Likelihood Ratio and Malliavin-Type Methods

Likelihood ratio methods move derivatives from the payoff to the distribution, enabling Greeks for nonsmooth payoffs.

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Score identity

If \(X^\theta\) has density \(p_\theta\),

\[ V(\theta)=\mathbb{E}_\theta[\Phi(X^\theta)] =\int \Phi(x)p_\theta(x)\,\mathrm{d}x, \]

then

\[ \boxed{ V'(\theta)=\mathbb{E}_\theta\!\left[\Phi(X^\theta)\frac{\partial}{\partial \theta}\log p_\theta(X^\theta)\right] } \]

This avoids \(\Phi'\). The quantity \(\frac{\partial}{\partial\theta}\log p_\theta\) is called the score function.

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Black–Scholes: Explicit LR estimators

In Black–Scholes, \(\log S_T = \log S + (r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z\) where \(Z \sim \mathcal{N}(0,1)\).

The density of \(Z\) is \(p(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\), so \(\frac{\partial}{\partial z}\log p(z) = -z\).

Delta via LR. Treating \(S\) as a location parameter:

\[ \boxed{\Delta = \mathbb{E}\!\left[e^{-r\tau}\Phi(S_T) \cdot \frac{Z}{S\sigma\sqrt{\tau}}\right]} \]

where \(Z = \frac{\log(S_T/S) - (r-\frac12\sigma^2)\tau}{\sigma\sqrt{\tau}}\) is the driving standard normal.

Verification. For a call, this yields \(\Delta = N(d_1)\) after evaluation.

Vega via LR. The volatility appears in both the drift and diffusion:

\[ \boxed{\nu = \mathbb{E}\!\left[e^{-r\tau}\Phi(S_T) \cdot \frac{Z^2 - 1 - Z\sigma\sqrt{\tau}}{\sigma}\right]} \]

Alternatively:

\[ \nu = \mathbb{E}\!\left[e^{-r\tau}\Phi(S_T) \cdot \left(\frac{Zd_1}{\sigma} - \sqrt{\tau}\right)\right] \]

Gamma via LR. The LR gamma estimator involves the second derivative of the log-density:

\[ \boxed{\Gamma = \mathbb{E}\!\left[e^{-r\tau}\Phi(S_T) \cdot \frac{Z^2 - Z\sigma\sqrt{\tau} - 1}{S^2\sigma^2\tau}\right]} \]

This formula works even for digital options where pathwise gamma is undefined.

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Variance comparison

LR estimators often have higher variance than pathwise estimators:

Greek	Pathwise variance	LR variance	Ratio
Delta (smooth)	\(\mathcal{O}(1)\)	\(\mathcal{O}(1/\tau)\)	LR worse for short \(\tau\)
Vega (smooth)	\(\mathcal{O}(\tau)\)	\(\mathcal{O}(1)\)	Comparable
Gamma (kinked)	N/A	\(\mathcal{O}(1/\tau)\)	Only LR available

Variance reduction. Control variates and antithetic sampling can significantly reduce LR variance.

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Malliavin calculus perspective (conceptual)

For diffusions driven by Brownian motion, Malliavin calculus provides a systematic framework for LR-type formulas.

Key objects: - Malliavin derivative \(D_s X_T\): sensitivity of \(X_T\) to perturbations of \(W_s\) - Skorokhod integral \(\delta(\cdot)\): adjoint of \(D\) - Malliavin covariance \(\langle DX_T, DX_T \rangle_{L^2[0,T]}\)

Integration by parts formula. For \(F = \Phi(X_T)\) and suitable \(u\):

\[ \mathbb{E}[\Phi'(X_T) \cdot u] = \mathbb{E}[\Phi(X_T) \cdot H(u)] \]

where \(H(u)\) is a Malliavin weight involving Skorokhod integrals and the inverse Malliavin covariance.

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Explicit Malliavin weights for Black–Scholes

Delta weight:

\[ H_\Delta = \frac{W_T - W_t}{S\sigma\tau} = \frac{\sqrt{\tau}Z}{S\sigma\tau} = \frac{Z}{S\sigma\sqrt{\tau}} \]

This recovers the LR delta formula.

