Skip to content

Mellin Transform for the Black-Scholes PDE

The Black-Scholes PDE has variable coefficients \(rS\frac{\partial V}{\partial S}\) and \(\frac{\sigma^2}{2}S^2\frac{\partial^2 V}{\partial S^2}\) that arise because stock prices evolve multiplicatively: a percentage return is the same whether the stock is at $10 or $1000. The Fourier transform is the natural tool for additive processes on \((-\infty, \infty)\), while the Mellin transform is the natural tool for multiplicative processes on \((0, \infty)\). Where the Fourier transform diagonalizes the constant-coefficient operators \(\frac{\partial}{\partial x}\) and \(\frac{\partial^2}{\partial x^2}\), the Mellin transform diagonalizes the variable-coefficient operators \(S\frac{\partial}{\partial S}\) and \(S^2\frac{\partial^2}{\partial S^2}\) that appear directly in the Black-Scholes equation. This allows us to work with the stock price \(S\) itself, without the logarithmic change of variable \(x = \ln S\) that other transform methods require.

This section develops the Mellin transform approach to solving the Black-Scholes PDE, derives the European call price via Mellin inversion, and establishes the precise duality between the Mellin and Fourier transforms.

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 \(s\).

Transforming the Black-Scholes PDE

The Black-Scholes PDE in original \((S, t)\) variables is

\[ \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{\sigma^2}{2}S^2\frac{\partial^2 V}{\partial S^2} - rV = 0 \]

Apply the Mellin transform in \(S\), 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

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

Mellin Solution for the European Call

Mellin Transform of the Call Payoff

For the European call payoff \(\Phi(S) = (S - K)^+\):

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

Expanding:

\[ = \int_K^{\infty}S^s\,dS - K\int_K^{\infty}S^{s-1}\,dS \]
\[ = \left[\frac{S^{s+1}}{s+1}\right]_K^{\infty} - K\left[\frac{S^s}{s}\right]_K^{\infty} \]

For convergence at infinity, we need \(\text{Re}(s+1) < 0\) and \(\text{Re}(s) < 0\), so \(\text{Re}(s) < -1\). Under this condition:

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

valid for \(\text{Re}(s) < -1\).

Option Value in Mellin Space

Combining the payoff transform with the general solution:

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

where \(\tau = T - t\) and \(\Lambda(s) = \frac{\sigma^2}{2}s^2 + \left(r - \frac{\sigma^2}{2}\right)s - r\).

Mellin Inversion via Residue Calculus

The call price is recovered by the inverse Mellin transform:

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

where \(c < -1\) places the contour in the strip of analyticity.

Finding the Poles

The integrand has simple poles at:

  • \(s = 0\), where \(\Lambda(0) = -r\)
  • \(s = -1\), where \(\Lambda(-1) = \frac{\sigma^2}{2} - r + \frac{\sigma^2}{2} - r = \sigma^2 - 2r\)

Residue at \(s = 0\)

\[ \text{Res}_{s=0} = \lim_{s \to 0}\,s \cdot \frac{K^{s+1}}{s(s+1)}\,e^{-\Lambda(s)\tau}\,S^{-s} \]
\[ = \frac{K}{1} \cdot e^{-(-r)\tau} \cdot 1 = Ke^{r\tau} \]

Residue at \(s = -1\)

\[ \text{Res}_{s=-1} = \lim_{s \to -1}\,(s+1) \cdot \frac{K^{s+1}}{s(s+1)}\,e^{-\Lambda(s)\tau}\,S^{-s} \]
\[ = \frac{K^0}{-1} \cdot e^{-(\sigma^2 - 2r)\tau} \cdot S = -S\,e^{(2r - \sigma^2)\tau} \]

Recovery of the Black-Scholes Formula

The standard approach decomposes the call price into two probability terms. Writing

\[ C(S,t) = S\,\Pi_1 - Ke^{-r\tau}\,\Pi_2 \]

where \(\Pi_1\) and \(\Pi_2\) are computed via separate Mellin inversions, one obtains after contour integration:

\[ \boxed{C(S,t) = S\,N(d_1) - Ke^{-r\tau}\,N(d_2)} \]

where \(d_1\) and \(d_2\) are the standard Black-Scholes quantities. The Mellin transform automatically generates the two-term structure of the Black-Scholes formula, with each term arising from a separate pole contribution.

Mellin-Fourier Duality

The Connection

The substitution \(S = e^x\) reveals the precise relationship between the Mellin and Fourier transforms:

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

Therefore:

\[ \text{Mellin in } S = \text{Fourier in } x = \ln S \text{ with } \omega = -is \]

The Mellin inverse correspondingly becomes

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

This duality explains why both transforms reduce the Black-Scholes PDE to an ODE: they are the same transform applied in different coordinates (multiplicative vs. additive).

Parseval's Theorem for the Mellin Transform

The Mellin transform preserves energy in the following sense:

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

The measure \(\frac{dS}{S}\) is the Haar measure on the multiplicative group \((0,\infty)\), confirming that the Mellin transform is the natural harmonic analysis on this group.

Convolution Theorem

The Mellin transform of a multiplicative convolution satisfies

\[ \mathcal{M}[V \cdot W](s) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\mathcal{M}[V](\xi)\,\mathcal{M}[W](s-\xi)\,d\xi \]

This is the analogue of the Fourier convolution theorem, adapted to the multiplicative structure.

Mellin Transforms for Exotic Payoffs

The Mellin transform extends naturally to other payoff structures. For power options with payoff \((S^n - K^n)^+\):

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

The solution structure parallels the European call but with modified parameters in \(\Lambda(s)\). More generally, any payoff that is piecewise polynomial in \(S\) admits a Mellin transform expressible in terms of rational functions of \(s\) and powers of \(K\), making the inversion tractable via residue calculus. In the operator framework of the introduction, the Mellin approach provides a multiplicative spectral representation of the pricing semigroup \(\mathcal{P}_\tau = e^{\tau\mathcal{L}}\), diagonalizing it in the basis of power functions \(S^s\).

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 \(\text{Re}(s) < -1\). \(\square\)