How Higher Volatility Increases Monte Carlo Option Pricing Error
Summary
The document explains why Monte Carlo estimates of a European option price can become less precise as volatility rises. The central point is that the estimate averages simulated payoffs, and its sampling error depends on payoff variance and the number of simulations. Greater volatility spreads the terminal asset price distribution, which can make option payoffs more variable and widen confidence intervals.
It relates this effect to the exponential growth of terminal-price variance in the stated model and notes that piecewise-linear payoffs may inherit substantial variance growth. A further answer illustrates why outcomes with very different payoff sizes require especially accurate estimates of the probability assigned to large payoffs. The discussion is qualitative: it gives no convergence study or variance-reduction comparison, and the claimed growth rate for the option payoff depends on the payoff and model assumptions. It does not establish that every option price's error grows exponentially at the same rate.
Key ideas
- Monte Carlo pricing error is governed by the variance of the simulated payoff and the number of simulations.
- Higher volatility broadens terminal-price outcomes and can increase payoff variance.
- For the stated model, terminal asset-price variance grows exponentially with volatility squared times maturity.
- Payoffs with rare, very large outcomes can require more precise estimation of their probabilities.
- The exact error behavior depends on the payoff structure and model assumptions.
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Full text
# Monte Carlo simulations with extremely high volatility
# Monte Carlo simulations with extremely high volatility
I am using monte Carlo simulations to price a forex option. This is a standard model and works very well with less than 1 % error from black scholes price for 10000 simulations. But, as I increase volatility, this error increases . Also, at extremely high volatility, like 100 percent and more, the error terms increase exponentially. Why does this happen?
## Answer by Johnny Iwash (score 2)
https://quant.stackexchange.com/a/79993
If you run $n$ simulations of your (suppose unbiased) estimator $X$ and take the average then the Monte-Carlo error is \begin{equation} \frac{1}{n}\sum_{i=1}^n X_{i}-\mathbb{E}(X)\xrightarrow{n\mapsto \infty}\mathcal{N}\left(0,\frac{{\rm Var}(X)}{n}\right) \end{equation} distributed in the limit, due to the Central Limit Theorem. If you augment the volatility of the model, then the random variable $S_T$ has more variance, and $X=f(S_T)$ too, meaning your Monte-Carlo error increments in variance. The map \begin{equation} \sigma \mapsto {\rm Var}(S^\sigma_T)=C(e^{\sigma^2 T}-1) \end{equation} is of exponential growth just see this, and if $X=f(S_T)$ is piecewise linear, then you can expect an exponential blow-up in the variance and the confidence intervals of your Monte-Carlo solution.
## Answer by KaiSqDist (score 1)
https://quant.stackexchange.com/a/79947
As the volatility increases, the spectrum of possible outcomes for $S_T$ increases (due to a more "choppy" evolution that fluctuates with a wider "up and down" range as well). As the output of the Monte Carlo simulation is an average of the intrinsic values at maturity discounted at the riskless rate to the current time, the same number of terminal intrinsic values (across a wider distribution) gives rise to a larger potential deviation from the true (Black-Scholes) price of an European option.
## Answer by Arshdeep (score 0)
https://quant.stackexchange.com/a/79950
When payoff is close [1,1.01,1.001,1.0001], you can actually use any probability distribution to find it's mean, the mean is going to be 1. Infact you don't even want to know the distribution of the underlying to evaluate this, it's going to be 1.
When payoff is apart [10,100,1000,10000],
you have to be very careful how much weight is put on each, as error will be quite large. You want to know the distribution exactly here.
Infact you will want to measure the mass at 10000 at 10 times more precision than mass of everything else.Shown 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.