guide:812f824494

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For a real-valued function [math]f(x)[/math] with [math]\int_{-\infty}^\infty |f(x)|^2dx \lt \infty[/math], Fourier transforms are defined as follows

[[math]] \begin{align} \label{eq:fourierIntegral_A} \widehat f(u)&=\int_{-\infty}^\infty e^{iux}f(x)dx\\ \label{eq:inverseFourierIntegral_A} f(x)&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat f(u)du\ . \end{align} [[/math]]

The Fourier transforms in \eqref{eq:fourierIntegral_A} and \eqref{eq:inverseFourierIntegral_A} are somewhat different from traditional definitions, which are

[[math]] \begin{align*} \widehat f(u)&=\int_{-\infty}^\infty e^{-2\pi iux}f(x)dx\\ f(x)&=\int_{-\infty}^\infty e^{2\pi iux}\widehat f(u)du\ . \end{align*} [[/math]]

The reason we choose \eqref{eq:fourierIntegral_A} in finance is because we work so much with probability theory, and as probabilists we like to have a Fourier transform of a density be equal to a characteristic function. However, nothing changes and we are able to carry out all the same calculations; the difference amounts merely to a change of variable inside the integrals.

Some Basic Properties

• Linearity: if [math]\psi(s) = af(s)+bg(s)[/math], then
  [[math]]\widehat \psi(u) = a\widehat f(u)+b\widehat g(u)\ .[[/math]]
• Translation/Time Shifting: if [math]f(s) = \psi(s-s_0)[/math], then
  [[math]]\widehat f(u)= e^{ius_0}\widehat \psi(u)\ .[[/math]]
• Convolution: for [math]f*g(x) = \int_{-\infty}^\infty f(y)g(x-y)dy[/math],
  [[math]] \begin{align*} \widehat {f*g}(u) &= \int_{-\infty}^\infty e^{iux} \int_{-\infty}^\infty f(y)g(x-y)dydx\\ &=\int_{-\infty}^\infty e^{iuy}f(y)\int_{-\infty}^\infty e^{iu(x-y)} g(x-y)dxdy\\ & = \int_{-\infty}^\infty e^{iuy}f(y)\widehat g(u)dy\\ &=\widehat f(u)\widehat g(u)\ . \end{align*} [[/math]]
• Reverse Convolution:
  [[math]] \begin{align*} \widehat{\widehat f*\widehat g}(x) &= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux} \int_{-\infty}^\infty \widehat f(v)\widehat g(u-v)dvdu\\ &= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-ivx} \widehat f( v) \int_{-\infty}^\infty e^{-i(u- v)x}\widehat g(u- v)dudv\\ &=g(x)\int_{-\infty}^\infty e^{-i vx} \widehat f( v)dv\\ &= 2\pi g(x)f(x)\ . \end{align*} [[/math]]

An important concept to realize about Fourier is that functions [math](e^{iux})_{u\in\mathbb R}[/math] can be thought of as orthonormal basis elements in a linear space. Similar to finite-dimensional vector spaces, we can write a function as a some inner products with the basis elements. Indeed, that is what we accomplish with the inverse Fourier transform; we can think of the integral as a sum,

[[math]]f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat f(u)du\approx\frac{1}{2\pi} \sum_{n=1}^N e^{-iu_nx}\widehat f(u_n) \ ,[[/math]]

where [math]u_n[/math] are discrete points [math]\mathbb C[/math] (this is a similar idea to Fast Fourier Transforms (FFT)). The basis elements are orthogonal in the sense that

[[math]]\frac{1}{2\pi}\int_{-\infty}^\infty e^{iux}e^{-ivx}dx = \delta_0(u-v)\ .[[/math]]

Finally, it needs to be pointed out that the function [math]\delta_0[/math] is something that defined under integrals, and is rather hard to formalize outside. For instance, in Parseval's identity, we used [math]\delta_0[/math], but only inside the integral:

Proposition

For a real-valued function [math]f(x)[/math] with [math]\int_{-\infty}^\infty |f(x)|^2dx \lt \infty[/math],

[[math]]\int_{-\infty}^\infty |f(x)|^2ds =\frac{1}{2\pi}\ \int_{-\infty}^\infty |\widehat f(u)|^2du\ .[[/math]]

