Merton Series Formula¶

The most celebrated result of Merton's 1976 paper is a closed-form series expansion for European option prices under jump-diffusion dynamics. The key insight is elegant: condition on the number of jumps $n$, and the resulting distribution is Gaussian, so each conditional price is a Black-Scholes formula with adjusted parameters. The unconditional price is then a Poisson-weighted sum of these Black-Scholes prices, converging rapidly due to the factorial decay of Poisson probabilities.

Learning Objectives

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

Derive the Merton series formula by conditioning on the number of jumps
Identify the adjusted volatility $\sigma_n$ and rate $r_n$ for each term
Establish convergence of the series and bound the truncation error
Price a European option numerically using the series expansion

Motivation¶

The Conditioning Trick¶

The log-return in the Merton model is:

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

Conditional on $N_T = n$, the jump sum $\sum_{i=1}^{n}\ln Y_i$ is a sum of $n$ independent $N(\mu_J, \sigma_J^2)$ random variables, hence $N(n\mu_J, n\sigma_J^2)$. Combined with the Gaussian diffusion term $\sigma W_T \sim N(0, \sigma^2 T)$, the conditional log-return is normal. A normal log-return means the conditional asset price is log-normal, and log-normal prices yield Black-Scholes formulas.

Why This Is Remarkable¶

Most extensions of Black-Scholes (stochastic volatility, local volatility) destroy the closed-form pricing property. The Merton model preserves it, at the cost of replacing a single Black-Scholes evaluation with a rapidly converging infinite series. In practice, 10--20 terms suffice for machine precision.

Derivation¶

Step 1: Conditional Distribution¶

Given $N_T = n$, the log-return is:

\[ \ln\frac{S_T}{S_0}\;\Big|\; N_T = n \;\sim\; N\!\left(\left(r - \lambda\bar{k} - \frac{1}{2}\sigma^2\right)T + n\mu_J,\; \sigma^2 T + n\sigma_J^2\right) \]

Define the conditional parameters:

\[ \sigma_n^2 = \sigma^2 + \frac{n\sigma_J^2}{T} \]

\[ r_n = r - \lambda\bar{k} + \frac{n\ln(1 + \bar{k})}{T} \]

where $\bar{k} = e^{\mu_J + \sigma_J^2/2} - 1$.

Interpreting the Adjusted Parameters

The conditional volatility $\sigma_n$ increases with $n$ because each additional jump adds variance $\sigma_J^2$ to the log-return. The conditional rate $r_n$ adjusts the forward price to account for the average cumulative jump effect of $n$ jumps.

Step 2: Conditional Option Price¶

Given $N_T = n$, the asset price is log-normal with parameters $(\sigma_n, r_n)$, so the European call price is the standard Black-Scholes formula:

\[ C_{\text{BS}}(S_0, K, T, r_n, \sigma_n) = S_0 e^{(r_n - r)T}N(d_1^{(n)}) - Ke^{-rT}N(d_2^{(n)}) \]

where:

\[ d_1^{(n)} = \frac{\ln(S_0/K) + (r_n + \frac{1}{2}\sigma_n^2)T}{\sigma_n\sqrt{T}} \]

\[ d_2^{(n)} = d_1^{(n)} - \sigma_n\sqrt{T} \]

and $N(\cdot)$ is the standard normal CDF.

Step 3: Poisson Averaging¶

The number of jumps $N_T$ follows a Poisson distribution. Define the adjusted intensity:

\[ \lambda' = \lambda(1 + \bar{k}) \]

Then the Poisson weight for $n$ jumps is:

\[ w_n = \frac{e^{-\lambda'T}(\lambda'T)^n}{n!} \]

Theorem: Merton Series Formula

The European call price under the Merton jump-diffusion model is

\[ C_{\text{Merton}} = \sum_{n=0}^{\infty} w_n \, C_{\text{BS}}(S_0, K, T, r_n, \sigma_n) \]

where:

$w_n = \frac{e^{-\lambda'T}(\lambda'T)^n}{n!}$ with $\lambda' = \lambda(1 + \bar{k})$
$\sigma_n^2 = \sigma^2 + n\sigma_J^2/T$
$r_n = r - \lambda\bar{k} + n\ln(1+\bar{k})/T$

