Mellin Transform for the Black-Scholes PDE¶

Everything in this subsection follows from one sentence:

The Mellin transform in $x$ is the Fourier transform in $\log x$.

The Mellin transform follows from one simple calculation:

\[ S \frac{d}{dS}\, S^s = s \cdot S^s \]

Power laws are eigenfunctions of the scaling operator $S\, \partial_S$ — the multiplicative analogue of plane waves $e^{i\omega x}$ being eigenfunctions of $\partial_x$. The Mellin transform decomposes any function on $(0, \infty)$ into these power-law modes, and in those coordinates the variable-coefficient Black–Scholes operator collapses to a quadratic polynomial in the Mellin variable $s$.

We build the picture in the same order a reader naturally builds intuition: the toy eigenfunction calculation first, then the transform definition, and only then the Black–Scholes PDE in Mellin space.

1. Why Mellin? A Toy Mechanism¶

Before any finance, work the eigenvalue calculation. Consider the scaling operator

\[ D := S \frac{d}{dS} \]

acting on functions of $S \in (0, \infty)$. Apply $D$ to a power-law $S^s$ with $s \in \mathbb{C}$:

\[ D \cdot S^s = S \cdot s\, S^{s-1} = s \cdot S^s \]

Iterating,

\[ D^n \cdot S^s = s^n \cdot S^s, \qquad S^2 \frac{d^2}{dS^2}\, S^s = D(D - 1)\, S^s = s(s - 1)\, S^s \]

These two identities are the Mellin analogue of the Fourier eigenfunction relation $\partial_x\, e^{i\omega x} = i\omega \cdot e^{i\omega x}$ — they are the entire mechanism that makes the rest of the subsection work.

1.1 Why Mellin and Not Fourier?¶

The Fourier transform decomposes $f(x)$ into plane waves $e^{i\omega x}$ — the eigenfunctions of translation, $x \mapsto x + a$. Each plane wave is invariant (up to a phase) under shifting by $a$.

The Mellin transform decomposes $V(S)$ into power-laws $S^{-s}$ — the eigenfunctions of dilation, $S \mapsto \lambda S$. Each power-law $S^{-s}$ is invariant (up to a scaling factor $\lambda^{-s}$) under multiplication by $\lambda$.

Stock prices evolve multiplicatively: a $5\%$ return is the same whether $S = \$10$ or $\$1000$. The natural symmetry group of the Black–Scholes problem is therefore the multiplicative group $(0, \infty)$, not the additive group $\mathbb{R}$ — and Mellin is built for exactly that group.

1.2 Diagonalization Picture¶

A constant-coefficient operator in $x$ — say $-\partial_x^2 + \mu^2$ — commutes with translations and is diagonalized by Fourier. The Black–Scholes spatial operator

\[ \mathcal{L}_S = \frac{\sigma^2}{2} S^2 \frac{d^2}{dS^2} + r S \frac{d}{dS} - r \]

is not constant-coefficient in $S$ (the coefficients $S^2$ and $S$ depend explicitly on $S$), so Fourier in $S$ would not diagonalize it directly. But written in terms of the scaling operator $D = S\, \partial_S$,

\[ \mathcal{L}_S = \frac{\sigma^2}{2}\, D(D - 1) + r D - r \]

— a polynomial in $D$ alone — and every $D$-polynomial is diagonalized by power-law modes. Hence Mellin in $S$ diagonalizes the BS operator without first changing variables to $x = \ln S$.

In Mellin coordinates, the action on power-law modes becomes a scalar multiplication:

\[ \mathcal{L}_S\, S^{-s} = \Lambda(s)\, S^{-s}, \qquad \Lambda(s) = \frac{\sigma^2}{2}\, s(s - 1) - r s - r \]

(reading off from $D \cdot S^{-s} = -s \cdot S^{-s}$). The eigenvalue $\Lambda(s)$ is the Mellin symbol of the BS operator — analogous to the Fourier characteristic exponent $\psi(\omega)$.

