Jump-Diffusion Models: Merton and Bates

If the smile of short-maturity options may reflect sudden price jumps, valuation staff need to check how much of it a diffusion-only model can explain.

Jumps added to a diffusion

The Black-Scholes model assumes that the underlying price moves continuously. A paper from the Bank of Japan's Institute for Monetary and Economic Studies shows cases in which the kurtosis of returns on the yen-dollar rate and on the Nikkei 225 exceeds the normal distribution's value of 3. It concludes that assuming prices follow a diffusion alone is not always appropriate¹. The jump-diffusion model adds occasional price jumps to this continuous movement (the diffusion). Merton gave the jump size a lognormal distribution, which let the European option price be written as a formula².

When a jump occurs, the price is multiplied by $J$. The average proportional price change from a jump is $k = E[J] - 1$. For a non-dividend-paying underlying asset, the price $S_t$ follows the equation below under the risk-neutral probability.

$$\frac{dS_t}{S_{t-}} = (r - \lambda k)\,dt + \sigma\,dW_t + (J - 1)\,dN_t$$

In this equation, $S_{t-}$ is the price just before a jump, and $r$ is the interest rate. The process $W_t$ is a standard Brownian motion that creates the diffusion. The process $N_t$ is a Poisson process with intensity $\lambda$ that counts the jumps. The jump size $J$ is drawn independently at each jump, and $W_t$, $N_t$, and $J$ are independent of one another³. For a Poisson process with intensity $\lambda$, the probability of exactly $n$ jumps in time $t$ is $e^{-\lambda t}(\lambda t)^n/n!$⁴.

Merton's model assumes that the logarithm of $J$ is normally distributed with mean $\mu_J$ and standard deviation $\delta$, so that $J$ follows a lognormal distribution. Then $k$ is given by the following formula².

$$k = E[J] - 1 = e^{\mu_J + \delta^2/2} - 1$$

The model has the following four parameters.

Direction can be described in two ways. The sign of $\mu_J$ shows which direction has more jumps, and the sign of $k$ shows whether jumps raise or lower the price on average. The condition $k < 0$ holds when $\mu_J < -\delta^2/2$, and $k > 0$ holds when $\mu_J = 0$.

The risk-neutral drift correction

The drift in the equation is $r - \lambda k$ rather than $r$ because the average amount by which the jump term moves the price is subtracted in advance. Since $J$ and $N_t$ are independent, $E[(J-1)\,dN_t] = \lambda k\,dt$. Subtracting this makes the expected growth of the price equal to the interest rate $r$, and the discounted price $e^{-rt}S_t$ becomes a martingale⁵.

If $k < 0$ (jumps lower the price on average), then $-\lambda k$ is positive. While no jump occurs, the price rises faster than $r$. It jumps down from time to time and, on average, grows at $r$.

An option on an asset with jumps generally cannot be replicated with the underlying asset and a deposit alone. There are two sources of uncertainty, the Brownian motion and the Poisson process, so the market is incomplete. Merton assumed that jump risk can be diversified and requires no risk premium. Whether this assumption is valid is debated⁶. When fitting to market prices, $\lambda$, $\mu_J$, and $\delta$ are treated as parameters under the risk-neutral probability⁷.

A pricing formula split by the number of jumps

Once the number of jumps is fixed at $n$, the log price at maturity is normally distributed. The conditional price is therefore a Black-Scholes formula with the volatility and the interest rate replaced. The price of a European call is the sum of these prices over $n$, weighted as shown below. The weights have the form of Poisson probabilities, but their intensity is $\lambda'$, not the actual jump intensity $\lambda$, so they differ from the actual probabilities⁸.

$$C = \sum_{n=0}^{\infty} \frac{e^{-\lambda' T}(\lambda' T)^n}{n!}\,C_{BS}(S_0, K, T, r_n, \sigma_n)$$

Here, $C_{BS}(S, K, T, r, \sigma)$ is the call price in the Black-Scholes model with interest rate $r$ and volatility $\sigma$. The symbol $K$ is the strike price, and $T$ is the time to maturity in years. The weight intensity $\lambda'$, and the interest rate $r_n$ and volatility $\sigma_n$ for $n$ jumps, are as follows.

$$\lambda' = \lambda(1+k)$$ $$r_n = r - \lambda k + \frac{n\ln(1+k)}{T}$$ $$\sigma_n^2 = \sigma^2 + \frac{n\delta^2}{T}$$

$n$ jumps add $n\delta^2$ to the variance of the log price and shift its mean by $n\mu_J$. The volatility $\sigma_n$ brings that variance into the Black-Scholes formula, and $r_n$ brings in the shift of the mean and the correction $-\lambda k$. Note that $\ln(1+k) = \mu_J + \delta^2/2$. The weights sum to 1. The difference between discounting at $r$ and at $r_n$, $e^{(r_n - r)T}$, equals $(1+k)^n e^{-\lambda k T}$. Multiplying the Poisson probability by this factor changes the intensity from $\lambda$ to $\lambda(1+k)$. Because the weights shrink quickly as $n$ grows, the sum can be truncated after a modest number of terms⁸.