Show Proof

[[math]] \begin{align*} \frac{1}{2\pi}\int_{-\infty}^\infty |\widehat f(u)|^2du&=\frac{1}{2\pi}\int_{-\infty}^\infty \widehat f(u)\overline{ \widehat f(u)}du\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty e^{iux}f(x)e^{-iuy}f(y)dxdydu\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty f(x)f(y) \left(\int_{-\infty}^\infty e^{iux}e^{-iuy}du\right)dxdy\\ &=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x)f(y) \delta_0(y-x)dxdy\\ &=\int_{-\infty}^\infty f(x)f(x) dx \ . \end{align*} [[/math]]

■

Regularity Strips

For some function [math]f:\mathbb R\rightarrow \mathbb R[/math] we modify \eqref{eq:fourierIntegral_A} and \eqref{eq:inverseFourierIntegral_A} to accommodate non-integrability and/or non-differentiability. We write

[[math]] \begin{align*} \widehat f(u)&=\int_{-\infty}^\infty e^{iux}f(x)dx\\ f(x)&=\frac{1}{2\pi}\int_{iz-\infty}^{iz+\infty} e^{-iux}\widehat f(u)du\ . \end{align*} [[/math]]

where [math]z=\Im(u)[/math]. The region of [math]z[/math] values where the Fourier transform is defined is called the regularity strip. This comes in handy for functions like the European call and put payoffs. Example For a call option, [math]f(x) = (e^s-K)^+[/math]. This function has [math]\int_{\log K}^\infty |f(s)|^2ds =\infty[/math] and non-differentiability at [math]s=\log(K)[/math]. However, we can still write a Fourier transform:

[[math]] \begin{align*} \widehat f(u)&= \int_{-\infty}^\infty e^{ius}(e^s-K)^+ds\\ &= \int_{\log K}^\infty e^{s+ius}ds-K\int_{\log K}^\infty e^{ius}ds\\ &= \frac{e^{s+ius}}{1+iu}\Big|_{s=\log K}^\infty -\frac{Ke^{ius}}{iu}\Big|_{s=\log K}^\infty\ . \end{align*} [[/math]]

where the anti-derivative of [math]e^{ius}[/math] and [math]e^{s+ius}[/math] are the same as they are for real variables because the function is analytic in the complex plain. For [math]z=\Im(u) \gt 1[/math], the anti-derivative will be zero when evaluated at [math]s=\infty[/math], will be infinite for [math]z \lt 1[/math], and undefined for [math]z=1[/math]. Taking [math]z \gt 1[/math], we have

[[math]] \begin{align*} \widehat f(u) &= -\frac{e^{(1+iu)\log(K)}}{1+iu}+\frac{Ke^{iu\log(K)}}{iu}\\ &= -\frac{\left(e^{\log(K)}\right)^{(1+iu)}}{1+iu}+\frac{K\left(e^{\log(K)}\right)^{iu}}{iu}\\ &= \frac{K^{1+iu}}{iu-u^2}\ , \end{align*} [[/math]]

as shown in Table of Chapter.

Example For a put option, [math]f(x) = (K-e^s)^+[/math]. This function has [math]\int_{\log K}^\infty |f(s)|^2ds \lt \infty[/math], like the put option is non-differentiability at [math]s=\log(K)[/math]. Repeating the steps from the previous example, it follows that the regularity strip is [math]z=\Im(u) \lt 0[/math] (see Table of Chapter).

The Wave Equation

Consider the wave (transport) equation,

[[math]] \begin{align*} \frac{\partial}{\partial t}V(t,x)&=a\frac{\partial}{\partial x}V(t,x)\\ V(0,x)&=f(x)\ . \end{align*} [[/math]]

The solution can be found with Fourier transforms:

[[math]] \begin{align*} \left(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x}\right)V(t,x) &=\frac{1}{2\pi}\int_{-\infty}^\infty \left(\frac{\partial}{\partial t}-a\frac{\partial}{\partial x}\right)e^{-iux}\widehat V(t,u)\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux} \left(\frac{\partial}{\partial t}+aiu\right)\widehat V(t,u)\ , \end{align*} [[/math]]

or simply

[[math]] \begin{align*} \frac{\partial}{\partial t}\widehat V(t,u)&=-aiu\widehat V(t,u)\ ,\\ \widehat V(0,u)&=\widehat f(0)\ . \end{align*} [[/math]]

