Characteristic Function¶

Fourier methods are among the most efficient tools for pricing options under models more complex than Black-Scholes. The starting point is the characteristic function of the log-price, which encodes the entire probability distribution in a single complex-valued function. For the Merton jump-diffusion model, the characteristic function has a closed-form expression that combines the Gaussian characteristic function of the diffusion with the Levy-Khintchine formula for the jump component.

Learning Objectives

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

Define the characteristic function and explain its role in Fourier-based option pricing
Derive the characteristic function of the Merton jump-diffusion log-return
Identify the Levy-Khintchine structure and interpret each component
Extract cumulants from the characteristic function and connect them to moments

Motivation¶

Why Characteristic Functions¶

The probability density of the Merton log-return is a Gaussian mixture (sum over Poisson-weighted normals). While the density itself is an infinite series, the characteristic function takes a compact closed form. This matters for two practical reasons:

Fourier pricing: The Carr-Madan formula and the COS method express option prices as integrals involving the characteristic function, which can be evaluated by FFT in $O(N \log N)$ operations.
Model comparison: Different jump-diffusion and stochastic volatility models are most easily compared through their characteristic functions, since the pricing machinery is the same regardless of the model.

Review of Characteristic Functions¶

For a real-valued random variable $X$, the characteristic function is:

\[ \phi_X(u) = \mathbb{E}[e^{iuX}], \quad u \in \mathbb{R} \]

Key properties:

$\phi_X(0) = 1$ and $|\phi_X(u)| \leq 1$ for all $u$
$\phi_X$ uniquely determines the distribution of $X$
If $X$ and $Y$ are independent, then $\phi_{X+Y}(u) = \phi_X(u) \cdot \phi_Y(u)$
The $n$-th moment (if it exists) satisfies $\mathbb{E}[X^n] = \frac{1}{i^n}\phi_X^{(n)}(0)$

Characteristic Function of the Merton Log-Return¶

Setup¶

From the Ito formula for jump processes, the log-return over $[0, T]$ is:

\[ x_T \equiv \ln\frac{S_T}{S_0} = \underbrace{\left(r - \lambda\bar{k} - \frac{1}{2}\sigma^2\right)T}_{\text{drift}} + \underbrace{\sigma W_T}_{\text{diffusion}} + \underbrace{\sum_{i=1}^{N_T}\ln Y_i}_{\text{jumps}} \]

The three components are mutually independent (Brownian motion, Poisson process, and jump sizes are independent by assumption).

Derivation¶

Theorem: Characteristic Function of the Merton Log-Return

The characteristic function of the log-return $x_T$ under $\mathbb{Q}$ is

\[ \phi_{x_T}(u) = \exp\!\left[iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T\!\left(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1\right)\right] \]

where $\mu' = r - \lambda\bar{k} - \frac{1}{2}\sigma^2$ is the adjusted drift.

Proof. By independence of the three components:

\[ \phi_{x_T}(u) = \mathbb{E}[e^{iux_T}] = e^{iu\mu'T} \cdot \mathbb{E}[e^{iu\sigma W_T}] \cdot \mathbb{E}\!\left[e^{iu\sum_{i=1}^{N_T}\ln Y_i}\right] \]

Factor 1 (drift): The deterministic drift contributes $e^{iu\mu'T}$.

Factor 2 (diffusion): Since $\sigma W_T \sim N(0, \sigma^2 T)$, the characteristic function of a normal is:

\[ \mathbb{E}[e^{iu\sigma W_T}] = e^{-\frac{1}{2}\sigma^2 u^2 T} \]

Factor 3 (jumps): This is the characteristic function of a compound Poisson process with jump sizes $\ln Y_i \sim N(\mu_J, \sigma_J^2)$. Using the compound Poisson MGF result (replacing the real argument by $iu$):

\[ \mathbb{E}\!\left[e^{iu\sum_{i=1}^{N_T}\ln Y_i}\right] = \exp\!\left[\lambda T\bigl(\phi_{\ln Y}(u) - 1\bigr)\right] \]

where $\phi_{\ln Y}(u) = \mathbb{E}[e^{iu\ln Y}] = e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$ is the characteristic function of a $N(\mu_J, \sigma_J^2)$ random variable.

