How does importance sampling reduce variance in Monte Carlo?

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AcadiFi · Accepted Answer

Importance sampling (IS) replaces the original density p(x) with a proposal q(x), then weights samples by p(x)/q(x). The trick: choose q(x) to put probability mass where the integrand matters most. E_p[h(X)] = integral of h(x) * p(x) dx = integral of h(x) * (p(x)/q(x)) * q(x) dx = E_q[h(X) * w(X)] where w(x) = p(x)/q(x). Annika Holmbergsson at Gothenburg Derivative Pricing prices a 10% OTM 3-month put on a diffusive asset. Under risk-neutral p, the payoff is non-zero only in the left tail. Crude MC with 100,000 paths: only ~2,500 paths land in the money, variance is huge. Importance sampling solution: shift drift to make the put ATM under q. Use Brownian motion with drift -mu instead of zero-drift. Now ~50,000 paths contribute. Variance drops 15x; standard error of price drops 4x. Equivalent to running 25x more paths under original measure. ```mermaid flowchart TD A[Crude MC] --> B[Few paths in tail] B --> C[High variance] D[Importance Sampling] --> E[Shift measure to tail] E --> F[Many paths contribute] F --> G[Lower variance] G --> H[Weight by p/q] ``` Optimal IS: q_optimal(x) = |h(x)| * p(x) / E_p[|h(X)|] — zero variance! But requires knowing the answer. In practice, use simple shifts (exponential tilting, drift change) that approximate optimal. Pitfalls: - Weights must have finite second moment: E_q[w(X)^2] < infinity. If q has thinner tails than p, weights explode. - Effective sample size: ESS = (sum w_i)^2 / sum(w_i^2). If ESS<

How does importance sampling reduce variance in Monte Carlo?

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