CS计算机代考程序代写 finance MATH11154 Stochastic Analysis in Finance

MATH11154 Stochastic Analysis in Finance
1. Let X and Y be random variables on a probability space (Ω, F , P ) such
that E|X| < ∞, and let G be σ-algebra contained in F. (a) Define precisely the notion of conditional expectations E(X|G) and E(X|Y ). [6 marks] (b) For a Wiener process (Wt)t≥0 calculate E(Wt|Ws), E(Wt|2Ws) and E(Ws|Wt) for 0 < s ≤ t. [Hint: Note that Wt−Ws is independent of Ws, and find a constant a such that Ws − aWt is independent of Wt.] [9 marks] (c) Suppose that X and Y are random variables such that EX2 < ∞ and EY 2 < ∞. Moreover, suppose that E(X|Y ) = Y (a.s.) Prove X = Y (a.s.). and E(Y |X) = X (a.s.) [ 10 marks] 1 [continued overleaf] MATH11154 Stochastic Analysis in Finance 2. Let W := (Wt)t≥0 be a Wiener process and let (Ft)t≥0 be its history, i.e.,Ft =σ(Ws :0≤s≤t)foreveryt≥0. (a) Show that W, X = (Wt2 − t)t≥0 and Y = (exp(−2t + Wt))t≥0 are martingales with respect to the filtration (Ft)t≥0. [10 marks] (b) Consider the equidistant partition {t0 , t1 , . . . , tn } of the interval [0,T] where tj = jTn, j ∈ I := {0,1,...,n} and n ∈ N. Moreover, consider the stochastic process f(n) := {f(n)} t t≥0 n−1 f(n):=􏰁W 1I (t), such that t tj (tj,tj+1] j=0 where W := {Wt}t≥0 is a Wiener process, and 1IA is the indicator function of a set A. (i) Prove that (ii) By observing that Wt2 −Wt2 =(∆Wtj+1)2+2Wtj∆Wtj+1, j+1 j where ∆Wtj+1 := Wtj+1 − Wtj , prove that n−1 n−1 WT2 = 􏰁(∆Wtj+1 )2 + 2 􏰁 Wtj ∆Wtj+1 lim E n→∞ 0 (ft − Wt) dt = 0. 􏰌T(n) 2 j=0 j=0 [5 marks] You may use without proof that the quadratic variation of W over (iii) Using the above results, prove that 􏰌T12 the interval [0,T] is equal to T. 2 [continued overleaf] WtdWt =2(WT −T). 0 [5 marks] foreveryj∈I, [5 marks] MATH11154 Stochastic Analysis in Finance 3. (a) Consider the stochastic differential equation (SDE): dXt =μXtdt+σdWt, forallt∈[0,T], for T > 0, where {Wt}t≥0 is a Wiener process, μ and X0 are constants.
(i) (ii)
(iii) (iv)
Explain why this stochastic differential equation has a unique
solution.
Show that the process
Xt =X0eμt+σeμt
solves the equation. Calculate EXt.
􏰌t 0
e−μrdWr,
[3 marks]
t∈[0,T]
[4 marks]
[2 marks] Calculate Cov(Xs, Xt) := E(X ̄sX ̄t), where X ̄r = Xr − EXr
for r ∈ [0,T].
τ be a a stopping time with respect to the filtration Ft =
(b) Let
σ(Wr : r ≤ t), t ≥ 0, such that Eτ < ∞. Show that (i) Xt = 1I{t≤τ} is Ft-measurable, (ii) EWτ = 0, [2 marks] [4 marks] [4 marks] (iii) EWτ2 = Eτ. 3 [6 marks] [continued overleaf] MATH11154 Stochastic Analysis in Finance 4. Assume that the price of a stock at time t is St = S0 exp(σWt + αt) forallt≥0,whereS0 >0,σ>0andαareconstants,andW =(Wt)t≥0 is a Wiener martingale with respect to a filtration (Ft)t≥0.
(a) Prove that S is a martingale with respect to (Ft)t≥0 if and only if α = −21σ2. [ 5 marks]
(b) State precisely Girsanov’s theorem. [ 5 marks]
(c) Let T > 0 be a fixed time. Using Girsanov’s theorem show the existence of a probability measure Q such that under Q the process (St)t∈[0, T ] is a martingale with respect to (Ft)t∈[0,T ].