Deep diveDerivation by conditioning on the number of jumps

The solution of the stochastic differential equation above can be written for the log price as follows.

$$\ln S_T = \ln S_0 + \left(r - \lambda k - \frac{\sigma^2}{2}\right)T + \sigma W_T + \sum_{i=1}^{N_T} \ln J_i$$

In this expression, $N_T$ is the number of jumps up to maturity, $J_i$ is the price multiplier of the $i$-th jump, and $W_T$ is the value of the Brownian motion at maturity. Conditional on $N_T = n$, $\ln S_T$ is normally distributed with mean $\ln S_0 + (r - \lambda k - \sigma^2/2)T + n\mu_J$ and variance $\sigma^2 T + n\delta^2$. In a Black-Scholes model with interest rate $r_n$ and volatility $\sigma_n$, the log price at maturity has mean $\ln S_0 + (r_n - \sigma_n^2/2)T$ and variance $\sigma_n^2 T$. Equating the two gives the $r_n$ and $\sigma_n$ above.

The conditional undiscounted expected payoff is then $e^{r_n T}\,C_{BS}(S_0, K, T, r_n, \sigma_n)$. Discounting to the present uses $r$, not $r_n$, so the factor $e^{-rT}e^{r_n T} = e^{-\lambda k T}(1+k)^n$ applies. Multiplying it by the Poisson probability $e^{-\lambda T}(\lambda T)^n/n!$ gives

$$e^{-\lambda T}\frac{(\lambda T)^n}{n!}\,e^{-\lambda k T}(1+k)^n = \frac{e^{-\lambda' T}(\lambda' T)^n}{n!}$$

so the weights become Poisson probabilities with intensity $\lambda'$. Splitting by the number of jumps is the same procedure as the risk-neutral method in Appendix 2 of the paper⁹.

Worked example

Consider a hypothetical example in which a call has $S_0 = 100$, $K = 100$, $T = 1$ year, interest rate $r = 1\%$, $\sigma = 15\%$, $\lambda = 1$, $\mu_J = -0.10$, and $\delta = 0.10$. Since $\mu_J + \delta^2/2 = -0.095$, the average jump is $k = e^{-0.095} - 1 = -0.0906$. The weight intensity is $\lambda' = 1 \times 0.9094 = 0.9094$, and $-\lambda k = 0.0906$. The values for each $n$ are in the table below, computed from $r_n = 1\% + 9.06\% - 9.5\%\,n$ and $\sigma_n^2 = 0.0225 + 0.01\,n$.

The sum through $n = 6$ is 8.4036. Through $n = 10$ it is 8.4037, and it does not change after that¹⁰. The rate $r_n$ is an adjustment value that aligns the discounting and the mean with the number of jumps, not an actual interest rate. The Black-Scholes price under the same conditions without jumps is 6.46 at a volatility of 15%. Adding jumps raises the option price by 1.94. Converting 8.4037 to a volatility with the Black-Scholes formula gives 19.92%. This is the diffusion's 15% plus the effect of an average of one jump per year.

Parameters and the shape of the smile

Converting Merton prices to implied volatility with the Black-Scholes formula gives a different value for each strike. The table below uses the conditions of the worked example and shows strikes from 80 to 120 for each maturity, in percent¹¹.

Because $\mu_J$ is negative and downward jumps are more common, the smile slopes downward: the lower the strike, the higher the implied volatility. The lower the strike, the more the price exceeds the Black-Scholes price (at 15% volatility). At a maturity of 0.25 years, strike 80 is 7.2 percentage points above strike 100. At 4 years, the gap is 0.4 percentage points. The smile flattens at longer maturities because, as the number of jumps grows, the distribution of the log return up to maturity approaches a normal distribution, and the difference between strikes shrinks. The difference remains even at 4 years. The level does not return to the diffusion's 15% either: at 4 years and strike 100, it is 20.16%, above the 1-year value of 19.92%. The Bank of Japan paper explains that, as more jumps occur, their effect is absorbed into the drift on average¹².

