7CCMFM02T (CMFM02) King’s College London University Of London This paper is part of an examination of the College counting towards the award of a degree. Examinations are governed by the College Regulations under the authority of the Academic Board. FOLLOW the instructions you have been given on how to upload your solutions MSc Examination 7CCMFM02T (CMFM02) Risk Neutral Valuation January 2021 Time Allowed: Two Hours Full marks will be awarded for complete answers to all FOUR questions. Within a given question, the relative weights of the different parts are indicated by a percentage figure. You may consult lecture notes. 2021 ?King’s College London 7CCMFM02T (CMFM02) 1. a. Consider a two-step standard binomial model with a riskless bond with initial price e?2r?T and a single risky asset S with initial price S0 = 1, with two equal time steps of length ?T = 0.5. Assume that at each time step, S is multiplied by u = 1.1 with probability p, or multiplied by d = .91 with probability 1 ? p and the bond price is multiplied by er?T , and the interest rate r = 0.02. Price an American put option with strike K = 1.01. [40%] b. Let X be a random variable with a Cauchy distribution which has density fX(x) = 1 pi(1+x2) for x ∈ R. Describe how we can simulate X (you may use that ∫ fX(x)dx = 1 pi tan?1(x) where tan?1 is the inverse of the tan function, and that tan(0) = 0). [30%] Solution. ∫ x ?∞ 1 pi(1 + z2) dz = 1 pi tan?1(x)? 1 pi tan?1(?∞) = 1 pi tan?1(x) + 1 2 SetX = F?1X (U), where FX(x) = 1 pi tan?1(x)+1 2 is the distribution function of X, so F?1X (x) = tan(pi(x? 12)). c. Let C(u, v) be a general copula on [0, 1]× [0, 1]. Explain how to use C(.) with two standard uniform random variables to simulate two correlated standard Normal random variables X and Y . Write down an integral expression for the correlation between X and Y . Solution. Let U ,V be two random variables with joint cdf C(u, v). Then from the definition of a copula, U and V are standard Uniform random variables. Then set X = Φ?1(U), Y = Φ?1(V ), and [30%] ρ = E(XY ) = ∫ 1 0 ∫ 1 0 Φ?1(u)Φ?1(u)Cuv(u, v)dudv where Cuv(u, v) is the joint density of U and V . - 2 - See Next Page 7CCMFM02T (CMFM02) 2. a. Let (Wt)t≥0 denote a standard Brownian motion on some probability space (?,F ,P) with filtration Ft satisfying the usual conditions. Which of these statements is true i. Wt is twice differentiable W is not differentiable (see notes) ii. A Ho¨lder continuous random process is continuous Yes, Holder con- tinuity implies continuity, see notes iii. Ws, Wt and Wu have a multivariate normal distribution Yes in gen- eral, Wt1 , ...,Wtn has a n-dimensional multivariate normal distribu- tion with joint pdf 1√ (2pi)ndet(Σ) e? 1 2 xΣ?1x, where Σi,j = E(WtiWtj) = min(ti, tj) iv. The conditional mean of the process Xt = ∫ t 0 (t ? u)H? 12dWu at time s < t is zero because stochastic integrals have zero expecta- tion E(Xt|Fs) = ∫ s 0 (t? u)H? 12dWu which is not zero in general v. |Wt| → ∞ as t→∞ No, since W returns to zero infinitely often vi. If Wt = x > 0, W will return to zero with probability 1 in finite time. [30%] b. Let Ui be a sequence of iid random variables which are ±1 with probabilty 1 2 , and let Sn = ∑n i=1 Ui. Consider the random function Xnt = S[nt]√ n (t ∈ [0, 1]) where [x] denotes the largest integer less than or equal to x. What can we say about Xn in the limit as n→∞? What is the maximum jump size of Xnt ? What is the quadratic variation process of X n t ? What is E(XnsXnt )? [35%] Solution. Xnt process tends to a Brownian motion from Donskers theo- rem. The jump sizes of Xnt are 1√ n . To compute its covariance, we note that XnsX n t = S[ns]√ n S[nt]√ n = 1 n (U1 + ...+ U[ns])(U1 + ...