Monte Carlo simulations with extremely high volatility

External question — Quantitative Finance Stack Exchange Author: vishal saharan Original post: https://quant.stackexchange.com/questions/79946 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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?

External answer — Quantitative Finance Stack Exchange Author: KaiSqDist Original post: https://quant.stackexchange.com/a/79947 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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.

External answer — Quantitative Finance Stack Exchange Author: Arshdeep Original post: https://quant.stackexchange.com/a/79950 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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.

External answer — Quantitative Finance Stack Exchange Author: Johnny Iwash Original post: https://quant.stackexchange.com/a/79993 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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.