Multiplying the three factors:

\[ \phi_{x_T}(u) = \exp\!\left[iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T\!\left(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1\right)\right] \]

$\square$

The Levy-Khintchine Representation¶

General Framework¶

The Levy-Khintchine theorem states that the characteristic function of any Levy process $L_t$ (a process with stationary, independent increments and cadlag paths) has the form:

\[ \phi_{L_t}(u) = \exp\!\left[t\left(iu\gamma - \frac{1}{2}\sigma^2 u^2 + \int_{\mathbb{R}\setminus\{0\}}\bigl(e^{iuy} - 1 - iuy\mathbf{1}_{|y|<1}\bigr)\nu(dy)\right)\right] \]

where:

$\gamma \in \mathbb{R}$ is the drift
$\sigma^2 \geq 0$ is the Gaussian variance
$\nu$ is the Levy measure, satisfying $\int \min(1, y^2)\,\nu(dy) < \infty$

The triple $(\gamma, \sigma^2, \nu)$ is called the Levy-Khintchine triplet and uniquely characterizes the process.

Merton Model as a Levy Process¶

The Merton log-return $x_T$ is a Levy process with triplet:

Component	Value
Drift $\gamma$	$r - \lambda\bar{k} - \frac{1}{2}\sigma^2 + \lambda\int_{
Gaussian variance $\sigma^2$	Diffusion variance
Levy measure $\nu(dy)$	$\lambda \cdot \frac{1}{\sigma_J\sqrt{2\pi}} e^{-(y-\mu_J)^2/(2\sigma_J^2)}\,dy$

The Levy measure $\nu$ is finite: $\nu(\mathbb{R}) = \lambda < \infty$. This means the Merton model is a finite-activity jump process --- there are finitely many jumps in any bounded time interval (almost surely). This is in contrast to infinite-activity models such as Variance Gamma or CGMY, where the Levy measure has infinite total mass and infinitely many small jumps accumulate in finite time.

Finite Activity Simplifies the Levy-Khintchine Formula

When $\nu(\mathbb{R}) < \infty$, the truncation indicator $\mathbf{1}_{|y|<1}$ in the Levy-Khintchine formula can be dropped (both integrals converge), and the drift term simplifies. The characteristic exponent becomes:

\[ \psi(u) = iu\mu' - \frac{1}{2}\sigma^2 u^2 + \lambda\bigl(\phi_{\ln Y}(u) - 1\bigr) \]

so that $\phi_{x_T}(u) = e^{T\psi(u)}$.

Cumulants¶

Definition¶

The cumulant generating function (for the characteristic function) is:

\[ \Psi(u) = \ln \phi_{x_T}(u) = iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T\!\left(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1\right) \]

The $n$-th cumulant $\kappa_n$ is extracted by Taylor-expanding $\Psi(u)$ around $u = 0$:

\[ \Psi(u) = \sum_{n=1}^{\infty} \kappa_n \frac{(iu)^n}{n!} \]

Explicit Cumulants¶

Expanding the exponential $e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$ and collecting powers of $iu$:

First cumulant (mean):

\[ \kappa_1 = \left(r - \lambda\bar{k} - \frac{1}{2}\sigma^2 + \lambda\mu_J\right)T \]

Second cumulant (variance):

\[ \kappa_2 = \left(\sigma^2 + \lambda(\mu_J^2 + \sigma_J^2)\right)T \]

Third cumulant:

\[ \kappa_3 = \lambda(\mu_J^3 + 3\mu_J\sigma_J^2)T \]

Fourth cumulant:

\[ \kappa_4 = \lambda(\mu_J^4 + 6\mu_J^2\sigma_J^2 + 3\sigma_J^4)T \]

The general pattern is that the $n$-th cumulant of the jump component equals $\lambda T$ times the $n$-th moment of $\ln Y \sim N(\mu_J, \sigma_J^2)$, while the diffusion contributes only to $\kappa_1$ and $\kappa_2$.

