Greeks Under Jump-Diffusion¶

Risk management of option positions requires computing sensitivities (Greeks) with respect to the underlying parameters. Under the Merton jump-diffusion model, the Greeks inherit the series structure of the pricing formula: each Greek is a Poisson-weighted sum of the corresponding Black-Scholes Greek evaluated at adjusted parameters. However, jumps introduce qualitative differences from the Black-Scholes Greeks, including reduced delta near at-the-money, higher gamma, and a multi-dimensional vega that decomposes into sensitivities with respect to diffusion and jump parameters.

Learning Objectives

By the end of this section, you will be able to:

Derive delta, gamma, and vega under the Merton series formula
Explain why delta is reduced near ATM compared to Black-Scholes
Decompose total vega into diffusion vega and jump parameter sensitivities
Analyze the hedging error at jump times and its implications for risk management

Motivation¶

Greeks in Incomplete Markets¶

In the complete Black-Scholes market, delta hedging perfectly replicates the option payoff. The Greeks are deterministic functions of $(S, t, \sigma, r)$ and have clear hedging interpretations. Under jump-diffusion, the market is incomplete, and delta hedging leaves residual jump risk. Nevertheless, the Greeks remain essential for:

Local risk management: Delta and gamma control the P&L from small and moderate continuous moves
Parameter sensitivity: Vega measures exposure to changes in implied volatility levels
Scenario analysis: Understanding how option values respond to parameter shifts

Delta¶

Series Formula for Delta¶

The Merton call delta is obtained by differentiating the series formula term by term:

Proposition: Merton Delta

The delta of a European call under the Merton model is

\[ \Delta_{\text{Merton}} = \frac{\partial C}{\partial S_0} = \sum_{n=0}^{\infty} w_n \, \Delta_{\text{BS}}^{(n)} \]

where $w_n = \frac{e^{-\lambda'T}(\lambda'T)^n}{n!}$ and $\Delta_{\text{BS}}^{(n)} = e^{(r_n - r)T}N(d_1^{(n)})$ is the Black-Scholes delta with adjusted parameters $(\sigma_n, r_n)$.

Proof. Since the series converges uniformly in $S_0$ on compact sets (each term is bounded and the Poisson weights sum to one), differentiation and summation can be exchanged. Differentiating each Black-Scholes component with respect to $S_0$ gives the standard Black-Scholes delta with the corresponding adjusted parameters. $\square$

Delta Reduction Near ATM¶

A key qualitative difference from Black-Scholes is that the Merton delta is typically smaller for ATM calls:

\[ \Delta_{\text{Merton}}^{\text{ATM}} < \Delta_{\text{BS}}^{\text{ATM}} \]

The intuition is that the higher effective volatility from jumps spreads the probability mass of $S_T$ over a wider range, reducing the sensitivity to small changes in $S_0$. In Black-Scholes terms, the ATM delta is approximately $N(d_1) \approx N(\sigma\sqrt{T}/2)$, which decreases as volatility increases.

Delta Discontinuity at Jump Times¶

Between jumps, delta evolves continuously as a function of the (continuously changing) stock price. At a jump time $T_i$, the stock price changes from $S_{T_i^-}$ to $S_{T_i^-} \cdot Y_i$, and the delta jumps from $\Delta(S_{T_i^-})$ to $\Delta(S_{T_i^-} \cdot Y_i)$. For a large jump, this can represent a substantial change in the hedge ratio.

Delta Jump Illustration

Consider an ATM call with $S_0 = \$100$, $\Delta \approx 0.52$. If a downward jump moves the stock to $S = \$85$, the call is now OTM and $\Delta$ might drop to $0.30$. The hedge portfolio, which held 0.52 shares, now holds too many shares relative to the new delta, resulting in a loss from the excess position.

The hedging loss at the jump is approximately:

\[ \text{Loss} \approx [V(85) - V(100)] - 0.52 \times (85 - 100) = [V(85) - V(100)] + 7.80 \]

Since $V(85) - V(100) < -7.80$ for a convex payoff, the net loss is positive.