The European put price follows by put-call parity:

\[ P_{\text{Merton}} = C_{\text{Merton}} - S_0 + Ke^{-rT} \]

Proof. By the law of total expectation:

\[ C_{\text{Merton}} = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] = e^{-rT}\sum_{n=0}^{\infty}\mathbb{P}(N_T = n)\,\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+ \mid N_T = n] \]

Conditional on $N_T = n$, the log-return is Gaussian, so the conditional expectation is a Black-Scholes price. The Poisson probabilities use the adjusted intensity $\lambda'$ because the change from $\lambda$ to $\lambda'$ absorbs the compensator into the weights, simplifying the conditional forward price. Specifically:

\[ \mathbb{E}^{\mathbb{Q}}[S_T \mid N_T = n] = S_0 e^{rT} \cdot \frac{(\lambda')^n e^{-\lambda T}}{(\lambda)^n e^{-\lambda' T}} \cdot (\text{normalizing factor}) \]

After careful algebraic manipulation (replacing $\lambda$ by $\lambda'$ in the Poisson weights and adjusting $r$ to $r_n$ in the Black-Scholes formula), the series takes the stated form. $\square$

Convergence Analysis¶

Rate of Convergence¶

The Poisson weights $w_n$ decay factorially:

\[ w_n = \frac{e^{-\lambda'T}(\lambda'T)^n}{n!} \]

For $n \gg \lambda'T$, Stirling's approximation gives $w_n \approx \frac{1}{\sqrt{2\pi n}}\left(\frac{e\lambda'T}{n}\right)^n$, which decreases faster than exponentially.

Truncation Error¶

Proposition: Truncation Error Bound

The truncation error from summing only $M$ terms satisfies

\[ \left|C_{\text{Merton}} - \sum_{n=0}^{M} w_n C_{\text{BS}}^{(n)}\right| \leq S_0 \sum_{n=M+1}^{\infty} w_n = S_0\left(1 - \sum_{n=0}^{M}w_n\right) \]

since each Black-Scholes price is bounded by $S_0$.

Proof. The Black-Scholes call price satisfies $0 \leq C_{\text{BS}} \leq S_0$. Therefore:

\[ \sum_{n=M+1}^{\infty} w_n C_{\text{BS}}^{(n)} \leq S_0 \sum_{n=M+1}^{\infty} w_n \]

The tail sum of Poisson probabilities can be bounded using the incomplete gamma function or computed directly. $\square$

Practical Guidelines¶

$\lambda'T$	Terms for $10^{-8}$ accuracy	Terms for $10^{-15}$ accuracy
0.5	8	15
1.0	10	18
5.0	16	25
10.0	22	32

The series converges rapidly for typical parameter values. Even for $\lambda = 5$ (five jumps per year on average), 25 terms provide double-precision accuracy.

Special Cases¶

Zero Jumps ($\lambda = 0$)¶

When $\lambda = 0$, the only nonzero weight is $w_0 = 1$, and $\sigma_0 = \sigma$, $r_0 = r$. The series reduces to a single Black-Scholes price, confirming consistency.

Pure Jump ($\sigma = 0$)¶

When $\sigma = 0$, the diffusion component vanishes, and $\sigma_n^2 = n\sigma_J^2/T$. The $n = 0$ term has $\sigma_0 = 0$, giving a degenerate Black-Scholes price (digital payoff behavior). The formula remains valid but the individual terms with small $n$ may require careful numerical treatment.

Large $\lambda T$¶

When $\lambda T$ is large, the Poisson distribution concentrates around its mean $\lambda'T$, and the series is dominated by terms near $n \approx \lambda'T$. The effective volatility approaches $\sqrt{\sigma^2 + \lambda\sigma_J^2}$ and the excess kurtosis diminishes, consistent with the central limit theorem.

Worked Example¶

Pricing a European Call

Parameters: $S_0 = \$100$, $K = \$100$, $T = 0.5$ years, $r = 0.05$, $\sigma = 0.20$, $\lambda = 1.0$, $\mu_J = -0.05$, $\sigma_J = 0.20$.

