Monte Carlo Greeks: Pathwise And Likelihood Ratio

The obvious way to get a Greek from a Monte Carlo pricer is to run it twice: once at today's stock price, once bumped up by a cent, and divide the price difference by the bump. This works, but it is expensive (two full simulations per Greek) and noisy (subtracting two noisy simulated prices can produce an even noisier difference, especially for a small bump). Two other techniques get the same sensitivity out of a single simulation, by differentiating something other than the price itself.

Reading the tape once, two different ways

Imagine measuring how a factory's output changes with conveyor belt speed. One way: run the factory once at the current speed, run it again slightly faster, and compare the two days' output — expensive, and noise in either day pollutes the comparison. A cleverer way: run the factory once, and for each item work out how much faster it would have come out at a slightly different speed, using the belt's own mechanics, then average those per-item sensitivities. Same answer from one run instead of two.

Pathwise derivative

The pathwise method differentiates the payoff itself with respect to the input, path by path, then averages:

$$\Delta \;=\; e^{-rT}\, \mathbb{E}\!\left[\frac{\partial \text{payoff}}{\partial S_T} \cdot \frac{\partial S_T}{\partial S_0}\right].$$

In plain English: for each simulated path, work out how that path's payoff would change if the starting stock price nudged up a tiny bit, using the chain rule through the simulation itself, then average across all paths — no second simulation needed. For a European call with $S_T = S_0 e^{(r-\sigma^2/2)T + \sigma\sqrt{T}Z}$, the derivative $\partial S_T/\partial S_0 = S_T/S_0$, and the payoff derivative is $\mathbb{1}[S_T > K]$ (1 if in the money, 0 otherwise), so pathwise delta is $e^{-rT}\,\mathbb{E}[\mathbb{1}[S_T>K]\cdot S_T/S_0]$, computed directly from the paths already used to price the option.

Likelihood ratio method

The likelihood ratio method instead differentiates the probability density of the outcome, leaving the payoff itself untouched:

$$\Delta \;=\; e^{-rT}\, \mathbb{E}\!\left[\text{payoff}(S_T) \cdot \frac{\partial \ln f(S_T; S_0)}{\partial S_0}\right],$$

where $f$ is the probability density of $S_T$ given $S_0$. In plain English: instead of asking how the payoff changes, ask how likely each outcome was to occur, and how that likelihood shifts with the starting price; weight each path's payoff by that shift and average. Because it never differentiates the payoff, this survives cases where pathwise fails — a digital option's discontinuous payoff, where $\partial\text{payoff}/\partial S_T$ is zero or an unusable spike.

Worked example 1: pathwise delta from three paths

A call, $K=100$, $S_0=100$, $r=0$, $T=1$. Three simulated terminal prices: $S_T = 95, 108, 121$ (from three random draws). Only the second and third finish in the money.

$$\Delta_{\text{pathwise}} = \frac{1}{3}\left[0 \cdot \frac{95}{100} + 1 \cdot \frac{108}{100} + 1 \cdot \frac{121}{100}\right] = \frac{1}{3}\left[0 + 1.08 + 1.21\right] = \frac{2.29}{3} = 0.763.$$

With only three paths this is a rough estimate (a real run uses tens of thousands), but it shows what's being averaged: each in-the-money path contributes $S_T/S_0$; out-of-the-money paths contribute nothing, since a tiny nudge to $S_0$ doesn't change whether they finish underwater.

Worked example 2: why the digital option needs likelihood ratio instead

A digital call pays exactly $1 if $S_T > K$ and $0 otherwise. Its payoff derivative is zero everywhere except exactly at $S_T = K$, where it is technically infinite — pathwise cannot average this, since almost every path contributes exactly zero. Likelihood ratio sidesteps the problem: it never touches the payoff's derivative. Using the same three paths, and a weight $\partial \ln f/\partial S_0 = Z/(S_0 \sigma \sqrt T)$ (with $\sigma=0.2$, weight $Z/20$, for draws $Z = -0.5, 0.8, 1.7$): weights are $-0.025, 0.040, 0.085$. Payoffs are $0, $1, $1.

$$\Delta_{\text{LR}} = \frac{1}{3}\left[0 \times (-0.025) + 1 \times 0.040 + 1 \times 0.085\right] = \frac{0.125}{3} = 0.0417.$$

The payoff's actual value (0 or 1) is used untouched; all the differentiation happens in the weight — exactly why this survives a discontinuous payoff that breaks pathwise.

The paths below are exactly the kind both methods work from: one batch of simulated outcomes, reused to extract a price and every Greek, rather than being thrown away and regenerated for each bumped input.

What this means in practice

Production Monte Carlo engines use pathwise derivatives wherever the payoff is smooth (vanilla and most path-dependent payoffs), and fall back to likelihood ratio for payoffs with kinks or jumps, like digitals and barriers. Getting Greeks from one simulation instead of two roughly halves or better the compute cost of a real-time risk system on a Monte Carlo book.