131eaZ minus
1
2a2
983133= S 20e2rteσ
2t
where we denote a = σradic
t Hence Var P(S t) = S 20e2rt1048616
eσ2t minus 1
1048617
3
7212019 MATH_3075_3975_s12
httpslidepdfcomreaderfullmath30753975s12 44
Exercise 4 (MATH3975) We denote by v the Black-Scholes call option pri-
cing that is the function v R+ times [0 T ] rarr R such that C t = v(S t t) for allt isin [0 T ]
1 We wish to check that the pricing function of the European call optionsatisfies the Black-Scholes PDE
partv
partt +
1
2 σ2s2
part 2v
parts2 + rs
partv
parts minus rv = 0 forall (s t) isin (0infin)times (0 T )
with the terminal condition v(s T ) = (sminusK )+
To this end we use the partial derivatives from Section 55 in the course
notes (note that v(s t) = c(s T minus t))
vs(s t) = N (d+(s T minus t))
vss(s t) = n(d+(s T minus t))
sσradic
T minus t
vt(s t) = minus sσ
2radic
T minus tn(d+(s T minus t))minusKreminusr(T minust)N (dminus(s T minus t))
where n is the density function of the standard normal distributionHence
minus sσ
2radic
T minus tn(d+(s T minus t)) minusKreminusr(T minust)N (dminus(s T minus t))
+ 1
2 σ2s2
n(d+(s T minus t))
sσradic
T minus t+ rsN (d+(s T minus t))minus rv(s t) = 0
where we have also used the equality
v(s t) = sN (d+(s T minus t))minusKeminusr(T minust)N (dminus(s T minus t))
2 We need to show that for every s isin R+
limtrarrT
v(s t) = (sminusK )+
For this purpose we observe that d+(s K ) and dminus(s K ) tend to infin(respectively minusinfin) when t rarr T and s gt K (respectively s lt K )Consequently N (d+(s K )) and N (dminus(s K )) tend to 1 (respectively 0)when t rarr T and s gt K (respectively s lt K ) This in turn impliesthat v(s T ) tends to either sminusK or 0 depending on whether s gt K ors lt K The case when s = K is also easy to analyse
4
7212019 MATH_3075_3975_s12
httpslidepdfcomreaderfullmath30753975s12 44
Exercise 4 (MATH3975) We denote by v the Black-Scholes call option pri-
cing that is the function v R+ times [0 T ] rarr R such that C t = v(S t t) for allt isin [0 T ]
1 We wish to check that the pricing function of the European call optionsatisfies the Black-Scholes PDE
partv
partt +
1
2 σ2s2
part 2v
parts2 + rs
partv
parts minus rv = 0 forall (s t) isin (0infin)times (0 T )
with the terminal condition v(s T ) = (sminusK )+
To this end we use the partial derivatives from Section 55 in the course
notes (note that v(s t) = c(s T minus t))
vs(s t) = N (d+(s T minus t))
vss(s t) = n(d+(s T minus t))
sσradic
T minus t
vt(s t) = minus sσ
2radic
T minus tn(d+(s T minus t))minusKreminusr(T minust)N (dminus(s T minus t))
where n is the density function of the standard normal distributionHence
minus sσ
2radic
T minus tn(d+(s T minus t)) minusKreminusr(T minust)N (dminus(s T minus t))
+ 1
2 σ2s2
n(d+(s T minus t))
sσradic
T minus t+ rsN (d+(s T minus t))minus rv(s t) = 0
where we have also used the equality
v(s t) = sN (d+(s T minus t))minusKeminusr(T minust)N (dminus(s T minus t))
2 We need to show that for every s isin R+
limtrarrT
v(s t) = (sminusK )+
For this purpose we observe that d+(s K ) and dminus(s K ) tend to infin(respectively minusinfin) when t rarr T and s gt K (respectively s lt K )Consequently N (d+(s K )) and N (dminus(s K )) tend to 1 (respectively 0)when t rarr T and s gt K (respectively s lt K ) This in turn impliesthat v(s T ) tends to either sminusK or 0 depending on whether s gt K ors lt K The case when s = K is also easy to analyse
4
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