[ 5 marks]
(d) For a given time T > 0 let τ denote the first time when the process
S = (St )t∈[0, T ] achieves its maximum over [0, T ], i.e, τ = inf{t ∈ [0, T] : St = S∗},
where S∗ = maxt∈[0, T ] St. Prove rigorously that τ is not a stopping time with respect to (Ft)t≥0. (You may use that by Doob’s theorem on optional sampling EMτ = EM0 for every martingale (Mt)t∈[0, T ] with respect to a filtration (Ft)t≥0 and stopping times τ ≤ T with respect to the same filtration.)
[ 5 marks]
(e) Consider a standard Black-Scholes market (B, S) with a stock S defined above and a bond B = (ert)t≥0. A cash-or-nothing option in this market pays 1 pound if the stock price exceeds a given value K at the maturity date T, and it pays nothing otherwise. Show that the fair price of this option at time t = 0 is
−rT 􏰅ln(S0/K)+(r−21σ2)T􏰆 eΦ√,
σT
where Φ is the distribution function of a standard normal random
variable.
[ 5 marks]
[continued overleaf]
4

MATH11154 Stochastic Analysis in Finance
5. Consider the standard Black-Scholes model with bond and stock prices,
Bt and St, satisfying
dSt =μStdt+σStdWt, S0 =s>0,
dBt =rBtdt, B0 =1,
for t ≥ 0, where r ≥ 0, σ > 0, s > 0 and μ are constants, and (Wt)t≥0
is a Wiener process.
(a) State precisely the Main Theorem on Pricing European Type
Options with payoff h at expiry date T .
(b) A naive approach to option pricing gives the value N = e−rT Eg(ST )
[5 marks]
for the price C at time t = 0 of European type options with pay-off h = g(ST ) at expiry date T , for non-negative functions g satisfying the linear growth condition.
(i) ShowthatN=Cifμ=r.
(ii) Assume that g is strictly increasing on (0, ∞) i.e., g(x) < g(y) forall0r.
[10 marks]
(c) Consider an option with payoff h = | ln(ST )|2 at time T .
(i) Show that EQh2 < ∞, where Q is the risk neutral measure. (ii) CalculatethepriceC oftheoptionatt=0whenr=σ=2 and S0 = 1. 5 [10 marks] [continued overleaf] MATH11154 Stochastic Analysis in Finance 6. Consider a financial market with a riskless asset of the price Bt = ert, and with two risky assets whose prices S(1) and S(2) satisfy tt dS(1) = rS(1)dt + σ S(1)dW (1), tt1tt dS(2) = rS(2)dt + σ S(2)dW (2), tt2tt where r, σ , σ , S(1) and S(2) are positive constants, and (W(1)) and 120 0 tt≥0 (W(2)) are Wiener processes under a ‘risk neutral measure’ Q, such t t≥0 that W(1) = ρW(2) +􏰑1−ρ2W(3) for some constant ρ ∈ [0,1] and a Wiener process (W(3)) , which is independent of (W(2)) under Q. t t≥0 t t≥0 Our aim is to calculate the price C at t = 0 of an exchange option that gives the holder the right to exchange the asset S(2) for the asset S(1) at expiry date T. This option has the payoff h = 􏰃S(1) −S(2)􏰄+, and by TT a multi-dimensional version of the Main Theorem on Pricing European type options one knows that C = e−rT EQh, where a+ := max(a, 0) for real numbers a. (a) Prove that 􏰒 σ2 (2)􏰃S(1) E h=S(2)erTE e− 2 T+σ2WT T Q0Q S(2) T (b) Use Girsanov’s theorem to prove that 􏰄+􏰓 −1 . 