Cumulants and Distribution Shape¶

The standardized cumulants give the skewness and excess kurtosis:

\[ \text{Skewness} = \frac{\kappa_3}{\kappa_2^{3/2}}, \qquad \text{Excess kurtosis} = \frac{\kappa_4}{\kappa_2^2} \]

Both quantities are determined entirely by the jump parameters $(\lambda, \mu_J, \sigma_J)$ and the diffusion volatility $\sigma$. As $T \to 0$, the skewness diverges as $T^{-1/2}$ and the kurtosis diverges as $T^{-1}$, reflecting the dominance of jumps at short time scales.

Connection to Fourier Pricing¶

The Carr-Madan Formula¶

The price of a European call with strike $K$ and maturity $T$ can be expressed as:

\[ C(K) = \frac{e^{-rT-\alpha k}}{\pi}\int_0^{\infty} e^{-ivk}\frac{\phi_{x_T}(v - (\alpha+1)i)}{(\alpha + iv)(\alpha + 1 + iv)}\,dv \]

where $k = \ln K$, $\alpha > 0$ is a damping parameter, and $\phi_{x_T}$ is the characteristic function derived above. This integral is evaluated numerically using the FFT, producing option prices across a grid of strikes in $O(N\log N)$ operations.

Why the Closed-Form Characteristic Function Matters¶

Without a closed-form characteristic function, Fourier pricing requires numerical evaluation of $\phi$ at each quadrature point, which itself might involve Monte Carlo or PDE solvers. The Merton model's analytical $\phi$ makes FFT pricing extremely fast --- typically under one millisecond for a full strike grid.

Worked Example¶

Computing the Characteristic Function

Consider the parameters $r = 0.05$, $\sigma = 0.20$, $\lambda = 0.5$, $\mu_J = -0.10$, $\sigma_J = 0.30$, $T = 1$.

Step 1: Adjusted drift.

\[ \bar{k} = e^{-0.10 + 0.045} - 1 \approx -0.0535 \]

\[ \mu' = 0.05 - 0.5(-0.0535) - \frac{1}{2}(0.04) = 0.05 + 0.0268 - 0.02 = 0.0568 \]

Step 2: Evaluate $\phi_{x_1}(u)$ at $u = 1$.

\[ \phi_{x_1}(1) = \exp\!\left[i(0.0568) - \frac{1}{2}(0.04) + 0.5\!\left(e^{-0.10i - 0.045} - 1\right)\right] \]

Computing the jump factor:

\[ e^{-0.10i - 0.045} = e^{-0.045}(\cos 0.10 - i\sin 0.10) \approx 0.956(0.995 - 0.0998i) \approx 0.951 - 0.0954i \]

\[ 0.5(0.951 - 0.0954i - 1) = 0.5(-0.049 - 0.0954i) = -0.0245 - 0.0477i \]

\[ \phi_{x_1}(1) = \exp\!\left[(0.0568i - 0.02 - 0.0245 - 0.0477i)\right] = \exp\!\left[-0.0445 + 0.0091i\right] \]

\[ |\phi_{x_1}(1)| = e^{-0.0445} \approx 0.9565 \]

Step 3: Cumulants.

\[ \kappa_1 = (0.0568 + 0.5 \times (-0.10)) \times 1 = 0.0068 \]

\[ \kappa_2 = (0.04 + 0.5(0.01 + 0.09)) \times 1 = 0.04 + 0.05 = 0.09 \]

\[ \kappa_3 = 0.5(-0.001 + 3(-0.10)(0.09)) \times 1 = 0.5(-0.001 - 0.027) = -0.014 \]

\[ \text{Skewness} = \frac{-0.014}{0.09^{3/2}} = \frac{-0.014}{0.027} \approx -0.519 \]

The negative skewness confirms the leftward asymmetry from downward jumps.

Summary¶

The characteristic function of the Merton log-return has the closed-form expression $\phi_{x_T}(u) = \exp[iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1)]$, combining a Gaussian factor from the diffusion with an exponential factor from the compound Poisson jumps. This structure is a special case of the Levy-Khintchine formula for finite-activity Levy processes. The cumulants extracted from the characteristic function reveal that jumps contribute all cumulants of order three and above, generating the negative skewness and excess kurtosis that distinguish the Merton model from Black-Scholes. The closed-form characteristic function enables efficient Fourier-based option pricing via the Carr-Madan formula and FFT methods.

