Numerical / Monte Carlo Greek Estimation
You are pricing a European call option under geometric Brownian motion and need to estimate the Greeks -- delta and vega -- by Monte Carlo simulation.
- Pathwise derivative estimator for delta. Write down the pathwise (infinitesimal perturbation analysis) estimator for $\Delta = \partial C / \partial S_0$. State the conditions under which this estimator is unbiased.
- Likelihood ratio estimator for vega. Write down the likelihood ratio (score function) estimator for vega, $\mathcal{V} = \partial C / \partial \sigma$. How does its variance compare to the pathwise estimator?
- Variance reduction and confidence intervals. Propose a combined estimator that uses the Black-Scholes closed-form Greek as a control variate. Explain how you would construct a confidence interval for each Greek estimate.
Hints
- Think about two different places you can push the derivative: onto the payoff function, or onto the probability density.
- For the pathwise method, use the chain rule: $\partial f(S_T)/\partial S_0 = f'(S_T) \cdot \partial S_T / \partial S_0$. What does $f'$ look like for a call payoff, and when does this break down?
- For variance reduction, the Black-Scholes closed-form Greek is perfectly correlated with the Monte Carlo Greek estimator on vanilla payoffs -- that makes it an ideal control variate.
Worked Solution
How to Think About It: When you have a closed-form price (like Black-Scholes), you also have closed-form Greeks, so Monte Carlo Greek estimation seems pointless. The reason this problem matters is that in practice, your payoff is rarely vanilla -- path-dependent exotics, barriers, and multi-asset deals all require numerical Greeks. The two main approaches are pathwise derivatives (differentiate through the payoff) and likelihood ratio / score function (differentiate the density). Each has strengths and failure modes, and a good quant knows when to use which.
Key Insight: Pathwise differentiation pushes the derivative inside the expectation onto the payoff function, so it needs the payoff to be differentiable (or at least a.e. differentiable with bounded derivative). The likelihood ratio method pushes the derivative onto the log-density, so it works for discontinuous payoffs but typically has higher variance because you are multiplying by a score function that can be noisy.
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Part (i): Pathwise Derivative Estimator for Delta
Under GBM, the terminal stock price is:
$$S_T = S_0 \exp\left[(r - \tfrac{1}{2}\sigma^2)T + \sigma \sqrt{T}\, Z\right], \quad Z \sim N(0,1)$$
The call payoff is $f(S_T) = (S_T - K)^+$ and the discounted price is $C = e^{-rT} E[f(S_T)]$. To get the pathwise estimator, differentiate inside the expectation:
$$\Delta = e^{-rT} E\left[\frac{\partial f}{\partial S_T} \cdot \frac{\partial S_T}{\partial S_0}\right]$$
Since $S_T$ is linear in $S_0$ up to a multiplicative constant (the exponential does not depend on $S_0$), we have $\partial S_T / \partial S_0 = S_T / S_0$. And $\partial (S_T - K)^+ / \partial S_T = \mathbf{1}_{\{S_T > K\}}$ a.e. So the pathwise estimator is:
$$\hat{\Delta}_{\text{path}} = e^{-rT} \cdot \mathbf{1}_{\{S_T > K\}} \cdot \frac{S_T}{S_0}$$
Average this over $N$ simulated paths to get your Monte Carlo estimate.
Conditions for unbiasedness: The interchange of differentiation and expectation is valid (by dominated convergence) when the payoff function is Lipschitz in $S_T$ -- i.e., the derivative $\partial f / \partial S_T$ exists a.e. and is bounded. For the vanilla call, $(S_T - K)^+$ has a kink at $K$ but is differentiable a.e., and the derivative is bounded by 1, so the pathwise estimator is unbiased. It would fail for a digital (binary) payoff where the derivative is a Dirac delta.
