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7CCMFM02T (CMFM02) Risk Neutral Valuation
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.
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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
for x ∈ R. Describe how we can simulate X (you may
tan−1(x) where tan−1 is the inverse of the tan
function, and that tan(0) = 0). [30%]
pi(1 + z2)
tan−1(−∞) = 1
SetX = F−1X (U), where FX(x) =
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 ) =
where Cuv(u, v) is the joint density of U and V .
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2. a. Let (Wt)t≥0 denote a standard Brownian motion on some probability space
(Ω,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√
xΣ−1x, where Σi,j = E(WtiWtj) =
iv. The conditional mean of the process Xt =
(t − u)H− 12dWu at
time s < t is zero because stochastic integrals have zero expecta-
tion E(Xt|Fs) =
(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.
b. Let Ui be a sequence of iid random variables which are ±1 with probabilty
, and let Sn =
i=1 Ui. Consider the random function
(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
t ? What is E(XnsXnt )?
Solution. Xnt process tends to a Brownian motion from Donskers theo-
rem. The jump sizes of Xnt are
. To compute its covariance, we note
(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]
) which tends to
min(s, t) as n→∞ (i.e. the Covariance of Brownian motion).
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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
dXt = βdSt = βσXtdWt .
So X is GBM with volatility βσ. Then
P(St ≤ S) = P(Xt ≤ βS + 1− β)
) + 1
) + 1
) = Φ(
) + 1
For the final part, we note that
P(St ≤ 0) = P(Xt ≤ 1− β) > 0
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3. Let (Wt)t≥0 denote a standard Brownian motion on some probability space
(Ω,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(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)+)
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where all expectations are under the risk-neutral measure Q.
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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
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
σ2p2 + λ
px − 1)ν(x)dx = µp + 1
λ(E(epξi)−1) = µp+ 1
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 ) =
e−rtfτa(t)dt + e
where fτa(t) is the hitting time density of
to the level a = logB−X0
and Xt = logSt.
c. If r − 1
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%]
dSt = St(rdt+ σdWt)
and recall that dXt = (r − 12σ2)dt + σdWt = σ(γdt + dWt), where γ =
(r − 1
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
P(S∞ < B) = P(log
< a) = P(X∞ −X0 < a) = P( ¯˜X∞ > −a) = e−
where a = log B
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