Gamma weight: The second-order Malliavin weight is:

\[ \boxed{H_\Gamma = \frac{1}{S^2\sigma^2\tau}\left(H_2(Z) - \frac{Z}{\sigma\sqrt{\tau}}\right)} \]

where \(H_2(z) = z^2 - 1\) is the second Hermite polynomial. This gives:

\[ \Gamma = \mathbb{E}\!\left[e^{-r\tau}\Phi(S_T) \cdot H_\Gamma\right] \]

Vega weight:

\[ H_\nu = \frac{Z^2 - 1}{\sigma} - \sqrt{\tau}Z \]

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Localization and truncation

Near maturity, LR weights can have heavy tails (e.g., \(Z/\sqrt{\tau}\) blows up). Practical implementations use:

1. Truncation: Cap \(|Z|\) at some threshold
2. Localization: Use pathwise methods away from kinks, LR near kinks
3. Smoothing: Replace digital payoffs with smooth approximations

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Applications beyond vanilla options

LR/Malliavin methods are essential for:

• Digital options: Pathwise delta is undefined
• Barrier options: Delta involves hitting probabilities
• Asian options: Multiple sources of non-smoothness
• Path-dependent exotics: Complex dependence on trajectory

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What to remember

• LR/Malliavin methods handle kinked payoffs well by moving derivatives to the density.
• Explicit LR formulas exist for all Black–Scholes Greeks.
• The cost is potentially high variance from heavy-tailed weights, especially for short maturities.
• Malliavin calculus provides the theoretical foundation; Hermite polynomials appear in higher-order weights.
• Practical implementations combine pathwise (away from kinks) with LR (near kinks).

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Exercises

Exercise 1. For the Black--Scholes model with \(S_T = S\exp((r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z)\), verify the LR delta formula \(\Delta = \mathbb{E}[e^{-r\tau}\Phi(S_T) \cdot Z/(S\sigma\sqrt{\tau})]\) by applying it to a European call \(\Phi(x) = (x-K)^+\) and showing it yields \(N(d_1)\).

Solution to Exercise 1

We have \(S_T = S\exp\!\left((r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z\right)\) with \(Z \sim \mathcal{N}(0,1)\). The LR delta formula is

\[ \Delta = \mathbb{E}\!\left[e^{-r\tau}(S_T - K)^+ \cdot \frac{Z}{S\sigma\sqrt{\tau}}\right] \]

The condition \(S_T > K\) is equivalent to \(Z > -d_2\) where \(d_2 = \frac{\ln(S/K) + (r - \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\). So:

\[ \Delta = \frac{e^{-r\tau}}{S\sigma\sqrt{\tau}}\int_{-d_2}^{\infty}(S e^{(r-\frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}z} - K)\,z\,\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz \]

Split into two integrals:

\[ I_1 = \frac{e^{-r\tau}}{S\sigma\sqrt{\tau}}\int_{-d_2}^{\infty} S e^{(r-\frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}z}\,z\,\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz \]

Completing the square: \((r - \frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}z - \frac{z^2}{2} = r\tau - \frac{1}{2}(z - \sigma\sqrt{\tau})^2\). Substituting \(u = z - \sigma\sqrt{\tau}\):

\[ I_1 = \frac{1}{\sigma\sqrt{\tau}}\int_{-d_1}^{\infty}(u + \sigma\sqrt{\tau})\frac{e^{-u^2/2}}{\sqrt{2\pi}}\,du = \frac{1}{\sigma\sqrt{\tau}}\!\left[N'(d_1) + \sigma\sqrt{\tau}\,N(d_1)\right] \]

using \(\int_{-d_1}^{\infty} u\,\phi(u)\,du = \phi(d_1) = N'(d_1)\) and \(\int_{-d_1}^{\infty}\phi(u)\,du = N(d_1)\).

For the second integral:

\[ I_2 = -\frac{Ke^{-r\tau}}{S\sigma\sqrt{\tau}}\int_{-d_2}^{\infty}z\,\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz = -\frac{Ke^{-r\tau}}{S\sigma\sqrt{\tau}}N'(d_2) \]

Since \(N'(d_2) = N'(d_1) \cdot \frac{S}{Ke^{-r\tau}}\) (which follows from \(d_1 = d_2 + \sigma\sqrt{\tau}\) and the identity \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\)), we get \(I_2 = -\frac{N'(d_1)}{\sigma\sqrt{\tau}}\).

Combining: \(\Delta = I_1 + I_2 = \frac{N'(d_1)}{\sigma\sqrt{\tau}} + N(d_1) - \frac{N'(d_1)}{\sigma\sqrt{\tau}} = N(d_1)\).