􏰒 σ2 (2)􏰃S(1) 􏰄+􏰓 􏰅S(1) E e−2T+σ2WT T −1 =E 0 e−2T+σWT−1 , σ2 where σ = 􏰑σ12 − 2ρσ1σ2 + σ2 and (Wt)t≥0 is a Wiener process Q S(2) Q S(2) T0 [2 marks] 􏰆+ under Q. (c) Prove that [15 marks] C = S(1)Φ(d ) − S(2)Φ(d ), 0102 where Φ is the probability distribution function of a standard normal random variable, and [8 marks] S(1) 2 ln(0 )+σT S(2)2 √ d1= 0√ ,d2=d1−σT. σT You may use without proof the Black-Scholes European option pricing formula provided that you state it accurately. 6 [End of Paper] MATH11154 Stochastic Analysis in Finance Solutions E(X|G) is a G-measurable random variable Z such that E(1GZ) = E(1GX) for all G ∈ G, and E(X|Y ) = E(X|σ(Y )). [6 marks] E(Wt|Ws) = E(Wt−Ws+Ws|Ws) = E(Wt−Ws|Ws)+E(Ws|Ws) = Ws due to the independence of Wt − Ws and Ws. E(Wt|2Ws) = E(Wt|Ws) = Ws, since σ(2Ws) = σ(Ws). Note that E{(Ws − aWt)Ws} = s − at. Consequently, Ws − st Wt and Wt are independent. Hence E ( W s | W t ) = E ( W s − st W t + st W t | W t ) = st W t . [9 marks] (c) NotethatE(XY)=EE(XY|Y)=EY2. Thus E(X − Y )2 = EX2 − EY 2, and by changing the role of X and Y we get E(X −Y)2 = −E(X −Y)2. Hence E(X − Y )2 = 0, which implies X = Y (a.s.). [10 marks] 2. (a) (i) RecalltheE|Wt|<∞forallt≥0,andfor0≤s≤twehave E(Wt|Fs) = E(Wt −Ws +Ws|Fs) = E(Wt −Ws)+Ws = Ws. [3 marks] (ii) We note first that Wt2 − t is measurable with respect to Ft = σ(Ws : 0 ≤ s ≤ t). Moreover, observe that E|Wt2 −t|≤EWt2 +t=t+t=2t<∞. For 0 ≤ s ≤ t E(Wt2 − t|Fs) = E((Wt − Ws + Ws)2 − t|Fs) =E((Wt −Ws)2 +2(Wt −Ws)Ws +Ws2 −t|Fs) =E(Wt −Ws)2 +2WsE(Wt −Ws)+Ws2 −t =t−s+0+Ws2 −t=Ws2 −s. Therefore (Wt2 − t)t≥0 is a martingale. [3 marks] 7 1. (a) (b) MATH11154 Stochastic Analysis in Finance (iii) We note first that e−2t+Wt is measurable with respect to Ft = σ(Ws : 0 ≤ s ≤ t) since the exponential function is a continuous function. Moreover, observe that t 1􏰌∞tx2 Ee−2+Wt =√ e−2+xe−2tdx 2πt −∞ 1 􏰌 ∞ x2−2xt+t2 =√ e− 2t dx (b) (i) [4 marks] (W −W)2dt tj t E(Wtj − Wt)2 dt (t−tj)dt =lim􏰁 (tj+1−tj)2 n→∞ j=0 2 T2 = lim = 0, Thus we obtain Consequently, (e− 2t +Wt )t≥0 is a martingale. 􏰌 T n−1 􏰌 t limE n→∞ t t n→∞ 0 n→∞ 2n 􏰌T 􏰌T n−1 2πt −∞ 1 􏰌∞ (x−t)2 e− 2t dx=1. 2s222 =√ E(e−t+Wt|F )=e−s+WsEe−t−s+Wt−Ws =e−s+Ws. 2πt −∞ (f(n)−W)2dt= limE n−1 􏰌 t which implies W dW = lim f(n) dW = lim 􏰁W ∆W , 􏰁 j+1 j=0 tj 􏰁 j+1 n→∞ j=0 tj n−1 􏰌 t 􏰁 j+1 = lim = lim n→∞ j=0 tj n−1 1 t t n→∞ t t n→∞ tj tj+1 0 0 j=0 where the limit is taken in mean square sense. [5 marks] (ii) Since W0 = 0, one obtains that n−1 WT2 =􏰁􏰅Wt2 j+1 j=0 n−1 n−1 −Wt2􏰆=􏰁(∆Wtj+1)2 +2􏰁Wtj∆Wtj+1. j 8 j=0 j=0 [5 marks] MATH11154 (ii) 3. (a) (i) Stochastic Analysis in Finance Thus (ii) (iii) (iv) 􏰌t 􏰌t e−μrdWr)dt+σeμtd 􏰌T n−1 Wt dWt = lim 􏰁 Wtj ∆Wtj+1 0 n→∞ j=0 1􏰅n−1 􏰆1 = lim WT2 −􏰁(∆Wtj+1)2 = (WT2 −T). n→∞ 2 j=0 2 [5 marks] The linear functions f (t, x) = μx and g(t, x) = σ satisfy the linear growth and Lipschitz conditions, which ensure that a unique solution exists. [3 marks] Remark: Full mark is given if Itoˆ’s formula is applied to Yt := e−μtXt to obtain d[e−μtXt] = σe−μtdWt and thus Xt = eμtX0 + σeμt for every t ∈ [0,T]. By Itoˆ’s formula dXt =μ(eμtX0+σeμt 􏰌t 0 e−μrdWr, 􏰋 t e−2μr dr < ∞. 