Exercises¶

Exercise 1. Derive the characteristic function of the Merton log-return by computing $\mathbb{E}[e^{iux_T}]$ as a product of three independent factors: the deterministic drift, the Gaussian diffusion, and the compound Poisson jumps. Verify that the jump factor is $\exp[\lambda T(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1)]$.

Solution to Exercise 1

The log-return decomposes as $x_T = \mu'T + \sigma W_T + \sum_{i=1}^{N_T}\ln Y_i$ where $\mu' = r - \lambda\bar{k} - \frac{1}{2}\sigma^2$. The three components are mutually independent, so

\[ \phi_{x_T}(u) = \mathbb{E}[e^{iux_T}] = e^{iu\mu'T} \cdot \mathbb{E}[e^{iu\sigma W_T}] \cdot \mathbb{E}\!\left[e^{iu\sum_{i=1}^{N_T}\ln Y_i}\right] \]

Factor 1 (drift): The deterministic drift contributes $e^{iu\mu'T}$.

Factor 2 (diffusion): Since $\sigma W_T \sim N(0, \sigma^2 T)$, the characteristic function of a normal random variable $X \sim N(0, \sigma^2 T)$ is $\mathbb{E}[e^{iuX}] = e^{-\frac{1}{2}\sigma^2 u^2 T}$.

Factor 3 (jumps): Condition on $N_T = n$:

\[ \mathbb{E}\!\left[e^{iu\sum_{i=1}^{N_T}\ln Y_i}\right] = \sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}e^{-\lambda T}\,\left[\phi_{\ln Y}(u)\right]^n \]

where $\phi_{\ln Y}(u) = \mathbb{E}[e^{iu\ln Y}]$. Since $\ln Y \sim N(\mu_J, \sigma_J^2)$, we have $\phi_{\ln Y}(u) = e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$. Summing the series:

\[ \sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}e^{-\lambda T}\left[\phi_{\ln Y}(u)\right]^n = e^{-\lambda T}\exp\!\left[\lambda T\,\phi_{\ln Y}(u)\right] = \exp\!\left[\lambda T\left(\phi_{\ln Y}(u) - 1\right)\right] \]

Substituting $\phi_{\ln Y}(u) = e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$ gives the jump factor $\exp[\lambda T(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1)]$.

Multiplying all three factors:

\[ \phi_{x_T}(u) = \exp\!\left[iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T\!\left(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1\right)\right] \]

Exercise 2. Extract the first four cumulants from the characteristic function $\Psi(u) = \ln \phi_{x_T}(u)$ by Taylor-expanding around $u = 0$. Verify that $\kappa_1 = (r - \lambda\bar{k} - \frac{1}{2}\sigma^2 + \lambda\mu_J)T$ and $\kappa_2 = (\sigma^2 + \lambda(\mu_J^2 + \sigma_J^2))T$.

Solution to Exercise 2

The cumulant generating function is $\Psi(u) = \ln\phi_{x_T}(u)$. We need to expand $\Psi(u) = \sum_{n=1}^{\infty}\kappa_n\frac{(iu)^n}{n!}$ by Taylor-expanding around $u = 0$.

Write $\Psi(u) = iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1)$.

Expand $e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$ around $u = 0$. Let $w = iu\mu_J - \frac{1}{2}\sigma_J^2 u^2$, so $e^w = 1 + w + \frac{w^2}{2} + \frac{w^3}{6} + \frac{w^4}{24} + \cdots$

Collecting powers of $(iu)$: $w = (iu)\mu_J + \frac{1}{2}\sigma_J^2(iu)^2$, so

\[ e^w - 1 = (iu)\mu_J + \frac{1}{2}(\sigma_J^2 + \mu_J^2)(iu)^2 + \frac{1}{6}(\mu_J^3 + 3\mu_J\sigma_J^2)(iu)^3 + \cdots \]

First cumulant ($\kappa_1$): From the coefficient of $iu$:

\[ \kappa_1 = \mu'T + \lambda T\mu_J = (r - \lambda\bar{k} - \tfrac{1}{2}\sigma^2 + \lambda\mu_J)T \]

