Monte Carlo Pricing and Finite-Difference Greeks for Asian Calls
Summary
The document describes pricing a European-style fixed-strike Asian call whose payoff depends on the discrete arithmetic average of the underlying asset at observation times. Under Black–Scholes assumptions, it simulates risk-neutral geometric Brownian motion across many paths, discounts each path’s payoff, and estimates the option value by averaging those discounted payoffs.
It then explains why estimating higher-order sensitivities can be expensive: a mixed spot-and-volatility Greek, for example, is approximated by repricing at combinations of perturbed spot and volatility inputs. Each repricing requires another simulation, and finer time steps add to the computational burden. The document poses the need for a more efficient approach but does not supply or compare alternatives, nor does it report numerical results. Its setup assumes an arbitrage-free complete market and Black–Scholes dynamics; no variance-reduction or pathwise Greek technique is discussed.
Key ideas
- A discrete arithmetic-average Asian call can be valued by simulating risk-neutral asset paths and averaging discounted payoffs.
- Finite differences estimate mixed Greeks by repricing under combinations of perturbed inputs.
- Repeated simulations for each parameter perturbation make higher-order finite-difference estimates costly.
- The document raises the efficiency problem but does not provide a proposed alternative.
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Full text
# Monte Carlo Greeks for Fixed Strike Asian Call
# Monte Carlo Greeks for Fixed Strike Asian Call
I am interested in pricing an European-style fixed strike asian call with payoff $\max(A(S)-K;0)$, where $A(S)=\frac{1}{n}\sum_{i=1}^nS(t_i)$ is a discrete arithmetic average and $K$ is the strike price.
Assuming an arbitrage-free and complete market, the fundamental theorem of asset pricing tells us that the arbitrage-free price at time $t=0$ is given by: $$ V(0)=E^{\mathbb{Q}}(\max(A(S)-K) \vert {\cal F}_0 ) $$ I have no idea whether there exists an analytic solution or not, so I decided to use MC by implementing the following pseudo-code in python (I omit the code so as not to prolong the question).
Under the Black-Scholes assumptions, let $m$ be the number of paths, $n$ be the number of intervals per path and $\delta t= \frac{T}{n}$, then:
- Simulate geometric brownian motion under $\mathbb{Q}$ measure $S_i(t+1)=S_i(t) \exp \left(\left(r-\frac{\sigma^2}{2}\right)\delta t+\sigma \sqrt{\delta t}Z_t\right)$ where $Z_t \sim{\cal N}(0,1)$ for $i \in [1,m]$ and $t \in [0,n]$.
- Calculate option payoff $X_i=\max(A_i(S_i)-K)$ and set $V_i(0,K,T,\sigma,r,S(0))=e^{-rT}X_i$
- Calculate sample average $V(0,K,T,\sigma,r,S(0))=\frac{1}{m}\sum_{i=1}^m V_i(0,K,T,\sigma,r,S(0))$
However, when I want to calculate higher order greeks, especially those involving mixed derivatives, things get messy because I have to repeat this process several times and change the parameters slightly. For instance, calculating DdeltaDvol using finite differences yields:
\begin{align*} DdeltaDvol &= \frac{1}{4 \Delta S \Delta \sigma} [V(0,K,T,\sigma+\Delta \sigma,r,S(0)+\Delta S)-V(0,K,T,\sigma-\Delta \sigma,r,S(0)+\Delta S) \\ &\quad -V(0,K,T,\sigma+\Delta \sigma,r,S(0)-\Delta S)+V(0,K,T,\sigma-\Delta \sigma,r,S(0)-\Delta S)] \end{align*} The fact that I have to simulate the whole asset path thousands of times and a with small $\delta t$ makes the whole approach computationally intensive.
Does anyone know an alternative approach which is not so computationally intensive ?
Thanks in advanceShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.