Step 1: Compute $\bar{k}$ and $\lambda'$.

\[ \bar{k} = e^{-0.05 + 0.02} - 1 = e^{-0.03} - 1 \approx -0.02955 \]

\[ \lambda' = 1.0 \times (1 - 0.02955) = 0.97045 \]

Step 2: Poisson weights for $T = 0.5$.

\[ \lambda'T = 0.4852 \]

$n$	$w_n$	$\sigma_n$	$r_n$
0	0.6157	0.2000	0.0796
1	0.2988	0.3111	0.0196
2	0.0725	0.4000	$-0.0404$
3	0.0117	0.4781	$-0.1004$
4	0.0014	0.5477	$-0.1604$

Step 3: Compute each $C_{\text{BS}}^{(n)}$.

For $n = 0$: $C_{\text{BS}}(100, 100, 0.5, 0.0796, 0.20) \approx 7.54$

For $n = 1$: $C_{\text{BS}}(100, 100, 0.5, 0.0196, 0.3111) \approx 9.28$

For $n = 2$: $C_{\text{BS}}(100, 100, 0.5, -0.0404, 0.40) \approx 10.03$

Step 4: Weighted sum.

\[ C_{\text{Merton}} \approx 0.6157(7.54) + 0.2988(9.28) + 0.0725(10.03) + 0.0117(9.83) + \cdots \]

\[ \approx 4.64 + 2.77 + 0.73 + 0.12 + \cdots \approx 8.27 \]

For comparison, the pure Black-Scholes price (no jumps) is approximately $7.54. The jump component adds about $0.73 to the option value, reflecting the additional risk from discontinuous moves.

Comparison with Black-Scholes¶

The Merton series formula reveals how jumps affect option prices across the strike spectrum:

Region	Effect of jumps	Reason
Deep ITM	Small impact	Payoff is nearly certain regardless of jumps
ATM	Moderate increase	Higher effective volatility from jumps
OTM puts	Large increase	Downward jumps ($\mu_J < 0$) increase crash probability
OTM calls	Moderate increase	Upward jump tail adds probability mass

This asymmetric impact across strikes is precisely what generates the implied volatility smile when Merton prices are inverted through the Black-Scholes formula.

Summary¶

The Merton series formula decomposes the European option price into a Poisson-weighted sum of Black-Scholes prices, each evaluated at adjusted volatility $\sigma_n^2 = \sigma^2 + n\sigma_J^2/T$ and rate $r_n = r - \lambda\bar{k} + n\ln(1+\bar{k})/T$. The series converges rapidly due to the factorial decay of Poisson weights, with 10--20 terms typically sufficient for machine precision. The formula reduces to Black-Scholes when $\lambda = 0$ and generates an implied volatility smile that steepens at short maturities, matching the empirical behavior that pure diffusion models cannot reproduce.

Exercises¶

Exercise 1. Verify that the Merton series formula reduces to the standard Black-Scholes formula when $\lambda = 0$. Specifically, show that $w_0 = 1$, $w_n = 0$ for $n \geq 1$, $\sigma_0 = \sigma$, and $r_0 = r$.

Solution to Exercise 1

When $\lambda = 0$, the adjusted intensity is $\lambda' = \lambda(1 + \bar{k}) = 0$, so $\lambda'T = 0$.

The Poisson weights become:

\[ w_n = \frac{e^{-0} \cdot 0^n}{n!} = \begin{cases} 1 & \text{if } n = 0 \\ 0 & \text{if } n \geq 1 \end{cases} \]

since $0^0 = 1$ by convention and $0^n = 0$ for $n \geq 1$.

For the $n = 0$ term, the conditional volatility is:

\[ \sigma_0^2 = \sigma^2 + \frac{0 \cdot \sigma_J^2}{T} = \sigma^2 \]

The conditional rate is:

\[ r_0 = r - \lambda\bar{k} + \frac{0 \cdot \ln(1+\bar{k})}{T} = r - 0 = r \]

since $\lambda = 0$ implies $\lambda\bar{k} = 0$.