Second cumulant ($\kappa_2$): From the coefficient of $(iu)^2/2!$, noting $-\frac{1}{2}\sigma^2 u^2 T = \frac{1}{2}\sigma^2 T(iu)^2$:

\[ \kappa_2 = \sigma^2 T + \lambda T(\sigma_J^2 + \mu_J^2) = (\sigma^2 + \lambda(\mu_J^2 + \sigma_J^2))T \]

Third cumulant ($\kappa_3$): From the coefficient of $(iu)^3/3!$:

\[ \kappa_3 = \lambda(\mu_J^3 + 3\mu_J\sigma_J^2)T \]

Fourth cumulant ($\kappa_4$): From the coefficient of $(iu)^4/4!$:

\[ \kappa_4 = \lambda(\mu_J^4 + 6\mu_J^2\sigma_J^2 + 3\sigma_J^4)T \]

The first two cumulants match the stated formulas.

Exercise 3. The Merton model is a finite-activity Levy process because $\nu(\mathbb{R}) = \lambda < \infty$. Explain the distinction between finite-activity and infinite-activity processes. Name one infinite-activity Levy model and describe how its characteristic function differs structurally from the Merton model's.

Solution to Exercise 3

A finite-activity Levy process has a Levy measure $\nu$ with $\nu(\mathbb{R}) < \infty$. This means there are finitely many jumps in any bounded time interval (almost surely). The Merton model has $\nu(\mathbb{R}) = \lambda < \infty$, so it is finite-activity. The jumps arrive as a Poisson process with rate $\lambda$, and the expected number of jumps in $[0, T]$ is $\lambda T$.

An infinite-activity Levy process has $\nu(\mathbb{R}) = \infty$, meaning infinitely many (small) jumps accumulate in any finite time interval. The truncation indicator $\mathbf{1}_{|y|<1}$ in the Levy-Khintchine formula is essential in this case because the integral $\int y\,\nu(dy)$ may diverge.

An example of an infinite-activity model is the Variance Gamma (VG) process, where $\nu(dy) = C\frac{e^{-G|y|}}{|y|}\mathbf{1}_{y<0}\,dy + C\frac{e^{-My}}{y}\mathbf{1}_{y>0}\,dy$ with $C, G, M > 0$. The Levy measure has a $1/|y|$ singularity at the origin, so $\int_{\mathbb{R}}\nu(dy) = \infty$. Structurally, the VG characteristic function is $\phi_{X_T}(u) = \left(\frac{GM}{(G+iu)(M-iu)}\right)^{CT}$, which involves a rational function raised to a power, rather than the exponential-of-exponential structure of the Merton model. The VG model has no Brownian component ($\sigma = 0$); all variation comes from the infinite accumulation of small jumps.

Exercise 4. For $r = 0.05$, $\sigma = 0.20$, $\lambda = 0.5$, $\mu_J = -0.10$, $\sigma_J = 0.30$, $T = 1$: (a) Compute the skewness and excess kurtosis of the log-return. (b) Repeat for $T = 0.1$ (one month). (c) Verify that the skewness scales as $T^{-1/2}$ and the kurtosis scales as $T^{-1}$.

Solution to Exercise 4

(a) Using the cumulant formulas with $r = 0.05$, $\sigma = 0.20$, $\lambda = 0.5$, $\mu_J = -0.10$, $\sigma_J = 0.30$, $T = 1$:

\[ \kappa_2 = (0.04 + 0.5(0.01 + 0.09)) \times 1 = 0.04 + 0.05 = 0.09 \]

\[ \kappa_3 = 0.5((-0.10)^3 + 3(-0.10)(0.09)) \times 1 = 0.5(-0.001 - 0.027) = -0.014 \]

\[ \kappa_4 = 0.5(0.0001 + 6(0.01)(0.09) + 3(0.0081)) \times 1 = 0.5(0.0001 + 0.0054 + 0.0243) = 0.0149 \]

\[ \text{Skewness} = \frac{\kappa_3}{\kappa_2^{3/2}} = \frac{-0.014}{0.09^{3/2}} = \frac{-0.014}{0.02700} \approx -0.519 \]

\[ \text{Excess kurtosis} = \frac{\kappa_4}{\kappa_2^2} = \frac{0.0149}{0.0081} \approx 1.840 \]