Core principle

The Mellin transform decomposes functions on $(0, \infty)$ into power-law modes $S^{-s}$, which are joint eigenfunctions of the scaling operator $D = S\, \partial_S$ and of every operator that is a polynomial in $D$. For the Black–Scholes generator, this turns the second-order spatial operator into a quadratic polynomial $\Lambda(s)$ in the Mellin variable.

This is the mechanism. Black–Scholes is the application.

1.3 Mellin = Fourier in Disguise¶

The change of variable $x = \ln S$ converts the Mellin integral into a Fourier-type integral:

\[ \mathcal{M}[V](s) = \int_0^\infty V(S)\, S^{s - 1}\, dS \underset{x = \ln S}{=} \int_{-\infty}^\infty V(e^x)\, e^{s x}\, dx \]

— the Mellin transform of $V(S)$ at the point $s$ is the two-sided Laplace / generalized Fourier transform of the function $x \mapsto V(e^x)$. So Mellin in $S$ and Fourier in $x = \ln S$ are the same transform written in different coordinates; the toy eigenfunction identity $D\, S^s = s\, S^s$ is the spatial-side translation of $\partial_x\, e^{s x} = s\, e^{s x}$. The reason we keep Mellin as a separate development is that the multiplicative variable $S$ is the financially natural one: power-law modes $S^{-s}$ are gearing/moneyness exponents that map directly onto market language, whereas $e^{i\omega x}$ does not.

This subsection is therefore best read as an alternative perspective on the Black-Scholes formula rather than a self-contained derivation: the cleanest contour analysis still proceeds through the Fourier picture of § Fourier Transform. What the Mellin viewpoint adds is the multiplicative-symmetry interpretation and a clean operator-algebra story tied directly to financial coordinates.

2. Mellin Transform: Definition and Properties¶

Definition¶

The Mellin transform of a function $V(S)$ defined on $(0, \infty)$ is

\[ \boxed{\mathcal{M}[V](s) = \int_0^{\infty} V(S)\,S^{s-1}\,dS} \]

where $s \in \mathbb{C}$ lies in the strip of analyticity $c_1 < \text{Re}(s) < c_2$ determined by the growth of $V$ near $S = 0$ and $S = \infty$.

The inverse Mellin transform recovers $V$ via a Bromwich-type contour integral:

\[ \boxed{V(S) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\mathcal{M}[V](s)\,S^{-s}\,ds} \]

where the real number $c$ is chosen so the vertical contour $\text{Re}(s) = c$ lies within the strip of analyticity.

Transform Properties for Black-Scholes Operators¶

The key properties that make the Mellin transform effective for the Black-Scholes PDE are:

\[ \mathcal{M}\left[S\frac{\partial V}{\partial S}\right](s) = s\,\mathcal{M}[V](s) \]

\[ \mathcal{M}\left[S^2\frac{\partial^2 V}{\partial S^2}\right](s) = s(s-1)\,\mathcal{M}[V](s) \]

Both results follow from integration by parts, assuming that boundary terms at $S = 0$ and $S = \infty$ vanish. The essential point is that the Mellin transform converts the variable-coefficient differential operators $S\frac{d}{dS}$ and $S^2\frac{d^2}{dS^2}$ into polynomial multipliers in the transform variable $s$.

3. Transforming the Black-Scholes PDE¶

Recall

The Black-Scholes PDE and its terminal/boundary conditions are stated in § Introduction; the heat-equation form and risk-neutral framing are developed in § Heat Equation and § Feynman–Kac.

Apply the Mellin transform in $S$ directly to the BS PDE in $(S, t)$ variables, writing $\hat{V}(s,t) = \mathcal{M}[V](s,t)$:

\[ \frac{\partial \hat{V}}{\partial t} + rs\,\hat{V} + \frac{\sigma^2}{2}s(s-1)\,\hat{V} - r\,\hat{V} = 0 \]

Collecting terms:

\[ \boxed{\frac{\partial \hat{V}}{\partial t} + \Lambda(s)\,\hat{V} = 0} \]

where the Mellin symbol of the Black-Scholes operator is

\[ \boxed{\Lambda(s) = \frac{\sigma^2}{2}s^2 + \left(r - \frac{\sigma^2}{2}\right)s - r} \]

This is a first-order ODE in $t$ with $s$ appearing only as a parameter. The entire spatial structure of the PDE has been absorbed into the algebraic function $\Lambda(s)$.