Therefore the Merton series reduces to:

\[ C_{\text{Merton}} = 1 \cdot C_{\text{BS}}(S_0, K, T, r, \sigma) + 0 + 0 + \cdots = C_{\text{BS}}(S_0, K, T, r, \sigma) \]

This confirms that the Merton formula is a genuine generalization of Black-Scholes: setting the jump intensity to zero recovers the classical formula exactly.

Exercise 2. For the parameters $S_0 = 100$, $K = 100$, $T = 0.5$, $r = 0.05$, $\sigma = 0.20$, $\lambda = 1.0$, $\mu_J = -0.05$, $\sigma_J = 0.20$: (a) Compute $\bar{k}$ and $\lambda'$. (b) Compute the Poisson weights $w_0, w_1, w_2, w_3$. (c) Compute the adjusted volatilities $\sigma_0, \sigma_1, \sigma_2$ and rates $r_0, r_1, r_2$. (d) Estimate the option price by summing the first 4 terms of the series.

Solution to Exercise 2

(a) Compute $\bar{k}$ and $\lambda'$.

\[ \bar{k} = e^{\mu_J + \sigma_J^2/2} - 1 = e^{-0.05 + 0.02} - 1 = e^{-0.03} - 1 \approx -0.02955 \]

\[ \lambda' = \lambda(1 + \bar{k}) = 1.0 \times (1 - 0.02955) = 0.97045 \]

(b) Poisson weights for $T = 0.5$. With $\lambda'T = 0.97045 \times 0.5 = 0.48523$:

\[ w_0 = e^{-0.48523} \approx 0.6157 \]

\[ w_1 = e^{-0.48523} \times 0.48523 \approx 0.2988 \]

\[ w_2 = e^{-0.48523} \times \frac{0.48523^2}{2} \approx 0.0725 \]

\[ w_3 = e^{-0.48523} \times \frac{0.48523^3}{6} \approx 0.0117 \]

(c) Adjusted volatilities and rates.

\[ \sigma_0 = \sqrt{0.04 + 0} = 0.200, \quad r_0 = 0.05 + 0.02955 + 0 = 0.07955 \]

\[ \sigma_1 = \sqrt{0.04 + 0.04/0.5} = \sqrt{0.12} \approx 0.3464, \quad r_1 = 0.07955 + \frac{\ln(0.97045)}{0.5} \approx 0.07955 - 0.05997 = 0.01958 \]

\[ \sigma_2 = \sqrt{0.04 + 0.08/0.5} = \sqrt{0.20} \approx 0.4472, \quad r_2 = 0.07955 + \frac{2\ln(0.97045)}{0.5} \approx 0.07955 - 0.11994 = -0.04039 \]

(d) Estimate the option price. Using the Black-Scholes formula for each term:

$n = 0$: $C_{\text{BS}}(100, 100, 0.5, 0.07955, 0.200) \approx 7.54$
$n = 1$: $C_{\text{BS}}(100, 100, 0.5, 0.01958, 0.3464) \approx 9.63$
$n = 2$: $C_{\text{BS}}(100, 100, 0.5, -0.04039, 0.4472) \approx 10.38$
$n = 3$: $C_{\text{BS}}(100, 100, 0.5, -0.10036, 0.5292) \approx 10.23$

The weighted sum:

\[ C_{\text{Merton}} \approx 0.6157(7.54) + 0.2988(9.63) + 0.0725(10.38) + 0.0117(10.23) \]

\[ \approx 4.64 + 2.88 + 0.75 + 0.12 = 8.39 \]

The remaining terms ($n \geq 4$) contribute less than 0.02, so the series converges very rapidly.

Exercise 3. The truncation error of the Merton series after $M$ terms is bounded by $S_0 \sum_{n=M+1}^{\infty} w_n$. For $\lambda' T = 2$, how many terms are needed to ensure truncation error below $10^{-8}$? Use the Poisson tail bound to estimate.

Solution to Exercise 3

We need the Poisson tail probability $\sum_{n=M+1}^{\infty} w_n < 10^{-8}/S_0$. Since $S_0$ varies, we bound $\sum_{n=M+1}^{\infty} w_n < 10^{-8}$ directly (a sufficient condition when $S_0 \leq 1$; for larger $S_0$, the bound is conservative).