(b) For $T = 0.1$: all cumulants scale linearly with $T$, so $\kappa_2 = 0.009$, $\kappa_3 = -0.0014$, $\kappa_4 = 0.00149$.

\[ \text{Skewness}_{T=0.1} = \frac{-0.0014}{0.009^{3/2}} = \frac{-0.0014}{0.000854} \approx -1.639 \]

\[ \text{Excess kurtosis}_{T=0.1} = \frac{0.00149}{0.009^2} = \frac{0.00149}{0.000081} \approx 18.40 \]

(c) Verification: The skewness ratio is $(-1.639)/(-0.519) \approx 3.162 \approx \sqrt{10} = (1/0.1)^{1/2}/(1/1)^{1/2}$, confirming the $T^{-1/2}$ scaling. The kurtosis ratio is $18.40/1.840 = 10 = T_1/T_2 = 1/0.1$, confirming the $T^{-1}$ scaling.

Exercise 5. The Carr-Madan formula $C(K) = \frac{e^{-rT-\alpha k}}{\pi}\int_0^{\infty} e^{-ivk}\frac{\phi_{x_T}(v - (\alpha+1)i)}{(\alpha + iv)(\alpha + 1 + iv)}\,dv$ requires evaluating the characteristic function at complex arguments. Show that $\phi_{x_T}(u)$ can be analytically continued to complex $u$ and remains well-defined for $\text{Im}(u) \in (-\alpha - 1, 0)$ provided $\alpha > 0$.

Solution to Exercise 5

The characteristic function $\phi_{x_T}(u) = \exp[iu\mu'T - \frac{1}{2}\sigma^2 u^2 T + \lambda T(e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2} - 1)]$ is defined for real $u$, but the expression makes sense for any complex $u$ as long as the exponent converges.

Let $u = a + ib$ where $a = \text{Re}(u)$ and $b = \text{Im}(u)$. The critical term for convergence is $e^{iu\mu_J - \frac{1}{2}\sigma_J^2 u^2}$. Substituting $u = a + ib$:

\[ iu\mu_J = i(a+ib)\mu_J = ia\mu_J - b\mu_J \]

\[ -\frac{1}{2}\sigma_J^2 u^2 = -\frac{1}{2}\sigma_J^2(a^2 - b^2 + 2iab) \]

The real part of the exponent of the jump term is $-b\mu_J - \frac{1}{2}\sigma_J^2(a^2 - b^2)$, which is finite for all finite $a, b$.

For the overall characteristic function to be well-defined, we also need $\mathbb{E}[e^{iuX_T}]$ to converge. With $u = a + ib$, this requires $\mathbb{E}[e^{-bX_T}] < \infty$. The exponential moment exists when $-b$ lies within the domain of the moment generating function of $x_T$. For the Merton model, we need the jump component $\mathbb{E}[e^{-b\ln Y}] = \mathbb{E}[Y^{-b}] = e^{-b\mu_J + b^2\sigma_J^2/2} < \infty$, which holds for all $b \in \mathbb{R}$ since $\ln Y$ is Gaussian.

In the Carr-Madan formula, the argument is $v - (\alpha+1)i$, so $b = -(\alpha+1)$. This requires $\mathbb{E}[S_T^{\alpha+1}] < \infty$, which holds for the Merton model for any finite $\alpha > 0$. Therefore $\phi_{x_T}(u)$ is well-defined for $\text{Im}(u) \in (-\alpha-1, 0)$ as required.

Component	Value
Drift \(\gamma\)	$r - \lambda\bar{k} - \frac{1}{2}\sigma^2 + \lambda\int_{
Gaussian variance \(\sigma^2\)	Diffusion variance
Levy measure \(\nu(dy)\)	\(\lambda \cdot \frac{1}{\sigma_J\sqrt{2\pi}} e^{-(y-\mu_J)^2/(2\sigma_J^2)}\,dy\)