136 F. D’Ippoliti et al.
4 Generating sample paths
Following [5, 6], we now give a Monte Carlo simulation estimator to compute option
price derivatives without discretising processesSandv. The main idea is that, by
appropriately conditioning on the paths generated by the variance and jump processes,
the evolution of the asset price can be represented as a series of log-normal random
variables. This method is calledExact Simulation Algorithm (ESA) for the SVCJ
Model. In Step 3 of this method, the sampling from a cdf is done through an inverse
transform method. Since the inverse function of the cdf is not available in closed form,
the authors apply a Newton method to obtain avalue of the distribution. To avoid the
inversion of the cdf, we use a rejection sampling whose basic idea is to sample from
a known distribution proportional to the real cdf (see [7]). This modification involves
an improvement of efficiency of the algorithm, as the numerical results in Table 2
show.
To price a path-dependent option whose payoff is a function of the asset price
vector(S(t 0 ),...,S(tM))(M=1 for a path-independent option), let 0=t 0 <t 1 <
...<tM=Tbe a partition of the interval [0,T]intoMpossibly unequal segments
of lengthti:=ti−ti− 1 ,fori= 1 ,...,M. Now consider two consecutive time
stepsti− 1 andtion the time grid and assumev(ti− 1 )is known. The algorithm can be
summarised as follows:
Step 1. Generate a Poisson random variable with meanλtiand simulateni,the
number of jumps. Letτi, 1 be the time of the first jump afterti− 1 .Setu:=ti− 1
andt:=τi, 1 (u<t). Ift>ti, skip Steps 5. and 6.
Step 2. Generate a sample from the distribution ofv(t)givenv(u), that is, a non-
central chi-squared distribution.
Step 3. Generate a sample from the distribution of
∫t
uv(q)dqgivenv(u)andv(t):
this is done by writing the conditional characteristic function of the integral
and then the density function. We simulate a value of the integral applying
the rejection sampling.
Step 4. Recover
∫t
u√
v(q)dW 2 (q)givenv(u),v(t)and∫t
uv(q)dq.
Step 5. Ift≤ti, generateJvby sampling from an exponential distribution with mean
jv. Update the variance value by setting ̃v(t)=v(t)+Jv(^1 ),whereJv(^1 )is the
first jump size of the variance.
Step 6. Ift<ti, determine the time of the next jumpτi, 2 afterτi, 1 .Ifτi, 2 ≤ti,set
u:=τi, 1 andt:=τi, 2. Repeat the iteration Steps 2–5. up toti.Ifτi, 2 >ti,
setu:=τi, 1 andt:=ti. Repeat once the iteration Steps 2–4.
Step 7. Define the average variance betweenti− 1 andtias
σ^2 i=niδ^2 S+σS^2 ti
ti, (8)
and an auxiliary variableβi=enilog(^1 +jS)−λjSti−ξ^2
2∫ti
ti− 1 v(q)dq+ξ∫ti
ti− 1√
v(q)dW 2 (q). (9)