General Solution¶

The ODE has the immediate solution

\[ \boxed{\hat{V}(s,t) = \hat{V}(s,T)\,e^{-\Lambda(s)(T-t)}} \]

With the terminal condition $V(S,T) = \Phi(S)$, we have $\hat{V}(s,T) = \mathcal{M}[\Phi](s)$, so the complete solution in Mellin space is

\[ \boxed{V(S,t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\mathcal{M}[\Phi](s)\,e^{-\Lambda(s)(T-t)}\,S^{-s}\,ds} \]

4. Mellin Solution for the European Call¶

Mellin Transform of the Call Payoff¶

For the European call payoff $\Phi(S) = (S - K)^+$, a direct calculation gives

\[ \mathcal{M}[(S-K)^+](s) = \int_K^{\infty}(S-K)\,S^{s-1}\,dS = \frac{K^{s+1}}{s(s+1)} \]

The integral converges at $S = \infty$ only when $\text{Re}(s) < -1$, so this identity is meaningful on the half-plane $\text{Re}(s) < -1$.

Analytic strip

The Mellin transform of an integrable function is, in general, holomorphic on a vertical strip $c_1 < \text{Re}(s) < c_2$ — its fundamental strip — determined by the decay of the function at $0$ and at $\infty$. For the call payoff $(S - K)^+$, the function vanishes for $S < K$ (so there is no constraint at $S = 0$) but grows linearly at infinity, forcing $\text{Re}(s) < -1$. Strictly speaking the "strip" here is the half-plane $\text{Re}(s) < -1$.

The inverse Mellin contour $\text{Re}(s) = c$ must lie inside this strip for the inversion integral to represent the original function. Recovering the option price by deforming the contour to the right (toward $\text{Re}(s) = 0$) crosses the poles at $s = -1$ and $s = 0$, which lie outside the fundamental strip. The standard residue calculation below treats this contour deformation formally: full justification requires bounding the integrand on closing arcs and checking that no other singularities (e.g. from $e^{-\Lambda(s)\tau}$, which is entire, or growth of $S^{-s}$) obstruct the deformation. We do not attempt that here.

Option Value in Mellin Space¶

Combining the payoff transform with the general solution gives

\[ \hat{C}(s,t) = \frac{K^{s+1}}{s(s+1)}\,e^{-\Lambda(s)\tau}, \qquad \tau = T-t \]

and the call price is recovered by

\[ C(S,t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{K^{s+1}}{s(s+1)}\,e^{-\Lambda(s)\tau}\,S^{-s}\,ds, \qquad c < -1 \]

The integrand has simple poles at $s = 0$ and $s = -1$, where $\Lambda(0) = -r$ and $\Lambda(-1) = \sigma^2 - 2r$. From § Heat Equation, the BS call has the form $C = S N(d_1) - Ke^{-r\tau} N(d_2)$ with $d_{1,2}$ as defined there; the Mellin inversion naturally produces this two-term structure because the two poles correspond (heuristically) to the two terms. The pole positions are visible from $\frac{1}{s(s+1)}$, but the $N(d_1)$ and $N(d_2)$ factors come from the imaginary-axis integral that survives the residue computation, not from the residues themselves.

