Mellin Transform

The Mellin transform and its inverse are unary function operations defined by

$M(f)(s) = \int_0^\infty f(x) x^{s-1} dx$

$M^{-1}(F)(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} F(s) x^{-s} ds$

where $f$ is a usually a real-valued function with domain $[0,\infty)$ , $s$ is a complex variable, $x$ is a real variable and $F$ is a complex-valued function. Under technical assumptions and a judicious choice of the complex value $c$ , these transforms exist.

A number of properties associated with the Mellin transform

linearity:

$M(af+bg)(s) = aM(f)(s) + bM(g)(s)$

linear scaling input:

$M(f(kx))(s) = k^{-s} M(f)(s)$

power scaling input:

$M( f(x^\alpha))(s) = |\alpha|^{-1} M(f)(s/\alpha)$

power multiplication output:

$M(x^\alpha f(x))(s) = M(f)(s+\alpha)$

differentiation:

$M((-D)^n f(x))(s) = s^n M(f)(s)$ where $D = x\frac{d}{dx}$

log-volution:

$M(f \star g) = M(f) M(g)$

where $(f \star g)(x) = \int_0^\infty f(y) g(xy^{-1}) y^{-1} dy$

log-normal:

$M(\phi(log x))(s) = e^{\frac{1}{2} s^2}$

option-related

$M(x^\gamma \mathbb{I}(x<k))(s) = k^{s+\gamma}(s + \gamma)^{-1}$

delta-function

$M(\delta(x-1)) = 1$

Transforming the BS-PDE using Mellin Transform

The Black Scholes PDE is given by

$v_\tau = -rv + rx v_x + \frac{1}{2} \sigma^2 x^2 v_xx$

$v(x,0) = f(x)$

where $v = V(x,\tau)$ is the option price expressed as a function of the underlying asset price $x$ and time-to-expiry $\tau$ and $f(x)$ is the payoff at expiry. Since the intention is to use the Mellin transform, note that

$x v_x = Dv$

$x^2 v_xx = D^2 v - Dv$

and so

$v_\tau = -rv - r(-D)v + \frac{1}{2} \sigma^2 (D^2 v - Dv) \\ = -rv - (r - \frac{1}{2} \sigma^2) (-D) v + \frac{1}{2} \sigma^2 D^2 v$

Applying the Mellin Transform to the equation and using linearity and differentiation properties,

$V_\tau = -rV - (r - \frac{1}{2} \sigma^2)sV + \frac{1}{2} \sigma^2 s^2 V$

where $V(s,\tau) = M(V(x,\tau))(s)$ .

Hence $V_\tau = Q(s) V$ where $Q(s) = -r - (r - \frac{1}{2} \sigma^2)s + \frac{1}{2} \sigma^2 s^2$

This is a simple linear ODE. Green’s function satisfies $f(x) = \delta(x-1)$ , and so the boundary condition is $V(s,0) = M(f)(s) = 1$ . The solution is given by

$V(s,\tau) = e^{Q(s) \tau} = e^{-r\tau - (r - \frac{1}{2} \sigma^2)\tau s + \frac{1}{2} \sigma^2 \tau s^2}$

and

$v(x,\tau) = M^{-1} (V(s,\tau))(x) = e^{-r\tau} M^{-1}(e^{- (r - \frac{1}{2} \sigma^2)\tau s + \frac{1}{2} \sigma^2 \tau s^2})(x)$

$e^{- (r - \frac{1}{2} \sigma^2)\tau s} = k^{-s}$ where $k = e^{ (r - \frac{1}{2} \sigma^2)\tau}$

and so

$v(x,\tau) = e^{-r\tau} M^{-1} (e^{\frac{1}{2} \sigma^2 \tau s^2})(kx)$

also

$e^{\frac{1}{2} \sigma^2 \tau s^2} = [\sigma \tau^{1/2}]^{-1} [[\sigma \tau^{1/2}]^{-1}]^{-1} e^{\frac{1}{2} (s[[\sigma \tau^{1/2}]^{-1}]^{-1})^2}$

and so

$M^{-1} (e^{\frac{1}{2} \sigma^2 \tau s^2})(x) = [\sigma \tau^{1/2}]^{-1} M^{-1} (e^{\frac{1}{2} s^2})(x^{[\sigma \tau^{1/2}]^{-1}]}) = [\sigma \tau^{1/2}]^{-1} \phi(\log(x^{[\sigma \tau^{1/2}]^{-1}]})$