The solution is

[[math]]\widehat V(t,u) = e^{-aiu t}\widehat f(u)\ ,[[/math]]

which we invert to obtain the solution to the wave equation,

[[math]] \begin{align*} V(t,x) &= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}e^{-aiu t}\widehat f(u)du\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-iu(x+ at)}\widehat f(u)du\\ &=f(x+at)\ . \end{align*} [[/math]]

The lines [math]x = c-at[/math] are the characteristic lines so that for any [math]c[/math] we have [math]v(t,c-at) = f(c)[/math]; the initial information [math]f(c)[/math] at point [math]c[/math] is propagated along the characteristic line.

The Heat Equation

Consider the heat equation

[[math]] \begin{align*} \frac{\partial}{\partial t}V(t,x)&=\frac12\Delta V(t,x)\\ V(0,x)&=f(x)\ , \end{align*} [[/math]]

where and [math]\int_{-\infty}^\infty |f(x)|^2dx \lt \infty[/math]. For scalar [math]x[/math] we have [math]\Delta =\frac{\partial^2}{\partial x^2}[/math], and the inverse Fourier transform yields an ODE under the integral sign,

[[math]] \begin{align*} \left(\frac{\partial}{\partial t}-\frac12\frac{\partial^2}{\partial x^2}\right)V(t,x)&= \frac{1}{2\pi}\int_{-\infty}^\infty \left(\frac{\partial}{\partial t}-\frac12\frac{\partial^2}{\partial x^2}\right)e^{-iux}\widehat V(t,u)du\\ &= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\left(\frac{\partial}{\partial t}+\frac{u^2}{2}\right)\widehat V(t,u)du \\ &=0\ , \end{align*} [[/math]]

or simply,

[[math]] \begin{align} \nonumber \frac{d}{dt}\widehat V(t,u)&=-\frac{u^2}{2}\widehat V(t,u)\\ \label{eq:Vhat_ODE} \widehat V(0,u)&=\widehat f(u) \ . \end{align} [[/math]]

The solution to \eqref{eq:Vhat_ODE} is

[[math]]\widehat V(t,u) = \widehat f(u) e^{-\frac{u^2}{2}t}\ .[[/math]]

Applying the inverse Fourier transform yields the solution,

[[math]] \begin{align*} V(t,x)& = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat V(t,u)du\\ & = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat f(u) e^{-\frac{u^2}{2}t}du\\ & = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\left(\int_{-\infty}^\infty e^{iuy}f(y) dy\right)e^{-\frac{u^2}{2}t}du\\ & = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty f(y) \left(\sqrt{\frac{t}{2\pi}}\int_{-\infty}^\infty e^{iu(y-x)} e^{-\frac{u^2}{2}t}du\right)dy\\ & = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty f(y)\left( \mathbb E e^{i\mathcal Z(y-x)/\sqrt t}\right) dy\qquad\hbox{where $\mathcal Z\sim N(0,1)$}\\ & = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty f(y) e^{-\frac{(y-x)^2}{2t}}dy\ . \end{align*} [[/math]]

In fact, it can be seen as the expectation of a Brownian motion,

[[math]]V(t,x) = \mathbb Ef(x+W_t)[[/math]]

where [math]W_t[/math] is a Brownian motion. If [math]f(x) = \delta_0(x)[/math], then the we have the fundamental solution,

[[math]]\Phi(t,x) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}\ ,[[/math]]

from which solutions for different initial conditions are convolutions,

[[math]]V(t,x) = f*\Phi(t,x) = \int_{-\infty}^\infty f(y)\Phi(t,x-y)\ .[[/math]]

The Black-Scholes Equation

The Black-Scholes model is

[[math]]\frac{dS_t}{S_t}=rdt+\sigma dW_t^Q[[/math]]

where [math]W^Q[/math] is a risk-neutral Brownian motion. Letting [math]X_t=\log(S_t)[/math] the Black-Scholes equation is