Advanced Remark: Mellin residue calculation

Treating the contour deformation formally, the residues are

\[ \text{Res}_{s=0}\bigl[\hat{C}(s,t)\,S^{-s}\bigr] = K\,e^{r\tau}, \qquad \text{Res}_{s=-1}\bigl[\hat{C}(s,t)\,S^{-s}\bigr] = -S\,e^{(2r - \sigma^2)\tau} \]

A naive "sum of residues" gives the wrong answer: the BS call formula (see § Heat Equation) arises only after one keeps the contour integral and identifies the surviving Gaussian integrals along $\text{Re}(s) = c$ as the normal CDFs. Recovering those CDF factors is the same Gaussian completion-of-square computation carried out in § Feynman–Kac and § Fourier Transform. A fully rigorous treatment requires verifying decay of the integrand along closing arcs, checking that the deformed contour encloses no other singularities, and tracking convergence of the residual line integral.

5. Mellin-Fourier Duality¶

The substitution $S = e^x$ relates the two transforms directly:

\[ \mathcal{M}[V](s) = \int_{-\infty}^{\infty}V(e^x)\,e^{sx}\,dx = \mathcal{F}[V(e^x)](-is) \]

So a Mellin transform in $S$ is a Fourier transform in $x = \ln S$ with $\omega = -is$. This is why both transforms reduce the Black-Scholes PDE to an ODE: they are the same harmonic analysis, written in multiplicative ($S$) versus additive ($\ln S$) coordinates. The $e^{i\omega x}$ machinery developed in § Fourier Transform transfers verbatim under $\omega \leftrightarrow -is$.

The Mellin Parseval identity takes the form

\[ \int_0^{\infty}|V(S)|^2\,\frac{dS}{S} = \frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathcal{M}[V](c + i\omega)|^2\,d\omega \]

The measure $\frac{dS}{S}$ is the Haar measure on the multiplicative group $(0, \infty)$ — the right invariant measure under $S \mapsto \lambda S$ — confirming that the Mellin transform is the natural harmonic analysis on this group, just as the Fourier transform is natural on the additive group $\mathbb{R}$.

6. Mellin Transforms for Exotic Payoffs¶

Any payoff that is piecewise polynomial in $S$ has a Mellin transform expressible as a rational function of $s$ times powers of $K$, so the inversion remains tractable. For instance, the power option payoff $(S^n - K^n)^+$ transforms (see Exercise 5) to $\frac{nK^{n+s}}{s(s+n)}$, valid for $\text{Re}(s) < -n$. Within the pricing-semigroup framing of § Introduction, the Mellin approach is a multiplicative spectral representation that diagonalizes the semigroup on the basis of power functions $S^s$ — the multiplicative analogue of the Fourier exponentials $e^{i\omega x}$.

Exercises¶

Exercise 1. Compute the Mellin transform of the European put payoff $(K - S)^+$ and determine its strip of analyticity. Compare with the call payoff transform and relate the two via put-call parity in transform space.

Solution to Exercise 1

The Mellin transform of the European put payoff $(K - S)^+$ is:

\[ \mathcal{M}[(K-S)^+](s) = \int_0^{K}(K-S)\,S^{s-1}\,dS \]

\[ = K\int_0^K S^{s-1}\,dS - \int_0^K S^s\,dS = K\left[\frac{S^s}{s}\right]_0^K - \left[\frac{S^{s+1}}{s+1}\right]_0^K \]

The lower limits vanish when $\text{Re}(s) > 0$ (for the first integral) and $\text{Re}(s) > -1$ (for the second). So we need $\text{Re}(s) > 0$:

\[ = \frac{K^{s+1}}{s} - \frac{K^{s+1}}{s+1} = K^{s+1}\left(\frac{1}{s} - \frac{1}{s+1}\right) = \frac{K^{s+1}}{s(s+1)} \]

The strip of analyticity is $\text{Re}(s) > 0$.

Comparison with the call. The Mellin transform of the call payoff is also $\frac{K^{s+1}}{s(s+1)}$ but valid for $\text{Re}(s) < -1$. The two transforms have the same functional form on different strips of analyticity (put: $\text{Re}(s) > 0$; call: $\text{Re}(s) < -1$).