For $\lambda'T = 2$, we compute the cumulative Poisson distribution. The Poisson CDF for $\text{Poisson}(2)$:

\[ w_n = \frac{e^{-2} \cdot 2^n}{n!} \]

Computing partial sums:

$M$	$\sum_{n=0}^{M} w_n$	Tail $= 1 - \sum w_n$
5	0.98344	0.01656
8	0.99983	$1.7 \times 10^{-4}$
10	0.999995	$5.0 \times 10^{-6}$
12	0.99999997	$3.0 \times 10^{-8}$
13	0.999999997	$3.0 \times 10^{-9}$

Therefore, $M = 13$ terms suffice to ensure the truncation error is bounded by $S_0 \times 3 \times 10^{-9} < S_0 \times 10^{-8}$. For $S_0 = 100$, the absolute truncation error is below $10^{-6}$.

A useful Poisson tail bound is: for $M > \lambda'T$,

\[ \sum_{n=M+1}^{\infty} w_n \leq \frac{w_M \cdot \lambda'T}{M + 1 - \lambda'T} \]

which gives the geometric-like decay once $M$ exceeds the mean.

Exercise 4. Explain the asymmetric impact of jumps across strikes: why do jumps increase OTM put prices more than OTM call prices when $\mu_J < 0$? Relate your answer to the probability of large downward moves under the conditional distribution $\ln(S_T/S_0) | N_T = n$.

Solution to Exercise 4

When $\mu_J < 0$, the conditional distribution $\ln(S_T/S_0) \mid N_T = n$ has mean:

\[ m_n = \left(r - \lambda\bar{k} - \frac{1}{2}\sigma^2\right)T + n\mu_J \]

Since $\mu_J < 0$, the conditional mean $m_n$ decreases with $n$. More jumps shift the conditional distribution further to the left, increasing the probability that $S_T$ falls significantly below $S_0$.

OTM puts (strike $K < S_0$) have payoff $(K - S_T)^+$ which benefits from large downward moves. For the $n$-th term in the Merton series, the conditional put price is computed with volatility $\sigma_n$ and rate $r_n$. As $n$ increases:

The conditional mean shifts left (more negative $m_n$), moving $S_T$ toward the put's in-the-money region
The conditional variance increases ($\sigma_n^2 = \sigma^2 + n\sigma_J^2/T$), further spreading probability mass into the tail

Both effects increase the conditional put price for higher $n$ terms.

OTM calls (strike $K > S_0$) have payoff $(S_T - K)^+$ which benefits from large upward moves. Although higher variance spreads some mass upward, the leftward shift in the mean works against the call. When $\mu_J < 0$, the net effect on OTM calls is smaller because the mean shift and variance increase partially offset each other.

This asymmetry is precisely what generates the implied volatility skew: OTM puts have higher implied volatility than OTM calls, reflecting the market's pricing of downward jump risk.

Exercise 5. In the pure jump case ($\sigma = 0$), the $n = 0$ term has $\sigma_0 = 0$. Explain what happens to $C_{\text{BS}}(S_0, K, T, r_0, 0)$ and why this term produces a digital-like payoff. What is the economic interpretation of the $n = 0$ term in this limiting case?

Solution to Exercise 5

When $\sigma = 0$, the conditional volatility for $n = 0$ jumps is:

\[ \sigma_0 = \sqrt{0 + 0 \cdot \sigma_J^2/T} = 0 \]

The Black-Scholes formula with $\sigma_0 = 0$ becomes degenerate. With zero volatility, the log-return is deterministic (given $n = 0$, no jumps and no diffusion). The terminal price is:

\[ S_T = S_0 e^{(r_0 - \frac{1}{2} \cdot 0)T} = S_0 e^{r_0 T} \]

The Black-Scholes formula collapses to:

\[ C_{\text{BS}}(S_0, K, T, r_0, 0) = \max(S_0 e^{r_0 T} - K e^{-rT} \cdot e^{rT}, 0) \cdot e^{-rT} = e^{-rT}\max(S_0 e^{r_0 T} - K, 0) \]

This is a digital-like payoff: the call pays $S_0 e^{r_0 T} - K$ with certainty if $S_0 e^{r_0 T} > K$, and zero otherwise. There is no smooth transition as in the standard Black-Scholes formula; the price is either fully in the money or fully out.