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Exercise 2. The LR vega estimator involves the weight \((Z^2 - 1 - Z\sigma\sqrt{\tau})/\sigma\). Compute the variance of this weight for \(\sigma = 0.20\) and \(\tau = 0.25\). Compare it to the variance of the pathwise vega weight \(S_T(\sqrt{\tau}Z - \sigma\tau)\). Which has higher variance?

Solution to Exercise 2

The LR vega weight is \(w_{\text{LR}} = \frac{Z^2 - 1 - Z\sigma\sqrt{\tau}}{\sigma}\).

Since \(Z \sim \mathcal{N}(0,1)\), we compute each moment:

• \(\mathbb{E}[Z^2] = 1\), \(\mathbb{E}[Z^4] = 3\), \(\mathbb{E}[Z^3] = 0\), \(\mathbb{E}[Z] = 0\)

The variance is \(\text{Var}(w_{\text{LR}}) = \mathbb{E}[w_{\text{LR}}^2] - (\mathbb{E}[w_{\text{LR}}])^2\).

First, \(\mathbb{E}[w_{\text{LR}}] = \frac{\mathbb{E}[Z^2] - 1 - \sigma\sqrt{\tau}\mathbb{E}[Z]}{\sigma} = \frac{1 - 1 - 0}{\sigma} = 0\).

Next, \(w_{\text{LR}}^2 = \frac{(Z^2-1)^2 - 2(Z^2-1)Z\sigma\sqrt{\tau} + Z^2\sigma^2\tau}{\sigma^2}\). Taking expectations:

• \(\mathbb{E}[(Z^2-1)^2] = \mathbb{E}[Z^4 - 2Z^2 + 1] = 3 - 2 + 1 = 2\)
• \(\mathbb{E}[(Z^2-1)Z] = \mathbb{E}[Z^3 - Z] = 0\)
• \(\mathbb{E}[Z^2] = 1\)

So \(\text{Var}(w_{\text{LR}}) = \frac{2 + \sigma^2\tau}{\sigma^2} = \frac{2}{0.04} + 0.25 = 50 + 0.25 = 50.25\).

For the pathwise vega weight \(w_{\text{PW}} = S_T(\sqrt{\tau}Z - \sigma\tau)\), the variance depends on the joint distribution of \(S_T\) and \(Z\). Since \(S_T = Se^{(r-\frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z}\):

\[ \text{Var}(w_{\text{PW}}) = S^2\,\text{Var}\!\left(e^{(r-\frac{1}{2}\sigma^2)\tau + \sigma\sqrt{\tau}Z}(\sqrt{\tau}Z - \sigma\tau)\right) \]

This requires computing \(\mathbb{E}[e^{2\sigma\sqrt{\tau}Z}(\sqrt{\tau}Z - \sigma\tau)^2]\) and similar terms. Using moment generating function techniques with \(S=100\), \(\sigma=0.20\), \(\tau=0.25\), \(r=0.05\):

The pathwise weight has variance of order \(S^2\tau \cdot e^{\sigma^2\tau} \approx 10000 \times 0.25 \times 1.01 \approx 2525\).

The LR weight variance is \(50.25\) (dimensionless), but it multiplies \(e^{-2r\tau}\mathbb{E}[\Phi(S_T)^2]\) in the full estimator variance. For comparable payoff scales, the LR estimator often has higher variance than pathwise because of the \(2/\sigma^2 = 50\) term, which dominates for small \(\sigma\).

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Exercise 3. The Malliavin delta weight is \(H_\Delta = Z/(S\sigma\sqrt{\tau})\). For \(\tau = 1/252\) (one day), \(\sigma = 0.20\), \(S = 100\), compute \(\mathbb{E}[H_\Delta^2]\) and explain why LR estimators have high variance for short maturities.

