python代写-7CCMFM02T

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.
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.
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 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
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.
- 3 - Final Page  