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. Construct a n-dimensional Gaussian copula with correlation matrix ρij and show how we can use this simulate n correlated standard Exponential random variables E1, ..., En. Solution. Recall that the joint density of n Normal random variables X1, ..., Xn is 1√ (2pi)ndet(Σ) e? 1 2 xTΣ?1x, where here Σi,j = E(XiXj), but in this case here the Zi’s will have variance of 1, so Σi,j = ρij = E(ZiZj). Σ is assumed to be given here, but has to positive definite to ensure that the Var of c1Z1 + ... + cnZn is non-negative for any vector c = (c1, .., cn) with ci ∈ R (this variance is given by cTΣc). We know from the Applied Probability Revision notes that Ui := Φ(Zi) ～ U([0, 1]). We then set Ei = F ?1(Ui) where F (x) = 1 ? e?x is the distribution function of an Exp(1) random variable. - 2 - See Next Page 7CCMFM02T (CMFM02) 2. a. Consider a jump-diffusion model for a log stock price process Xt as in the lecture notes with negative only jumps. Explain how we can price a call option on SˉT , where Sˉt = max0≤u≤tSu is the running maximum process of S (example of a lookback option). [30%] Solution. For pricing options, recall we first have to choose μ so that E(St) = E(eXt) = eV (1)t = ert (so as to make the discounted stock price a martingale) so we must impose that V (1) = r. To see this note that E(e?rtSt|Fs) = e?rtE(eXt |Fs) = e?rtE(e(Xs+Xt?Xs)|Fs) = e?rteXsE(eXt?Xs|Fs) = e?rteXsE(eXt?Xs) = e?rteXsE(eXt?s) = Sse ?rteV (1)(t?s) = e?rsSs where we have used that X has independent increments i.e. Xt ? Xs is independent of (Xu)0≤u≤s, and X is a stationary process, i.e. Xt?Xs has the same distribution as Xt?s. From the notes we also know that E(e?qτa) = e?aΦ(q) for q > 0, where Φ(q) is the largest inverse of V (p) = logE(epX1) = 1 t logE(epXt). To compute the density of τa, recall that we set ?q = ik for k ∈ R, and compute the inverse Fourier transform of E(eikτa) as fτa(t) = 1 2pi ∫ ∞ k=?∞ e?ikt E(eikτa)dk = 1 2pi ∫ ∞ k=?∞ e?ikte?aΦ(?ik)dk . Then P(Xˉt ≥ a) = P(τa ≤ t) so the density fXˉt(a) of Xˉt is given by fXˉt(a) = ? d da P(Xˉt ≥ a) = ? d da P(τa ≤ t) = ? d da ∫ t 0 fτa(s)ds e?rTE((SˉT ?K)+) = e?rTE((eXˉT ?K)+) = e?rT ∫ ∞ a=0 fXˉT (a)(a?K)+da assuming X0 = 0. - 3 - Final Page 學霸聯盟:"http://aessay.org"