Solution to Exercise 3

The Malliavin delta weight is \(H_\Delta = \frac{Z}{S\sigma\sqrt{\tau}}\), so

\[ H_\Delta^2 = \frac{Z^2}{S^2\sigma^2\tau} \]

Since \(Z \sim \mathcal{N}(0,1)\), \(\mathbb{E}[Z^2] = 1\). Therefore:

\[ \mathbb{E}[H_\Delta^2] = \frac{1}{S^2\sigma^2\tau} \]

Substituting \(S = 100\), \(\sigma = 0.20\), \(\tau = 1/252\):

\[ \mathbb{E}[H_\Delta^2] = \frac{1}{100^2 \times 0.04 \times (1/252)} = \frac{1}{10000 \times 0.04/252} = \frac{252}{400} = 0.63 \]

While \(\mathbb{E}[H_\Delta^2] = 0.63\) may seem moderate, the issue is the \(1/\tau\) scaling. The variance of the full LR delta estimator is

\[ \text{Var}\!\left(e^{-r\tau}\Phi(S_T) H_\Delta\right) \geq \text{Var}(H_\Delta) \cdot (\text{typical payoff scale})^2 \]

and \(\text{Var}(H_\Delta) = \frac{1}{S^2\sigma^2\tau} \propto \frac{1}{\tau}\). As \(\tau \to 0\):

• The weight \(H_\Delta = Z/(S\sigma\sqrt{\tau})\) has standard deviation \(1/(S\sigma\sqrt{\tau})\), which blows up as \(\sqrt{252} \approx 15.87\) for one-day maturity compared to \(1\) for one-year maturity.
• Individual sample contributions \(\Phi(S_T) \cdot H_\Delta\) can be very large in magnitude, causing the Monte Carlo average to converge slowly.
• The number of paths needed for a given accuracy scales as \(O(1/\tau)\), making LR estimation impractical for very short maturities without variance reduction.

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Exercise 4. Consider a digital call with payoff \(\Phi(S_T) = \mathbf{1}_{S_T > K}\). Explain why pathwise differentiation fails for this payoff (what is \(\Phi'\)?), while the LR method produces a valid estimator. Write down the LR delta estimator for the digital call.

Solution to Exercise 4

The digital call payoff is \(\Phi(S_T) = \mathbf{1}_{S_T > K}\).

Why pathwise differentiation fails: The pathwise delta formula requires \(\Phi'(S_T)\). For the digital payoff, \(\Phi\) is discontinuous at \(S_T = K\):

\[ \Phi'(x) = \delta(x - K) \quad \text{(Dirac delta, in the distributional sense)} \]

This is not a function, and \(\Phi\) is not Lipschitz continuous (the essential requirement for pathwise methods). The dominated convergence theorem fails because the difference quotients \(\frac{\Phi(x+h) - \Phi(x)}{h}\) do not converge to an integrable function near \(x = K\): they oscillate wildly (being \(\pm 1/h\) in a neighborhood of \(K\)).

The LR method produces a valid estimator because it avoids differentiating \(\Phi\) entirely. Using the score function \(\frac{Z}{S\sigma\sqrt{\tau}}\):

\[ \Delta_{\text{digital}} = \mathbb{E}\!\left[e^{-r\tau}\mathbf{1}_{S_T > K} \cdot \frac{Z}{S\sigma\sqrt{\tau}}\right] \]

This is well-defined because \(\mathbf{1}_{S_T > K}\) is bounded and \(Z/(S\sigma\sqrt{\tau})\) has finite moments. The estimator involves only the payoff itself (not its derivative) multiplied by the score weight.

Expanding, with \(Z = \frac{\log(S_T/S) - (r-\frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\):

\[ \Delta_{\text{digital}} = \frac{e^{-r\tau}}{S\sigma\sqrt{\tau}}\mathbb{E}\!\left[\mathbf{1}_{S_T > K} \cdot Z\right] = \frac{e^{-r\tau}}{S\sigma\sqrt{\tau}}\int_{-d_2}^{\infty} z\,\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz = \frac{e^{-r\tau} N'(d_2)}{S\sigma\sqrt{\tau}} \]

This equals the known digital delta \(\frac{e^{-r\tau}N'(d_2)}{S\sigma\sqrt{\tau}}\).

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Exercise 5. The Malliavin gamma weight involves the second Hermite polynomial \(H_2(z) = z^2 - 1\). Show that \(\mathbb{E}[H_2(Z)] = 0\) and \(\mathbb{E}[H_2(Z)^2] = 2\) for \(Z \sim \mathcal{N}(0,1)\). Why does the appearance of Hermite polynomials in higher-order weights increase the variance of the estimator?

Solution to Exercise 5

Let \(Z \sim \mathcal{N}(0,1)\) and \(H_2(z) = z^2 - 1\).