Figure description

The chart shows implied volatilities backed out from the Merton prices under the conditions of the worked example, for strikes from 75 to 125. The 0.25-year curve slopes downward, with higher values at low strikes, and is steeper on the low side. Near the right end it turns slightly upward again. The 1-year curve slopes the same way but is flatter. The two curves cross between strikes 95 and 97.5: below that, the 0.25-year curve is higher, and above it, the 1-year curve is higher. The vertical line marks the current price of 100. Without jumps, the 15% line is horizontal and does not depend on the strike. The shorter the maturity, the larger the slope and curvature that the jumps create.

The table below shows the implied volatility at a maturity of 0.25 years when one parameter is moved at a time, with the rest as in the worked example. Changing $\mu_J$ or $\delta$ also changes $k$, and with it the correction $-\lambda k$¹¹.

A larger $\lambda$ raises the value at every strike. A larger $\delta$ lifts both ends and strengthens the curvature. The sign of $\mu_J$ sets the direction of the slope. With $\mu_J = 0$, the smile is nearly symmetric, though not perfectly. With $\mu_J = +0.10$, upward jumps are more common, so the smile slopes upward.

Bates's model

Bates's model adds jumps of the same form as Merton's to the stochastic volatility of the Heston model¹³. Under the risk-neutral probability, the equations are as follows.

$$\frac{dS_t}{S_{t-}} = (r - \lambda k)\,dt + \sqrt{v_t}\,dW_t^{S} + (J - 1)\,dN_t$$ $$dv_t = \kappa(\bar v - v_t)\,dt + \xi\sqrt{v_t}\,dW_t^{v}, \qquad dW_t^{S}\,dW_t^{v} = \rho\,dt$$

The symbols $v_t$, $\kappa$, $\bar v$, $\xi$, and $\rho$ are the same as in the Heston model. The jump symbols $\lambda$, $J$, $N_t$, and $k$ are the same as in Merton's model. The average jump $k$ is a different quantity from the speed of mean reversion $\kappa$. The number and size of the jumps are assumed to be independent of the two Brownian motions. The Heston equations, characteristic function, and semi-analytical solution are covered in The Heston Model.

Under this assumption, the log stock price at maturity is the sum of a Heston part and a jump part, and the two are independent. The characteristic function of a sum of independent random variables is the product of their characteristic functions, so the characteristic function of Bates's model has the following form¹⁴.

$$\varphi_{\text{Bates}}(u, T) = \varphi_{\text{Heston}}(u, T)\cdot\exp\Bigl(\lambda T\bigl(e^{iu\mu_J - \delta^2u^2/2} - 1 - iuk\bigr)\Bigr)$$

Here, $\varphi_{\text{Heston}}$ is the characteristic function (without dividends) given on that page. In the factor on the right, $e^{iu\mu_J - \delta^2u^2/2} - 1$ is the compound Poisson part, and $-iuk$ corresponds to the correction $-\lambda k$. The price comes from the same integral formula as in the Heston model, $C = S_0P_1 - Ke^{-rT}P_2$, with this characteristic function inserted. With $\lambda = 0$, the model is the Heston model. If $v_0 = \bar v$ is kept while $\xi \to 0$, so that the variance is constant, the model is Merton's model with $\sigma^2 = v_0 = \bar v$.

Consider a hypothetical example in which the Heston part has $v_0 = \bar v = 0.0225$ (volatility 15%), $\kappa = 2$, $\xi = 0.2$, and $\rho = -0.5$, the jumps are the same as in the worked example ($\lambda = 1$, $\mu_J = -0.10$, $\delta = 0.10$), $S_0 = 100$, and the interest rate is 1%. The implied volatilities (%) with the Heston model alone and with jumps added are as follows¹⁵.

Adding jumps raises the implied volatility at every strike, and the lower the strike, the more it rises. At a maturity of 0.25 years, the rise is 8.50 percentage points at strike 80 and 2.73 percentage points at strike 120. The Bank of Japan paper cites earlier studies finding that adding jumps improves the model's explanatory power for short-term options, and that a stochastic volatility model can be considered for long-term options¹². Bates's model combines both. With the symbols used on this page, it has eight parameters ($v_0$, $\bar v$, $\kappa$, $\xi$, $\rho$, $\lambda$, $\mu_J$, and $\delta$).

Assumptions and limits

The parameter values move the result a lot. In the worked example, raising $\lambda$ from 1 to 2 lifts the 1-year call price from 8.40 to 10.03. It also lifts the implied volatility at a 1-year maturity and strike 80 from 21.68% to 25.95%¹¹.