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Part (ii): Likelihood Ratio Estimator for Vega
The likelihood ratio method differentiates the density rather than the payoff. Let $p(S_T; \sigma)$ be the lognormal density. Then:
$$\mathcal{V} = e^{-rT} \int f(S_T) \frac{\partial}{\partial \sigma} p(S_T; \sigma)\, dS_T = e^{-rT} E\left[f(S_T) \cdot \frac{\partial \log p(S_T; \sigma)}{\partial \sigma}\right]$$
The score function with respect to $\sigma$ can be computed from the standard normal representation. Writing $S_T = S_0 \exp[(r - \sigma^2/2)T + \sigma \sqrt{T} Z]$, the log-density contribution from $Z$ is $-Z^2/2 - \tfrac{1}{2}\log(2\pi)$, but we also need the Jacobian. Working through the algebra (or differentiating the lognormal log-density directly), the score is:
$$\frac{\partial \log p}{\partial \sigma} = \frac{(\log(S_T/S_0) - (r - \sigma^2/2)T)^2 - \sigma^2 T}{\sigma^3 T} - \frac{1}{\sigma}$$
Equivalently, in terms of $Z$:
$$\frac{\partial \log p}{\partial \sigma} = \frac{Z^2 - 1}{\sigma} - Z\sqrt{T}$$
So the likelihood ratio estimator for vega is:
$$\hat{\mathcal{V}}_{\text{LR}} = e^{-rT} (S_T - K)^+ \left(\frac{Z^2 - 1}{\sigma} - Z\sqrt{T}\right)$$
Variance comparison: The pathwise estimator for vega (if applicable) would involve differentiating the payoff and the path w.r.t. $\sigma$, giving $\hat{\mathcal{V}}_{\text{path}} = e^{-rT} \mathbf{1}_{\{S_T > K\}} \cdot S_T \cdot (-\sigma T + \sqrt{T}\, Z)$. This has lower variance than the likelihood ratio estimator because it avoids multiplying by the noisy score function. The LR estimator's variance grows because the score term $(Z^2 - 1)/\sigma$ has heavy tails. In general, pathwise estimators have lower variance whenever they are applicable. The LR method's advantage is generality -- it works even when the payoff is discontinuous.
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Part (iii): Control Variate and Confidence Intervals
Since we know the Black-Scholes Greeks in closed form, we can use them as control variates. For delta, define:
$$\hat{\Delta}_{\text{CV}} = \hat{\Delta}_{\text{path}} - \beta\left(\hat{C}_{\text{MC}} - C_{\text{BS}}\right)$$
where $\hat{C}_{\text{MC}}$ is the Monte Carlo price estimate, $C_{\text{BS}}$ is the Black-Scholes price (known), and $\beta$ is the optimal control variate coefficient:
$$\beta^{*} = \frac{\text{Cov}(\hat{\Delta}_{\text{path}},\, \hat{C}_{\text{MC}})}{\text{Var}(\hat{C}_{\text{MC}})}$$
In practice, estimate $\beta^{*}$ from a pilot run or use the full sample. The idea is that the Monte Carlo price error is highly correlated with the Greek estimation error (they use the same paths), so subtracting off the known pricing error removes much of the Greek estimation noise.
Alternatively, you can directly use the closed-form Greek as a control variate. Simulate a vanilla call alongside your exotic, compute pathwise delta for both, and subtract:
$$\hat{\Delta}_{\text{exotic,CV}} = \hat{\Delta}_{\text{exotic}} - \beta(\hat{\Delta}_{\text{vanilla,MC}} - \Delta_{\text{BS}})$$
Confidence interval construction: For $N$ paths, compute the sample mean $\bar{G}$ and sample standard deviation $s_G$ of the Greek estimator (with or without control variate). The $(1-\alpha)$ confidence interval is:
$$\bar{G} \pm z_{\alpha/2} \cdot \frac{s_G}{\sqrt{N}}$$
For small $N$ or when the estimator has heavy tails (common with LR), use a $t$-distribution or increase $N$ until CLT kicks in. The control variate typically reduces $s_G$ by a factor of 3-10x for vanilla-like payoffs, meaning you need 10-100x fewer paths for the same CI width.
Answer: The pathwise delta estimator is $e^{-rT} \mathbf{1}_{\{S_T > K\}} S_T / S_0$ (unbiased when the payoff is Lipschitz). The likelihood ratio vega estimator multiplies the payoff by the score $(Z^2-1)/\sigma - Z\sqrt{T}$; it is more general but higher variance than pathwise. A control variate using the known Black-Scholes Greek dramatically reduces variance, and confidence intervals follow from standard CLT on the sample mean of the per-path Greek estimates.
Intuition
The core tension in Monte Carlo Greeks is differentiating something you can only estimate stochastically. There are two philosophies: differentiate the payoff (pathwise) or differentiate the measure (likelihood ratio). Pathwise is the workhorse -- it is low variance and intuitive (just propagate sensitivities through the simulation, like backpropagation in ML). But it breaks when the payoff has jumps, like digitals or barriers, because you cannot differentiate a step function. The likelihood ratio method handles discontinuities gracefully because it never touches the payoff -- it reweights the probability instead. The price you pay is variance, sometimes a lot of it.
In practice on a derivatives desk, you almost always start with pathwise (or its cousin, adjoint algorithmic differentiation) and only switch to likelihood ratio for specific legs with discontinuous payoffs. The control variate trick -- using Black-Scholes Greeks on a matched vanilla as a baseline -- is not just a textbook exercise. It is standard practice in production Monte Carlo engines. When your exotic is "close" to a vanilla (e.g., a slightly path-dependent option), the correlation between the MC Greek and the vanilla Greek is very high, and the variance reduction can be enormous. This is the same principle behind importance sampling and stratification: use what you know analytically to reduce the work the simulation has to do.