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Consider a Monte Carlo (MC) approximation to a European call with BS parameters $r = 0.05, \sigma = 0.4, T = 10, S_0 = 50$ and $K = 95$. Consider the following results, each using 1M points:

  1. plain MC: $\$21.6901 \pm \$0.1735$
  2. importance sampling with the sampling mean of $S_T = K$: $\$21.7161 \pm \$0.1511$
  3. importance sampling with the sampling median of $S_T = K$: $\$21.8104 \pm \$0.0650$
  4. importance sampling with the sampling mode of $S_T = K$: $\$21.7801 \pm \$0.0210$

where the $\pm$ is a 95% confidence interval. The BS price is $\$21.7766$.

It seems that setting the sampling mode of $S_T$ to $K$ offers the greatest variance reduction, but is this a general rule?

I am actually a bit suspicious of importance sampling because when I use more "reasonable" parameters, importance sampling sometimes increases the variance. Indeed, again with 1M points but using $r = 0.05, \sigma = 0.2, T = 1, S_0 = 50$ and $K = 50$ I get

  1. plain MC: $\$5.2192 \pm \$0.0144$
  2. importance sampling with the sampling mean of $S_T = K$: $\$5.2162 \pm \$0.0197$
  3. importance sampling with the sampling median of $S_T = K$: $\$5.2133 \pm \$0.0173$
  4. importance sampling with the sampling mode of $S_T = K$: $\$5.2207 \pm \$0.0136$

with BS price $\$5.2253$.

For plain vanillas, is there a good rule on how to pick the sampling distribution? (plain vanillas because they are more tractable for me :))

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    $\begingroup$ usually people carry through with the distributional assumption that underlies BS. Some seem to apply exponential change of measure via a cumulant generating function (but I can't comment on it as I have never used it). Have you considered control variate methods and otherwise QMC? $\endgroup$ – Matthias Wolf Jun 16 '15 at 6:03
  • $\begingroup$ I would recommend this paper arxiv.org/abs/math/0702473 . It makes a link between importance sampling and large deviations, and give some applications to finance. $\endgroup$ – lehalle Jun 17 '15 at 6:14
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importance sampling is well known to be tricky. See the extensive discussion in Glasserman's book.

I presume that you are simply meanshifting and multiply by the ratio of normal densities. For this sort of problem, I'd use a more stratified algorithm instead and force every path to end in the money. To do this I'd compute the uniform that goes to the strike and rescale uniforms to be above it. I'd then multiply by the probability of this event to compensate.

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  • $\begingroup$ Would you elaborate a bit more? $\endgroup$ – bcf Jun 14 '16 at 13:11
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Given two representations: $$ C = E_f[\varphi(X)] = \int \varphi(x) f(x)dx = \int \varphi(x) \frac{f(x)}{g(x)}g(x)dx = E_g[\varphi(X)\frac{f(X)}{g(X)}] $$ The difference of the variances of the MC estimators associated with the two expression is $$ Var[\widehat{C}^f_N] - Var[\widehat{C}^g_N] = \frac{1}{N}\int \varphi(x)^2 \left(1 - \frac{f(x)}{g(x)}\right) f(x)dx $$ We want this to be positive for importance sampling to improve estimation.

In your case, you can compute explictly this variance gain by setting $\varphi(x) = S_0(e^x - e^k)_+$ (where $k =\ln(K/S_0e^{rT})$ is the log-moneyness forward) and $f$, $g$ gaussian densities of parameters $(-\sigma^2/2,\sigma^2)$ and $(\mu-\tau^2/2,\tau^2)$.

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