Show \(\mathbb{E}[H_2(Z)] = 0\):

\[ \mathbb{E}[H_2(Z)] = \mathbb{E}[Z^2 - 1] = \mathbb{E}[Z^2] - 1 = 1 - 1 = 0 \]

Show \(\mathbb{E}[H_2(Z)^2] = 2\):

\[ \mathbb{E}[H_2(Z)^2] = \mathbb{E}[(Z^2 - 1)^2] = \mathbb{E}[Z^4 - 2Z^2 + 1] \]

Using \(\mathbb{E}[Z^4] = 3\) (the fourth moment of a standard normal):

\[ \mathbb{E}[H_2(Z)^2] = 3 - 2(1) + 1 = 2 \]

These results are special cases of the orthogonality relation for Hermite polynomials: \(\mathbb{E}[H_m(Z)H_n(Z)] = n!\,\delta_{mn}\). For \(m = n = 2\): \(\mathbb{E}[H_2(Z)^2] = 2! = 2\). For \(m = 2, n = 0\): \(\mathbb{E}[H_2(Z) \cdot 1] = 0\).

Why Hermite polynomials increase variance of higher-order estimators:

The \(n\)-th order Malliavin weight involves \(H_n(Z)\), whose variance is \(n!\). Since \(n!\) grows rapidly:

• \(\text{Var}(H_1) = 1! = 1\) (delta weights)
• \(\text{Var}(H_2) = 2! = 2\) (gamma weights)
• \(\text{Var}(H_3) = 3! = 6\) (third-order speed/color weights)
• \(\text{Var}(H_n) = n!\) (general \(n\)-th order)

The factorial growth means that LR/Malliavin estimators for higher-order Greeks have inherently higher variance. Moreover, the weights are multiplied by \(1/(S^n\sigma^n\tau^{n/2})\) factors, compounding the variance issue. This is why gamma LR estimators require significantly more paths than delta estimators, and why third-order Greeks via LR methods are rarely practical without aggressive variance reduction.

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Exercise 6. A practitioner wants to estimate gamma for a barrier option near the barrier. She has access to both finite-difference and LR methods. The finite-difference method uses step \(h = 1\), and the LR method uses \(N = 100{,}000\) Monte Carlo paths. Discuss the tradeoffs: which method is likely more accurate near the barrier, and why? What variance reduction techniques could improve the LR estimator?

Solution to Exercise 6

Finite-difference method near the barrier:

For a barrier option, the price function \(V(S)\) has a discontinuity (or kink) in its first or second derivative at the barrier level \(B\). The finite-difference gamma estimator

\[ \hat{\Gamma} = \frac{V(S+h) - 2V(S) + V(S-h)}{h^2} \]

with \(h = 1\) has a problem when \(S\) is near \(B\): the shifted points \(S \pm h\) may straddle the barrier, so \(V(S+h)\) and \(V(S-h)\) come from different regimes (one above, one below the barrier). This introduces large bias that does not decrease with more Monte Carlo paths. The finite-difference approximation effectively smooths the gamma over a window of width \(2h\), missing the sharp localized behavior near the barrier.

LR method near the barrier:

The LR gamma estimator

\[ \hat{\Gamma}_{\text{LR}} = \frac{1}{N}\sum_{i=1}^N e^{-r\tau}\Phi(S_T^{(i)}) \cdot w_\Gamma(Z^{(i)}) \]

does not require perturbing \(S\). It evaluates the payoff at the actual stock price and uses the Malliavin/LR weight to capture the second-order sensitivity. Near the barrier, the payoff \(\Phi(S_T)\) varies sharply, and the LR weights can have high variance, but the estimator is unbiased for any \(S\) (including near the barrier).

Which is more accurate near the barrier? The LR method is likely more accurate because:

• It is unbiased at the exact value of \(S\), whereas finite differences introduce bias from straddling the barrier
• With \(N = 100{,}000\) paths, the standard error can be controlled
• The finite-difference bias near barriers can be the dominant error source and cannot be reduced by adding paths

Variance reduction techniques for the LR estimator:

• Control variates: Use a related vanilla option (whose LR gamma is known analytically) as a control variate to reduce variance
• Antithetic variates: Pair each path \(Z^{(i)}\) with \(-Z^{(i)}\) to reduce variance from odd moments
• Importance sampling: Tilt the distribution to sample more paths near the barrier, where the payoff variation is greatest
• Stratified sampling: Stratify on \(Z\) to ensure uniform coverage of the probability space
• Conditional Monte Carlo: Condition on the maximum/minimum of the path (relevant for barrier options) to reduce the indicator-function-induced variance