+ U[nt]) Taking expectations of this expression, all cross terms vanish since all Ui’s are independent with zero mean and all diagonal terms are 1, so we get [ns]/n if s ≤ t, hence the covariance is min( [ns] n , [nt] n ) which tends to min(s, t) as n→∞ (i.e. the Covariance of Brownian motion). - 3 - See Next Page 7CCMFM02T (CMFM02) c. Consider the following stock price process: dSt = σ(βSt + 1? β)dWt with S0 > 0, β ∈ (0, 1) and σ > 0. By setting Xt = βSt + 1 ? β and applying Ito’s lemma, compute the exact distribution of St. Can S go negative? [35%] Solution. dXt = βdSt = βσXtdWt . So X is GBM with volatility βσ. Then P(St ≤ S) = P(Xt ≤ βS + 1? β) = P( log(Xt X0 ) + 1 2 β2σ2t βσ √ t ≤ log( βS+1?β X0 ) + 1 2 β2σ2t βσ √ t ) = Φ( log(βS+1?β X0 ) + 1 2 β2σ2t βσ √ t ) For the final part, we note that P(St ≤ 0) = P(Xt ≤ 1? β) > 0 - 4 - See Next Page 7CCMFM02T (CMFM02) 3. Let (Wt)t≥0 denote a standard Brownian motion on some probability space (?,F ,P) with filtration Ft. Consider the Black-Scholes model dSt = St(μdt + σdWt) for a Stock price process S. a. Write down the Black-Scholes PDE and boundary condition for the price of an option which pays 1 at time T if S hits a barrier level B > S0 at any time t ∈ [0, T ], and otherwise pays 1 at time T . What is the no-arbitrage price of this option? [30%] Solution. P satisfies usual Black-Scholes PDE, but with boundary condi- tions P (B, t) = e?r(T?t) and P (S, T ) = 1. Option pays 1 in both scenarios at time T , so its value is just e?r(T?t). b. Assume r = 0, and consider an up-and-out put option with strike K and barrier level B = K > S0 which pays max(K ? ST , 0) at time T if S stays below B for all t ∈ [0, T ], and pays zero otherwise. Show that the no-arbitrage price of this contract at time at time t ∈ [0, T ] is K ? St. What is the vega of this option? [35%] Solution. P (S, t) = K?S satisfies the BS PDE with boundary condition P (B, t) = K ? B = 0 and P (S, T ) = K ? S. Vegas is zero since K ? S does not depend on σ. c. Show that the price of a European call option under the Black-Scholes model is increasing in the maturity T (Hint: use conditional Jensen’s inequality and the tower property). Solution. Let 0 < T1 < T2. Then e?rT2E((ST2 ?K)+) = e?rT2E(E(ST2 ?K)+|ST1) (from the tower property from FM01) ≥ e?rT2E(E(ST2 ?K|ST1)+) (from conditional Jensen inequality with convex function f(S) = (S ?K)+) = e?rT2E((ST1er(T2?T1) ?K)+) = e?rT1E((ST1 ?Ke?r(T2?T1))+) ≥ e?rT1E((ST1 ?K)+) - 5 - See Next Page 7CCMFM02T (CMFM02) where all expectations are under the risk-neutral measure Q. [35%] - 6 - See Next Page 7CCMFM02T (CMFM02) 4. Consider a jump-diffusion model where the log stock price Xt = μt + σWt +∑Nt i=1 ξi, where Wt is standard Brownian motion, the ξi’s are i.i.d N(α, δ 2) ran- dom variables and Nt is a Poisson process with N0 = 0 and arrival rate λ > 0, and W , the ξi’s and Nt are all independent of each other. a. Compute E(epXt). [30%] Solution. V (p) = μp + 1 2 σ2p2 + λ ∫∞ ?∞(e px ? 1)ν(x)dx = μp + 1 2 σ2p2 + λ(E(epξi)?1) = μp+ 1 2 σ2p2 +λ(eαp+ 1 2 δ2p2?1), where ν(x) is the (Normal) jump size density. Then E(epXt) = etV (p). This model is known as the Merton jump diffusion model b. For the Black-Scholes model, compute the no-arbitrage price of an option which pays 1 at the hitting time HB if S hits a barrier level B > S0 at any time t ∈ [0, T ], and otherwise pays 1 at time T . [40%] Solution. No-arbitrage price is E(e?rτa1τa≤T ) + e?rTE(τa > T ) = ∫ T 0 e?rtfτa(t)dt + e ?rT ∫ ∞ T fτa(t)dt where fτa(t) is the hitting time density of Xt?X0 σ to the level a = logB?X0 σ , and Xt = logSt. c. If r ? 1 2 σ2 > 0 for part b), compute the probability that S does not hit B < S0 in finite time, where B > 0. What is the distribution of the ultimate minimum: min0≤u<∞St? [30%] Solution. dSt = St(rdt+ σdWt) and recall that dXt = (r ? 12σ2)dt + σdWt = σ(γdt + dWt), where γ = (r ? 1 2 σ2)/σ. Then Xt → ∞ as t → ∞ and may not ever hit a lower barrier a < X0. Moreover, we note that ?min0≤s≤t(Xs ?X0) ～ max0≤s≤t(X?s ? X?0) where X?t = σ(?γt+Wt) and recall that γ > 0. Then S∞ := min0≤u<∞Su and P(S∞ < B) = P(log S∞ S0 < a) = P(X∞ ?X0 < a) = P( ˉ?X∞ > ?a) = e? 2|γ| σ2 |a| where a = log B S0 < 0. - 7 - Final Page 學霸聯盟:"http://aessay.org"