python代写-7CCMFM02T
时间:2020-12-30
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.
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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 S¯T , where S¯t = 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(X¯t ≥ a) = P(τa ≤ t)
so the density fX¯t(a) of X¯t is given by
fX¯t(a) = −
d
da
P(X¯t ≥ a) = − d
da
P(τa ≤ t) = − d
da
∫ t
0
fτa(s)ds
e−rTE((S¯T −K)+) = e−rTE((eX¯T −K)+) = e−rT
∫ ∞
a=0
fX¯T (a)(a−K)+da
assuming X0 = 0.
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