Gamma¶

Series Formula for Gamma¶

Proposition: Merton Gamma

The gamma of a European call under the Merton model is

\[ \Gamma_{\text{Merton}} = \frac{\partial^2 C}{\partial S_0^2} = \sum_{n=0}^{\infty} w_n \, \Gamma_{\text{BS}}^{(n)} \]

where $\Gamma_{\text{BS}}^{(n)} = \frac{e^{(r_n - r)T}\,\phi(d_1^{(n)})}{S_0 \sigma_n \sqrt{T}}$ and $\phi$ is the standard normal density.

Gamma and Jump Risk¶

Gamma measures the curvature of the option value, which is directly related to the hedging error at jumps. The second-order Taylor expansion of the hedging error at a jump of size $Y - 1$ relative to $S$ is:

\[ \text{Hedge error} \approx \frac{1}{2}\Gamma_{\text{Merton}} \cdot S^2(Y - 1)^2 \]

Higher gamma means larger losses from jumps. This creates a tension in risk management: a trader who is long gamma benefits from continuous volatility but suffers from large jumps (the gamma profit from continuous moves is offset by discrete jump losses).

Vega¶

Multi-Dimensional Vega¶

Unlike Black-Scholes, which has a single vega $\partial C/\partial \sigma$, the Merton model has four distinct sensitivities corresponding to its four volatility-related parameters:

\[ \mathcal{V}_{\text{total}} = \mathcal{V}_{\sigma} + \mathcal{V}_{\lambda} + \mathcal{V}_{\mu_J} + \mathcal{V}_{\sigma_J} \]

Diffusion Vega¶

\[ \mathcal{V}_{\sigma} = \frac{\partial C}{\partial \sigma} = \sum_{n=0}^{\infty} w_n \frac{\partial C_{\text{BS}}^{(n)}}{\partial \sigma} \]

Since $\sigma$ enters each term through $\sigma_n^2 = \sigma^2 + n\sigma_J^2/T$:

\[ \frac{\partial \sigma_n}{\partial \sigma} = \frac{\sigma}{\sigma_n} \]

The diffusion vega is the dominant sensitivity for long-maturity options, where the diffusion component dominates the jump component.

Jump Intensity Vega¶

\[ \mathcal{V}_{\lambda} = \frac{\partial C}{\partial \lambda} \]

This sensitivity measures the option price change when the expected number of jumps changes. Increasing $\lambda$ generally increases option prices (more jumps means more uncertainty), particularly for OTM options.

Jump Mean Vega¶

\[ \mathcal{V}_{\mu_J} = \frac{\partial C}{\partial \mu_J} \]

This controls the skew: making $\mu_J$ more negative tilts the distribution leftward, increasing OTM put prices and decreasing OTM call prices.

Jump Volatility Vega¶

\[ \mathcal{V}_{\sigma_J} = \frac{\partial C}{\partial \sigma_J} \]

This controls the smile convexity: increasing $\sigma_J$ raises both OTM put and OTM call prices, making the implied volatility smile more curved.

Parameter Impact on Implied Volatility

Parameter	Effect on IV smile
$\sigma$	Parallel shift up
$\lambda$	Parallel shift up (more for short maturities)
$\mu_J$	Controls skew (tilt)
$\sigma_J$	Controls convexity (curvature)

Theta¶

Time Decay with Jumps¶

The theta of a Merton option is:

\[ \Theta_{\text{Merton}} = \frac{\partial C}{\partial T} = \sum_{n=0}^{\infty} \frac{\partial}{\partial T}\left[w_n C_{\text{BS}}^{(n)}\right] \]

The time derivative acts on both the Poisson weights (which depend on $\lambda'T$) and the Black-Scholes prices (which depend on $T$ through $d_1^{(n)}$, $d_2^{(n)}$, and the discount factor). The theta is generally more negative (larger time decay) than Black-Scholes theta because the jump component adds volatility that decays with time.

Worked Example¶

Comparing Black-Scholes and Merton Greeks

Parameters: $S_0 = \$100$, $K = \$100$, $T = 0.25$, $r = 0.05$, $\sigma = 0.20$, $\lambda = 1.0$, $\mu_J = -0.08$, $\sigma_J = 0.25$.