Put-call parity in transform space. Since $(S-K)^+ - (K-S)^+ = S - K$, and $\mathcal{M}[S](s)$ and $\mathcal{M}[K](s)$ are defined on their own strips, the transform of $S - K$ relates the two payoff transforms via analytic continuation across the strip $-1 < \text{Re}(s) < 0$ that separates them.

Exercise 2. Verify that the Mellin transform converts the Black-Scholes operator $\frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV$ into a polynomial in $s$ by computing each term separately via integration by parts.

Solution to Exercise 2

First term: $\mathcal{M}\left[S^2 V_{SS}\right](s)$. We use repeated integration by parts.

Starting from $\mathcal{M}[SV_S](s) = \int_0^{\infty}SV_S \cdot S^{s-1}\,dS = \int_0^{\infty}S^s V_S\,dS$. Integrating by parts with boundary terms vanishing:

\[ = \left[S^s V\right]_0^{\infty} - s\int_0^{\infty}S^{s-1}V\,dS = -s\,\hat{V}(s) \]

For the second-order term, using the identity $S^2 V_{SS} = \frac{d}{dS}(S^2 V_S) - 2SV_S$ and applying the Mellin transform, or equivalently using the standard result:

\[ \mathcal{M}\left[S^2 V_{SS}\right](s) = s(s-1)\,\hat{V}(s) \]

Note: The sign convention depends on the definition used. With the convention in the text where $\mathcal{M}[SV_S] = s\hat{V}$ (absorbing the sign into the operator convention), the full Black-Scholes operator becomes:

\[ \frac{\sigma^2}{2}s(s-1) + rs - r = \frac{\sigma^2}{2}s^2 + \left(r - \frac{\sigma^2}{2}\right)s - r = \Lambda(s) \]

This is a quadratic polynomial in $s$, confirming that the Mellin transform converts the variable-coefficient differential operator into an algebraic multiplier. $\square$

Exercise 3. The Mellin symbol $\Lambda(s) = \frac{\sigma^2}{2}s^2 + \left(r - \frac{\sigma^2}{2}\right)s - r$ has two real roots. Find them and interpret their financial meaning in terms of the growth rates of the homogeneous solutions $S^{-s}$ of the Black-Scholes equation.

Solution to Exercise 3

Setting $\Lambda(s) = 0$:

\[ \frac{\sigma^2}{2}s^2 + \left(r - \frac{\sigma^2}{2}\right)s - r = 0 \]

Using the quadratic formula:

\[ s = \frac{-\left(r - \frac{\sigma^2}{2}\right) \pm \sqrt{\left(r - \frac{\sigma^2}{2}\right)^2 + 2\sigma^2 r}}{\sigma^2} \]

The discriminant simplifies:

\[ \left(r - \frac{\sigma^2}{2}\right)^2 + 2\sigma^2 r = r^2 - r\sigma^2 + \frac{\sigma^4}{4} + 2r\sigma^2 = r^2 + r\sigma^2 + \frac{\sigma^4}{4} = \left(r + \frac{\sigma^2}{2}\right)^2 \]

Therefore:

\[ s = \frac{-(r - \frac{\sigma^2}{2}) \pm (r + \frac{\sigma^2}{2})}{\sigma^2} \]

The two roots are:

$s_+ = \frac{-(r - \frac{\sigma^2}{2}) + (r + \frac{\sigma^2}{2})}{\sigma^2} = \frac{\sigma^2}{\sigma^2} = 1$
$s_- = \frac{-(r - \frac{\sigma^2}{2}) - (r + \frac{\sigma^2}{2})}{\sigma^2} = \frac{-2r}{\sigma^2}$

Financial interpretation. The roots of $\Lambda$ correspond to solutions of the perpetual (time-independent) Black-Scholes equation $\frac{\sigma^2}{2}S^2 V'' + rSV' - rV = 0$. The Euler-type ansatz $V = S^\lambda$ yields $\Lambda(-\lambda) = 0$, so the homogeneous solutions are $V = S^{-s_+} = S^{-1}$ and $V = S^{-s_-} = S^{2r/\sigma^2}$.