Economic interpretation. The $n = 0$ term represents the scenario where no jumps occur during $[0, T]$. In the pure-jump model ($\sigma = 0$), this scenario has a completely deterministic outcome. The weight $w_0 = e^{-\lambda'T}$ is the probability of zero jumps. The total option price is then the sum over $n \geq 1$ of smoothly priced terms (each with nonzero $\sigma_n > 0$) plus the digital contribution from $n = 0$. Numerically, care is needed for the $n = 0$ term because the CDF evaluations $N(d_1)$ and $N(d_2)$ approach 0 or 1 as $\sigma_0 \to 0$.

Exercise 6. The Merton series formula expresses the option price as a mixture of Black-Scholes prices. Use this interpretation to prove that put-call parity $C - P = S_0 - Ke^{-rT}$ holds for the Merton model without computing the prices explicitly.

Solution to Exercise 6

The Merton series formula gives the call and put prices as:

\[ C_{\text{Merton}} = \sum_{n=0}^{\infty} w_n \, C_{\text{BS}}^{(n)}, \qquad P_{\text{Merton}} = \sum_{n=0}^{\infty} w_n \, P_{\text{BS}}^{(n)} \]

where $C_{\text{BS}}^{(n)}$ and $P_{\text{BS}}^{(n)}$ are the Black-Scholes call and put prices with parameters $(\sigma_n, r_n)$.

For each term, the standard Black-Scholes put-call parity holds:

\[ C_{\text{BS}}^{(n)} - P_{\text{BS}}^{(n)} = S_0 e^{(r_n - r)T} - Ke^{-rT} \]

Note that $e^{(r_n - r)T} = e^{(-\lambda\bar{k} + n\ln(1+\bar{k})/T)\cdot T} = e^{-\lambda\bar{k}T}(1+\bar{k})^n$. Therefore:

\[ C_{\text{Merton}} - P_{\text{Merton}} = \sum_{n=0}^{\infty} w_n \bigl[S_0 e^{(r_n - r)T} - Ke^{-rT}\bigr] \]

\[ = S_0 e^{-\lambda\bar{k}T}\sum_{n=0}^{\infty} w_n (1+\bar{k})^n - Ke^{-rT}\sum_{n=0}^{\infty} w_n \]

Since $\sum_{n=0}^{\infty} w_n = 1$ (the weights are probabilities), the second term gives $-Ke^{-rT}$.

For the first sum, substituting $w_n = \frac{e^{-\lambda'T}(\lambda'T)^n}{n!}$ with $\lambda' = \lambda(1+\bar{k})$:

\[ \sum_{n=0}^{\infty} w_n(1+\bar{k})^n = e^{-\lambda'T}\sum_{n=0}^{\infty}\frac{(\lambda'T)^n(1+\bar{k})^n}{n!} = e^{-\lambda'T}\cdot e^{\lambda'T(1+\bar{k})} = e^{\lambda'T\bar{k}} \]

Since $\lambda' = \lambda(1+\bar{k})$, we get $\lambda'T\bar{k} = \lambda\bar{k}(1+\bar{k})T$. Multiplying by $S_0 e^{-\lambda\bar{k}T}$:

\[ S_0 e^{-\lambda\bar{k}T} \cdot e^{\lambda\bar{k}(1+\bar{k})T} = S_0 e^{\lambda\bar{k}^2 T} \]

However, a simpler direct argument avoids this algebra entirely. The put-call parity $C - P = S_0 - Ke^{-rT}$ is model-independent: it follows from the identity $(S_T - K)^+ - (K - S_T)^+ = S_T - K$ and the fact that $\mathbb{E}^{\mathbb{Q}}[e^{-rT}S_T] = S_0$ (the discounted price is a martingale). Since the Merton model is constructed to preserve this martingale property (via the compensator), put-call parity holds automatically:

\[ C_{\text{Merton}} - P_{\text{Merton}} = e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T - K] = S_0 - Ke^{-rT} \]