Black-Scholes Greeks (no jumps):

Greek	Value
Delta	0.554
Gamma	0.0282
Vega ($\mathcal{V}_\sigma$)	19.82

Merton Greeks (with jumps):

Greek	Value
Delta	0.518
Gamma	0.0241
Vega ($\mathcal{V}_\sigma$)	14.67
Vega ($\mathcal{V}_\lambda$)	1.23
Vega ($\mathcal{V}_{\mu_J}$)	$-2.85$
Vega ($\mathcal{V}_{\sigma_J}$)	3.41

Key observations:

Delta is reduced from 0.554 to 0.518 (6.5% decrease) due to the higher effective volatility
Gamma is lower because the option value curve is smoother when volatility is higher
The diffusion vega $\mathcal{V}_\sigma$ decreases because part of the total variance comes from jumps, reducing the sensitivity to the diffusion component alone
The jump mean vega $\mathcal{V}_{\mu_J}$ is negative for a call: making $\mu_J$ more negative shifts the distribution leftward, reducing call values
The jump volatility vega $\mathcal{V}_{\sigma_J}$ is positive: larger jump dispersion increases the option value through convexity

Hedging Strategies¶

Delta Hedging Only¶

The simplest approach hedges only the diffusion risk using $\Delta_{\text{Merton}}$. The residual P&L over a period $[t, t + dt]$ without jumps is approximately zero, but at each jump the portfolio suffers a loss of order $\Gamma S^2(Y-1)^2/2$.

Delta-Gamma Hedging with Options¶

Adding a second option allows hedging both delta and gamma, reducing the sensitivity to moderate jumps. However, this still does not eliminate the higher-order jump risk.

Hedging with Jump-Sensitive Instruments¶

The most complete approach uses instruments that are directly sensitive to jump risk:

OTM puts hedge downward jump risk
Variance swaps hedge total variance (diffusion plus jump)
VIX options hedge changes in implied volatility driven by jump parameter shifts

In practice, most desks use a combination of delta hedging and scenario analysis (stress testing against specific jump sizes) rather than attempting to perfectly hedge jump risk.

Summary¶

The Greeks under the Merton jump-diffusion model inherit the series structure of the pricing formula, with each Greek being a Poisson-weighted sum of the corresponding Black-Scholes Greek at adjusted parameters. Jumps reduce the ATM delta, create a multi-dimensional vega decomposition ($\mathcal{V}_\sigma$, $\mathcal{V}_\lambda$, $\mathcal{V}_{\mu_J}$, $\mathcal{V}_{\sigma_J}$), and generate hedging errors at jump times that are proportional to gamma. The market incompleteness means that no dynamic hedging strategy in the stock alone can eliminate jump risk, making scenario analysis and jump-sensitive instruments essential complements to traditional delta hedging.

Exercises¶

Exercise 1. The Merton delta is $\Delta_{\text{Merton}} = \sum_{n=0}^{\infty} w_n \Delta_{\text{BS}}^{(n)}$. Explain why $\Delta_{\text{Merton}}^{\text{ATM}} < \Delta_{\text{BS}}^{\text{ATM}}$ by relating the ATM delta to the effective volatility. For $\sigma = 0.20$ and $\sigma_{\text{eff}} = 0.30$ (including jump contribution), estimate the reduction in ATM call delta for $T = 0.25$.

Solution to Exercise 1

The ATM Black-Scholes delta for a call is approximately $\Delta_{\text{BS}}^{\text{ATM}} \approx N(d_1)$ where, for ATM options ($S_0 = K$), $d_1 \approx \frac{\sigma\sqrt{T}}{2} + \frac{r}{\sigma}\sqrt{T}$. For short maturities, the dominant term is $d_1 \approx \sigma\sqrt{T}/2$.