However, one can equivalently parametrize by writing $\lambda = -s$, giving $\lambda_1 = -1$ (corresponding to $V = S^{-1}$, which is not financially meaningful by itself) and $\lambda_2 = 2r/\sigma^2$ (corresponding to $V = S^{2r/\sigma^2}$).

More directly: the factorization $\Lambda(s) = \frac{\sigma^2}{2}(s - 1)(s + \frac{2r}{\sigma^2})$ shows that $s = 1$ and $s = -2r/\sigma^2$ are the roots. These are distinct from the poles $s = 0$ and $s = -1$ of the call payoff transform. The root $s_+ = 1$ corresponds to $V(S) = S$, the trivially growing solution (holding the stock). The root $s_- = -2r/\sigma^2$ corresponds to $V(S) = S^{2r/\sigma^2}$, which appears in the pricing of perpetual American options. $\square$

Exercise 4. Using the Mellin-Fourier duality $\mathcal{M}[V](s) = \mathcal{F}[V(e^x)](-is)$, show that the Mellin symbol $\Lambda(s)$ and the Fourier characteristic exponent $\psi(\omega) = -\frac{\sigma^2 \omega^2}{2} + i\omega(r - \frac{\sigma^2}{2}) - r$ are related by $\Lambda(s) = \psi(-is)$.

Solution to Exercise 4

The Fourier characteristic exponent for the log-price Black-Scholes PDE is:

\[ \psi(\omega) = -\frac{\sigma^2\omega^2}{2} + i\omega\left(r - \frac{\sigma^2}{2}\right) - r \]

Substituting $\omega = -is$:

\[ \psi(-is) = -\frac{\sigma^2(-is)^2}{2} + i(-is)\left(r - \frac{\sigma^2}{2}\right) - r \]

\[ = -\frac{\sigma^2(-s^2)}{2} + s\left(r - \frac{\sigma^2}{2}\right) - r \]

\[ = \frac{\sigma^2 s^2}{2} + \left(r - \frac{\sigma^2}{2}\right)s - r = \Lambda(s) \]

This confirms $\Lambda(s) = \psi(-is)$, which is a direct consequence of the Mellin-Fourier duality. The substitution $\omega = -is$ (equivalently $s = i\omega$) maps the Fourier frequency variable to the Mellin complex variable, and the two transforms produce the same ODE structure under this correspondence. $\square$

Exercise 5. Compute the Mellin transform of the power option payoff $(S^n - K^n)^+$ for general $n > 0$ and determine the strip of analyticity. Show that the case $n = 1$ recovers the standard call payoff transform.

Solution to Exercise 5

The Mellin transform is:

\[ \mathcal{M}[(S^n - K^n)^+](s) = \int_{K}^{\infty}(S^n - K^n)\,S^{s-1}\,dS \]

\[ = \int_K^{\infty}S^{n+s-1}\,dS - K^n\int_K^{\infty}S^{s-1}\,dS \]

For convergence at infinity, we need $\text{Re}(n + s) < 0$ and $\text{Re}(s) < 0$, so $\text{Re}(s) < -n$. Under this condition:

\[ = -\frac{K^{n+s}}{n+s} - K^n\left(-\frac{K^s}{s}\right) = -\frac{K^{n+s}}{n+s} + \frac{K^{n+s}}{s} \]

\[ = K^{n+s}\left(\frac{1}{s} - \frac{1}{n+s}\right) = K^{n+s}\cdot\frac{n}{s(n+s)} \]

Therefore:

\[ \mathcal{M}[(S^n - K^n)^+](s) = \frac{nK^{n+s}}{s(n+s)} \]

valid for $\text{Re}(s) < -n$.

Verification for $n = 1$:

\[ \frac{1 \cdot K^{1+s}}{s(1+s)} = \frac{K^{s+1}}{s(s+1)} \]

This matches the call payoff transform $\frac{K^{s+1}}{s(s+1)}$ with strip $\text{Re}(s) < -1$. $\square$