0 ⇒ eμtXt = X0 + σ 􏰌t 0 e−μr dWr e−μrdWr =μXtdt+σeμte−μtdWt =μXtdt+σdWt. 00 EXt = eμtX0 + σeμtE since the expectation of stochastic integral is zero due to Wemayassumethats≤t. Thenforμ̸=0 􏰇􏰌s 􏰌t 􏰈􏰌s 1−e−2μs E e−μr dWr eμr dWr = E e−2μr dr = 2μ , 000 9 􏰌t e−μr dWr = eμtX0, [3 marks] 0 [4 marks] MATH11154 (b) (i) (ii) and Stochastic Analysis in Finance ̄ ̄ 2 μ(s+t)1−e−2μs E(XsXt) = σ e 2μ (iii) By Itoˆ’s identity EWτ2 =E [5 marks] 􏰌 ∞ =E Xt2dt=Eτ. . For μ = 0 we have E(X ̄sX ̄t) = σ2s = σ2 min(s, t). [3 marks] Let us define X := 1I (indicator function of the event t t≤τ [t ≤ τ]). Note that [Xt =0]=[τ0,σ>0andαareconstants,andW =(Wt)t≥0 isaWiener martingale with respect to a filtration (Ft)t≥0.
(a) ESt = S0eαtEeσWt = S0eαteσ2t/2 < ∞, and St is Ft-measurable, since it is a (continuous) function of Wt. For 0 ≤ s < t E(St|Fs) = Sseα(t−s)Eeσ(Wt−Ws) = Sseα(t−s)+σ2(t−s)/2, which shows that E(St|Fs) = Ss if and only if α = −21σ2. [5 marks] 10 [5 marks] MATH11154 Stochastic Analysis in Finance (b) Girsanov’s theorem: On a probability space (Ω, F , P ) let (Wt )t∈[0,T ] be a Wiener martingale with respect to a filtration (F)t∈[0,T], and consider a process 􏰌t 0 where (bs)s∈[0,T] ∈ S([0,T]). Define the measure Q, by Q(F)=E(1FγT), F∈F, whereγT =exp(−􏰋TbsdWs−1􏰋Tb2sds).AssumethatEγT =1. 020 Then Q is a probability measure, and under Q the process X is a Wiener martingale with respect to (F)t∈[0,T]. [5 marks] (c) By (a) under a probability measure Q the process St = S0eσW ̃t−σ2t/2, t ∈ [0,T] is a martingale with respect to (Ft)t≥0, if under Q the process W ̃t=Wt+(ασ+σ2)t, t∈[0,T] is a Wiener martingale with respect to (Ft)t≥0, that by virtue of Girsanov’s theorem holds for the measure Q with dQ = γT dP , γT=exp(−θWT−θ2/2), θ:=ασ+σ2, since by (a) we have EγT = 1 satisfied. [5 marks] (d) By (c) there is a probability measure Q such that under Q the process (St)t∈[0,T] is a martingale with respect to (Ft)t∈[0,T]. Assume that τ is a stopping time. Then by Doob’s optional sampling EQ(Sτ − S0) = 0. Hence Sτ = S0 (a.s.), that gives σWt ≤ −α (a.s.) for all t ∈ [0, T ], which contradicts the fact that P(Wt >−α/σ)>0foreveryt>0.
[5 marks]
(e) The pay-off is 1ST>K, and by the Main Theorem on Pricing
European Type options the price at t = 0 of the option is N0 =e−rTEQ(1ST>K),
where Q is the risk neutral probability. Recall that ST = S0e(r−σ2/2)T+σW ̃T ,
11
Xt =
bsds+Wt, t∈[0,T],

MATH11154 Stochastic Analysis in Finance where W ̃T ∼ N(0,T) under Q. Hence
5. (a)
Main Theorem on Pricing European Type Options: Let h be a nonnegative σ(Wr : r ≤ T )-measurable random variable such that EQh < ∞, where Q is the risk neutral measure. Then there is a replicable strategy (ψt, φt)t∈[0,T ] for the European type option with pay-off h at expiry date T, and the price at t ∈ [0,T] of this option is the value of the replicating portfolio at t, that equals e−r(T −t)EQ(h|Ft), whereFt =σ(Ws :s≤t). [5marks] EQ(1ST >K) = P(S0e(r−σ2/2)T+σWT > K)
= P􏰃WT > ln(K/S0)+(σ2/2−r)T 􏰄 = Φ􏰃ln(S0/K)+(r−σ2/2)T 􏰄.