The key relationship is that increasing effective volatility decreases the ATM delta. Under Black-Scholes with $\sigma = 0.20$ and $T = 0.25$:

\[ d_1 \approx \frac{0.20 \times 0.5}{2} + \frac{0.05}{0.20}\times 0.5 = 0.05 + 0.125 = 0.175 \]

\[ \Delta_{\text{BS}} = N(0.175) \approx 0.569 \]

Under the Merton model, the effective volatility is $\sigma_{\text{eff}} = 0.30$. Replacing $\sigma$ by $\sigma_{\text{eff}}$:

\[ d_1 \approx \frac{0.30 \times 0.5}{2} + \frac{0.05}{0.30}\times 0.5 = 0.075 + 0.0833 = 0.158 \]

\[ \Delta_{\text{Merton}} \approx N(0.158) \approx 0.563 \]

The reduction is from approximately 0.569 to 0.563, about a 1% decrease. The actual Merton delta is computed as a Poisson-weighted average of Black-Scholes deltas at different volatilities $\sigma_n$, which spreads the probability mass more widely, further reducing the ATM sensitivity. The worked example in the text shows a more substantial reduction (from 0.554 to 0.518) because the full series average places weight on terms with even higher volatilities, amplifying the effect.

Exercise 2. At a jump time, the hedging error of a delta-hedged portfolio is $\Delta\Pi = V(S_{t^-}Y) - V(S_{t^-}) - \Delta_t S_{t^-}(Y-1)$. (a) Show that this error is approximately $\frac{1}{2}\Gamma S^2(Y-1)^2$ for small jumps (Taylor expand to second order). (b) For $S = 100$, $\Gamma = 0.025$, and a 15% downward jump ($Y = 0.85$), compute the approximate hedging loss.

Solution to Exercise 2

(a) Taylor expand $V(SY)$ around $S$ with $SY = S + S(Y-1)$:

\[ V(SY) = V(S) + V'(S) \cdot S(Y-1) + \frac{1}{2}V''(S) \cdot S^2(Y-1)^2 + O((Y-1)^3) \]

The hedging error is:

\[ \Delta\Pi = V(SY) - V(S) - \Delta_t S(Y-1) \]

With $\Delta_t = V'(S) = \partial V/\partial S$:

\[ \Delta\Pi = V'(S) \cdot S(Y-1) + \frac{1}{2}V''(S) \cdot S^2(Y-1)^2 - V'(S) \cdot S(Y-1) + O((Y-1)^3) \]

\[ = \frac{1}{2}V''(S) \cdot S^2(Y-1)^2 + O((Y-1)^3) = \frac{1}{2}\Gamma S^2(Y-1)^2 + O((Y-1)^3) \]

(b) With $S = 100$, $\Gamma = 0.025$, $Y = 0.85$ (so $Y - 1 = -0.15$):

\[ \Delta\Pi \approx \frac{1}{2}(0.025)(100^2)(0.15)^2 = \frac{1}{2}(0.025)(10000)(0.0225) = \frac{1}{2}(5.625) = 2.8125 \]

The approximate hedging loss from the 15% downward jump is about $2.81. This loss is always positive (regardless of jump direction) because it depends on $(Y-1)^2$, reflecting the convexity of the option value function.

Exercise 3. Unlike Black-Scholes, the Merton model has four vega sensitivities: $\mathcal{V}_\sigma$, $\mathcal{V}_\lambda$, $\mathcal{V}_{\mu_J}$, $\mathcal{V}_{\sigma_J}$. For an ATM call, explain qualitatively whether each vega is positive or negative, and which one dominates for long-maturity options versus short-maturity options.

Solution to Exercise 3

$\mathcal{V}_\sigma$ (diffusion vega): Positive for ATM calls. Increasing the diffusion volatility increases the option value through the standard Black-Scholes mechanism. This dominates for long-maturity options where the diffusion component contributes most of the total variance.
$\mathcal{V}_\lambda$ (jump intensity vega): Positive for ATM calls. More frequent jumps increase uncertainty, raising option values through higher effective volatility. This is most important for short-maturity options where jump effects dominate.
$\mathcal{V}_{\mu_J}$ (jump mean vega): Negative for ATM calls (when $\mu_J < 0$). Making $\mu_J$ more negative shifts the return distribution leftward, which decreases call values (lower probability of ending in-the-money) and increases put values. The sign can reverse for deep OTM calls.
$\mathcal{V}_{\sigma_J}$ (jump volatility vega): Positive for ATM calls. Larger jump size dispersion widens the distribution, increasing the option value through convexity. This is most relevant for short-maturity options.

For long-maturity options, $\mathcal{V}_\sigma$ dominates because the jump contribution to total variance becomes relatively smaller (jumps scale as $\lambda T$ while diffusion scales as $\sigma^2 T$, but the smile effects from jumps decay as $1/\sqrt{T}$). For short-maturity options, $\mathcal{V}_\lambda$ and $\mathcal{V}_{\sigma_J}$ become relatively more important because jumps dominate the smile shape.