Merton's model keeps $\sigma$, $\lambda$, $\mu_J$, and $\delta$ constant. As a result, the smile flattens as maturity lengthens, and the model may fail to explain the long-maturity smiles seen in the market¹². Adding stochastic volatility, as in Bates's model, captures the effect of volatility that changes over time.

Fitting to market prices is not stable. In a study that applied least squares to DAX option prices, starting the fit at a jump intensity of 1 instead of 5 produced a different shape for the fitted jump distribution. Even in Merton's model, increasing $\lambda$ while lowering $\sigma$ leaves the error at about the same level¹⁶. The tables above also show that $\lambda$, $\delta$, and $\mu_J$ each move the level and the shape of the smile together. The fitting procedure and overfitting are covered in Model Calibration.

Notes

1. 久田祥史 (Yoshifumi Hisata), "ジャンプ拡散過程を用いたオプション価格付けモデルについて" (external site) (On option pricing models using jump-diffusion processes), Kinyu Kenkyu, Institute for Monetary and Economic Studies, Bank of Japan (June 2003), section 2 (pages 53–54), Figure 2 (in Japanese). The kurtosis is 12.01 for the yen-dollar rate (August 1998 to July 1999) and 4.93 for the Nikkei 225 (October 2000 to September 2001). ↩

2. Hisata (external site) (in Japanese), section 3(2) (page 57). The case in which the log of the jump ratio is normally distributed is the example for which Merton (1976) obtained an analytical solution, with $k = \exp(\mu_J + \delta^2/2) - 1$ (the source's symbols are $m_J$ and $s_J$). ↩ ↩²

3. Hisata (external site) (in Japanese), section 3(1) (pages 55–56), equation (1). The source writes the drift with the expected return $m$, and the text replaces $m$ with the risk-neutral $r$. ↩

4. Hisata (external site) (in Japanese), Appendix 1 (page 70), equation (A-3). ↩

5. Hisata (external site) (in Japanese), note 10 (page 56). It subtracts the expected value of the compound Poisson process, $\lambda k\,dt$, so that the effect on the drift is zero on average. Cont and Tankov (external site), section 2.2 (page 5), equation (8), gives the condition that the discounted price is a martingale as a constraint on the drift. ↩

6. Hisata (external site) (in Japanese), section 5 (page 66). Two sources of uncertainty make the market incomplete. Merton assumed that jump risk can be diversified and requires no premium; whether this is valid is debated. ↩

7. Cont and Tankov (external site), section 2.2 (page 5). The price is set as $S_t = e^{rt + X_t}$ under the risk-neutral probability and fitted to option prices. ↩

8. Hisata (external site) (in Japanese), section 3(2) (page 57), equation (2); the symbols are aligned with the text. Cont and Tankov (external site), section 5 (page 24), also say that a Merton call price expands into a weighted sum of Black-Scholes prices whose weights converge exponentially to zero. ↩ ↩²

9. Hisata (external site) (in Japanese), Appendix 2 (pages 72–73). In the risk-neutral method, the expected value for each number of jumps is summed with the probability of that number as the weight. How the mean and variance conditional on the number are matched is a calculation from the formulas in the text. ↩

10. Calculation. The series values (up to $n = 40$) were checked against numerical integration of the characteristic function. ↩

11. Calculation. Implied volatilities backed out from the series prices with the Black-Scholes formula. ↩ ↩² ↩³

12. Hisata (external site) (in Japanese), section 3(2) (page 58), the explanation of Figure 6. It includes the finding of Bakshi, Cao and Chen (1997). The explanation that the distribution of the total approaches a normal distribution comes from the calculation in the text (the central limit theorem), not from the source. ↩ ↩² ↩³

13. Bates working paper (external site) (December 1993), section I, equation (1) (page 4). The $\alpha - \beta V$ and $\sigma_v$ in the variance equation correspond to $\kappa(\bar v - v)$ and $\xi$ in the text. The published version is in the Review of Financial Studies in 1996. ↩

14. Bates working paper (external site), equations (9)–(10) (page 10). The exponent of the characteristic function is the sum of the Heston-type terms and the jump term $\lambda T[\ldots - 1]$, and the probabilities come from the Fourier inversion in equation (13). The independence of the jumps from the two Brownian motions is a model assumption of the text, consistent with the exponent in equations (9)–(10) being a sum. For the Heston part, see Albrecher et al. (external site), section 3 (page 4). ↩

15. Calculation. Prices from numerical integration of the characteristic function were compared with Monte Carlo (the Heston part approximated with 400 steps per year). ↩

16. Cont and Tankov (external site), section 3.1 (pages 13–14), Figures 1 and 2. ↩