[ 5 marks]
√√√
TσTσT
(b) (i) By Itˆo’s formula we can check that St = S0e(μ−σ2/2)t+σWt) for t ≥ 0. Hence for the discounted stock price S ̃t = e−rtSt we have
S ̃t = S0eσWt+(μ−r−σ2/2)t
Ifμ=r,thenS ̃t =S0eσWt−σ2t/2 fort≥0,thatisaFt- martingale under P. Thus P is a risk neutral measure, and therefore
C = e−rT Eh = N.
(ii) Recall that ST = S0er−σ2/2)T+σW ̃T where W ̃T ∼ N(0,T) under
Q. Thus
C = e−rT EQg(S0e(r−σ2/2)T +σW ̃ T ) = e−rT Eg(S0e(r−σ2/2)T +WT ).
If μ > r then
S0e(r−σ2/2)T +WT < S0e(μ−σ2/2)T +WT for all ω ∈ Ω. Consequently, C = e−rT Eg(S0e(r−σ2/2)T +WT ) < e−rT Eg(S0e(μ−σ2/2)T +WT ) = N. 12 [5 marks] [5nmarks] MATH11154 Stochastic Analysis in Finance (c) (i) Note that ln(ST) = ln(S0)+(r−σ2/2)T +σW ̃T, where W ̃T ∼ 6. (a) Using the notation a+ := max(a, 0) we have (1) (2) + (2) S(1) + N(0,T) under the risk neutral measure Q. Thus EQh2 =E|ln(S0)+(r−σ2/2)T +σWT|4 ≤8|ln(S0)+(r−σ2/2)T|4 +8σ4EWT4 <∞. [5 marks] (ii) IfS0 =1andr=σ=2,thenh=|ln(ST)|2 =σ2W ̃T2,and C = e−rT EQh = e−rT σ2EWT2 = 4e−2T T. [5 marks] E(S−S)=E{S(T −1)} QT T QTS(2) σ2 (2) (1) =S(2)erTE {e− 2 T+σ2WT (ST −1)+}. 0Q S(2) T [2 marks] (b) Introduce a new probability measure Q ̃ by dQ ̃ = γdQ, σ2 (2) γ=e−2T+σ2WT . Then by Girsanov’s theorem the process (W ̃ (2), W (3)) , t t t∈[0,T] W ̃(2) =W(2) −σ t, W ̃(3) =W(3) tt2tt is a two-dimensional Wiener process. Thus (1) (1) (1) S S 2 2 (1) (2) S T = 0 e(σ2−σ1)T+σ1WT −σ2WT = 0 exp(−σ2T/2 + W ̃T ) S(2) S(2) S(2) T00 with σW ̃ := (σ ρ−σ )W ̃ (2)+σ 􏰑1 − ρ2W ̃ (3), T σ2 = σ2−2σ σ ρ+σ2. T12T1 2121 Notice that (W ̃ t)t∈[0,T ] is a Wiener process under Q ̃. Consequently, σ2 (2) (1) E {e− 2 T+σ2WT (ST (1) − 1)+} = E (S0 e−σ2T/2+σWT − 1)+ Q S(2) T0 Q S(2) where (Wt)t≥0 is a Wiener process under Q. [15 marks] 13 MATH11154 (c) Stochastic Analysis in Finance −rT (2) S(1) −σ2T/2+σW + C=e Eh=SE(0 e T−1) Q 0QS(2) 0 = E (S(1)e−σ2/2+σWT − S(2))+, Q00 that is the same as the price of the European Call option in the standard (B, S) market with stock price S = S(1)eσWt−σ2/2, B = t0t 1, strike price K := S(2) at maturity T, when (W ̃ ) is a 0 t t∈[0,T] Wiener process under the probability measure Q. Thus by the Black-Scholes formula C = S(1)Φ(d ) − S(2)Φ(d ), where 0102 S(1) 2 ln(0 )+σT S(2)2 √ d1= 0√ ,d2=d1−σT, σT and Φ is the probability distribution function of a standard normal random variable. [8 marks] 14 [End of Solutions]