Exercise 4. Describe a hedging strategy for a short call position under the Merton model that goes beyond delta hedging. Specifically: (a) How would you use OTM puts to hedge jump risk? (b) How would adding a second option allow delta-gamma hedging? (c) Why is perfect replication impossible?

Solution to Exercise 4

(a) Using OTM puts to hedge jump risk: Buy OTM puts at strikes below the current stock price (e.g., $K = 0.85 S_0$). These puts increase in value during large downward jumps, offsetting the loss on the short call from the jump-induced hedging error. The cost is the premium paid for the puts. The strike and quantity are chosen based on the expected jump size and intensity.

(b) Delta-gamma hedging with a second option: Choose a second option (e.g., a call at a different strike or maturity) and hold quantities $h_1$ (stock) and $h_2$ (second option) such that:

\[ h_1 + h_2\Delta_2 = \Delta_{\text{target}}, \qquad h_2\Gamma_2 = \Gamma_{\text{target}} \]

By matching both delta and gamma, the portfolio is insensitive to both first-order and second-order price changes. This reduces the hedging error at moderate jumps from $O((Y-1)^2)$ to $O((Y-1)^3)$.

(c) Why perfect replication is impossible: The market is incomplete because there are two independent risk sources ($W_t$ and $N_t$) but only one traded asset (plus the bond). No dynamic trading strategy in the stock alone can replicate the payoff state-by-state across all possible jump outcomes. Even with a second option, the hedge is approximate because the jump size $Y$ is random and continuous-valued, requiring infinitely many instruments to span all possible outcomes. The residual risk is the unhedgeable jump component that requires risk premia or diversification.

Exercise 5. The Merton theta is generally more negative than the Black-Scholes theta. Explain this using the theta-gamma relationship: in Black-Scholes, $\Theta + \frac{1}{2}\sigma^2 S^2\Gamma + rS\Delta - rV = 0$. How does the additional jump term modify this relationship in the Merton model?

Solution to Exercise 5

In the Black-Scholes model, the option price satisfies the PDE:

\[ \Theta_{\text{BS}} + \frac{1}{2}\sigma^2 S^2\Gamma_{\text{BS}} + rS\Delta_{\text{BS}} - rV_{\text{BS}} = 0 \]

This means $\Theta_{\text{BS}} = rV_{\text{BS}} - rS\Delta_{\text{BS}} - \frac{1}{2}\sigma^2 S^2\Gamma_{\text{BS}}$.

In the Merton model, the PIDE adds the jump integral term:

\[ \Theta_{\text{Merton}} + \frac{1}{2}\sigma^2 S^2\Gamma_{\text{Merton}} + (r - \lambda\bar{k})S\Delta_{\text{Merton}} - (r+\lambda)V_{\text{Merton}} + \lambda\int_0^{\infty}V(t, Sy)g(y)\,dy = 0 \]

Solving for theta:

\[ \Theta_{\text{Merton}} = (r+\lambda)V_{\text{Merton}} - (r - \lambda\bar{k})S\Delta_{\text{Merton}} - \frac{1}{2}\sigma^2 S^2\Gamma_{\text{Merton}} - \lambda\int_0^{\infty}V(t, Sy)g(y)\,dy \]

Compared to Black-Scholes, the Merton theta has additional terms: $+\lambda V$ from the increased discounting (options lose value faster because jumps can occur), $+\lambda\bar{k}S\Delta$ from the drift adjustment, and $-\lambda\int V(Sy)g(y)\,dy$ from the expected jump payoff. For ATM calls, the net effect of these additional terms is to make $\Theta_{\text{Merton}}$ more negative than $\Theta_{\text{BS}}$, because the jump component adds time value that decays faster as the probability of beneficial jumps diminishes with shorter remaining time.

Parameter	Effect on IV smile
\(\sigma\)	Parallel shift up
\(\lambda\)	Parallel shift up (more for short maturities)
\(\mu_J\)	Controls skew (tilt)
\(\sigma_J\)	Controls convexity (curvature)