[[math]] \begin{align*} \left(\frac{\partial}{\partial t}+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}+\left(r-\frac{\sigma^2}{2}\right)\frac{\partial}{\partial x}-r\right)V(t,x)&=0\\ V(T,x)&=\psi(x) \ , \end{align*} [[/math]]

where [math]\psi(X_T)[/math] is the claim, e.g. a call option [math]\psi(x) = (e^x-K)^+[/math]. Using the same Fourier techniques as we did with the wave and heat equations, we have

[[math]] \begin{align*} \left(\frac{\partial}{\partial t}-\frac{\sigma^2u^2}{2}-iu\left(r-\frac{\sigma^2}{2}\right)-r\right)\widehat V(t,u)&=0\\ \widehat V(T,u)&=\widehat \psi(u) \ , \end{align*} [[/math]]

which has solution

[[math]]\widehat V(t,u)=\widehat\psi(u)e^{-(T-t)\left(\frac{\sigma^2u^2}{2}+iu\left(r-\frac{\sigma^2}{2}\right)+r\right)}\ .[[/math]]

Applying the inverse Fourier transform, we arrive at the risk-neutral pricing formula (i.e. Feynman-Kac),

[[math]] \begin{align*} V(t,x)&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat V(t,u)du\\ &=\frac{e^{-r(T-t)}}{2\pi}\int_{-\infty}^\infty e^{-iux}\widehat\psi(u)e^{-(T-t)\left(\frac{\sigma^2u^2}{2}+iu\left(r-\frac{\sigma^2}{2}\right)\right)}du\\ &=\frac{e^{-r(T-t)}}{2\pi}\int_{-\infty}^\infty e^{-iux}\left(\int_{-\infty}^\infty e^{iuy}\psi(y)dy\right)e^{-(T-t)\left(\frac{\sigma^2u^2}{2}+iu\left(r-\frac{\sigma^2}{2}\right)\right)}du\\ &=\frac{e^{-r(T-t)}}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty \psi(y) e^{iu\left(y-x-(T-t)\left(r-\frac{\sigma^2}{2}\right)\right)}e^{-(T-t)\frac{\sigma^2u^2}{2}} dudy\ . \end{align*} [[/math]]

Take change of variable [math]y'=y-x-(T-t)\left(r-\frac{\sigma^2}{2}\right)[/math] such that [math]dy'=dy[/math], then

[[math]] \begin{align*} V(t,x)&=\frac{e^{-r(T-t)}}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty \psi\left(y'+x+(T-t)\left(r-\frac{\sigma^2}{2}\right)\right) e^{iuy'}e^{-(T-t)\frac{\sigma^2u^2}{2}} dudy'\\ &=\frac{e^{-r(T-t)}}{2\pi}\int_{-\infty}^\infty \psi\left(y'+x+(T-t)\left(r-\frac{\sigma^2}{2}\right)\right)\left( \int_{-\infty}^\infty e^{iuy'}e^{-(T-t)\frac{\sigma^2u^2}{2}} du\right)dy'\\ &=\frac{e^{-r(T-t)}}{\sqrt{2\pi\sigma^2(T-t)}}\int_{-\infty}^\infty \psi\left(y'+x+(T-t)\left(r-\frac{\sigma^2}{2}\right)\right)\left(\mathbb E e^{i\frac{\mathcal Z}{\sigma\sqrt{T-t}}y'}\right)dy'\\ &=\frac{e^{-r(T-t)}}{\sqrt{2\pi\sigma^2(T-t)}}\int_{-\infty}^\infty \psi\left(y'+x+(T-t)\left(r-\frac{\sigma^2}{2}\right)\right)e^{-\frac12\frac{\left(y'\right)^2}{\sigma^2(T-t)}}dy'\\ &=e^{-r(T-t)}\mathbb E^Q\psi\left(x+(T-t)\left(r-\frac{\sigma^2}{2}\right)+\sigma (W_T^Q-W_t^Q)\right)\\ &=e^{-r(T-t)}\mathbb E^Q\left[\psi\left(X_T\right)\Big|X_t=x\right]\ . \end{align*} [[/math]]

General references

Papanicolaou, Andrew (2015). "Introduction to Stochastic Differential Equations (SDEs) for Finance". arXiv:1504.05309 [q-fin.MF].