Finally,

$v(x,\tau) = e^{-r\tau} [\sigma \tau^{1/2}]^{-1} \phi(\log((kx)^{[\sigma \tau^{1/2}]^{-1}]})$

$v(x,\tau) = e^{-r\tau} [\sigma \tau^{1/2}]^{-1} \phi\left( \frac{(r - \frac{1}{2} \sigma^2)\tau + \ln x}{\sigma \tau^{1/2}}\right)$

Building Solutions from Solutions

Properties of the Log Volution

As mentioned earlier, the log-volution is defined by

$(f \star g)(x) = \int_0^\infty f(y) g(xy^{-1}) y^{-1} dy$

A number of properties are associated with the operation:

linearity:

$(a f + bg) \star h = a f \star h + b g \star h$

commutativity:

$(f \star g) = (g \star f)$

associativity:

$(f \star g) \star h = f \star (g \star h)$

identity:

$(f(x) \star \delta(x-1)) = f$

These are plainly evident on applying the Mellin Transform, since log-volution corresponds with scalar multiplication and this operation is commutative, associative.

some operations distribute across one function:

input scaling:

$(f \star g)(kx) = (f(kx) \star g(x))(x)$

differentiation:

$D(f \star g) = (Df) \star g$

other operations distribute across both functions:

power output:

$(x^a f(x)) \star (x^a g(x)) = x^a (f * g)(x)$

inverse:

$f(x^{-1}) \star g(x^{-1}) = (f \star g)(x^{-1})$

again evident on applying the Mellin transform.

lemma: Let $P(s)$ be a polynomial in $s$ . Then $M(P(-D) f)(s) = P(s) f(s)$ .

proof: Since $P(-D) f(x) = \sum_{i=1}^n p_i (-D)^i f(x)$ and since the Mellin operation is linear,

$M(P(-D)f)(s) = \sum_{i=1}^n p_i M((-D)^i f)(s) = \sum_{i=1}^n p_i s^i M(f)(s) = P(s) M(f)(s)$ .

lemma: Let $P(s)$ be a polynomial in $s$ . Then $M(P(-D) (f \star g)) = M((P(-D) f)\star g)$

proof: $M(P(-D) (f \star g)) = P(s) M(f \star g)(s) = P(s) f(s) g(s) = (P(s) Mf(s)) Mg(s) = M(P(-D) f)(s) Mg(s) = M(P(-D) f \star g)(s)$ .

theorem: If $U = U(x,\tau)$ is a zero of the operator $O(-D) = \partial_\tau - Q(-D)$ where $Q(s)$ is a constant-coefficient polynomial in $s$ , then for any $F = F(x)$ for which $G(x,\tau) = (F \star U(\tau))(x)$ exists, $G$ is a zero of $O(-D)$ .

proof:

Since $O(-D)(F \star U(\tau)) = F \star O(-D) U(\tau) = F \star 0 = 0$ it follows that $F \star U(\tau)$ is a zero of $O(-D)$ .

Example Application

A standard call option has payoff

$f(x) = (x - k)^+ = (x-k)\mathbb{I}_{x > k} = x \mathbb{I}_{x > k} - k \mathbb{I}_{x > k}$

Since $(f(x) * \delta(x-1))(x) = \int_0^\infty \delta(y - 1) f(x/y) dy/ y = f(x)$ , and since the solution to $Q(-D)v = 0$ with $v(x,0) = \delta(x-1)$ was determined and since $f$ does not depend on time, it follows that the solution to BS-PDE boundary value problem is $f(x) * v(x,\tau)$ .

The rest of the analysis : to be completed.

Mellin Transform

Transforming the BS-PDE using Mellin Transform

Building Solutions from Solutions

Properties of the Log Volution

Example Application

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