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I realize that path integral techniques can be applied to quantitative finance such as option value calculation. But I don't quite understand how this is done.

Is it possible to explain this to me in qualitative and quantitative terms?

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closed as too broad by ACuriousMind, Kyle Kanos, Brandon Enright, John Rennie, JamalS Feb 9 at 7:35

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs. If this question can be reworded to fit the rules in the help center, please edit the question.

Although some physicists are interested in quantitative finance, this question is off-topic here. Unfortunately I can't point you to the appropriate forum off the top of my head, but I'm sure quantitative finance forums exist if you poke around. –  Mark Eichenlaub Dec 13 '10 at 9:30
@Mark, I do not think so; I think econophysics questions should be allowed here, much like mathematical physics, or physics questions with engineering bent are allowed here. –  Graviton Dec 13 '10 at 9:36
I will have to agree with @Ngu on this one. Path integrals are pretty much a modern physicist's bread and butter. So if someone asked "can you apply path integrals to understanding how to butter bread" I'd say that was a question for physicists :-) And a lot more physicists are entering this field now. One prominent example is Lee Smolin (reference) who is also one of the BIG names in quantum gravity. –  user346 Dec 13 '10 at 9:51
I voted to reopen. (I couldn't figure out how to redact my vote to close.) I don't have a strong personal stake in this question - my initial vote to close was just my immediate reaction because I hadn't seen such questions here before and I didn't know about any significant ties to physics. It now seems that a majority of users think the question is appropriate, and I'm happy to go along with the majority. –  Mark Eichenlaub Dec 14 '10 at 4:51
@marek I don't know about string theory but the complete works of shakespeare should be mandatory reading for all experimentalists :-) Theorists are just born with it ! –  user346 Dec 14 '10 at 20:23

1 Answer 1

up vote 22 down vote accepted

The fundamental equation which serves as the basis for the path-integral formulation of finance and many physical problems is the Chapman-Kolmogorov equation.

$$p(X_f|X_i)=\int p(X_f|X_k)p(X_k|X_i) dX_k $$

This is analogous to the following equation for amplitudes in quantum mechanics

$$\langle X_f|X_i \rangle=\int \langle X_f|X_k\rangle\langle X_k|X_i\rangle dX_k $$

That's right, it's the same form, but the interpretation of the basic entities changes. In the former, they are probability densities and thus real and positive, in the latter they are probability amplitudes and thus complex.

The class of physical problems that can be tackled with the first type of equation are called Markov processes, their characteristic is that the state of the system depends only on its previous state. Despite its seeming limitedness, this comprises many phenomena since any process with a long but finite memory can be mapped onto a Markov process provided the state space is enlarged appropriately. On the other hand, the second equation is pretty natural and general in quantum mechanics. It is basically stating that the unity operator can always be decomposed into a, possibly overcomplete, sum of pure states

$$\mathbb{I}=\int |X_k\rangle\langle X_k| dX_k \; .$$

Now, constructing a path integral is done by slashing up the path from $X_i$ to $X_k$ into ever smaller components. Let's suppose that the endpoints are fixed, then we might assume that to go from one endpoint to another, the system has to go through paths $(X_i,X_1(t_1),X_2(t_2),\ldots,X_n(t_n),X_f)$. This leads to the following integral

$$p(X_f|X_i)=\int\cdots\int \prod_{k=0}^n p(X_{k+1}(t_{k+1})|X_k(t_k)) \prod_{k=1}^n dX_k(t_k) $$

where I put $X_0(t_0)=X_i$ and $X_{n+1}(t_{n+1})=X_f$. The tricky part is now to see if the limit can be defined meaningfully. This can be very problematic, especially in the quantum case. Ironically, the cases that are used for finance and statistical mechanics are often much more well-behaved. This is again related to one integral being over complex numbers and the other over real numbers, but it's not the only reason. Up till now, I have not been specific about the kind of system I want to study, this will play an important role as well.

So, let's take an option which is a financial security of which the price is dependent on the price of the underlying stock and time. So we can write $O(X,t)$ for the price of the option and we'll assume the underlying stock follows a geometric brownian motion:

$$\frac{dX}{X}=\mu dt + \sigma dW$$

where $W$ represents a Wiener process with increments $dW$ having mean zero and variance $dt$. Also assume that the pay-off of the option at the expiration time $T$ is:


with $F$ a given function of the terminal stock price.

Then, Fisher Black and Myron Scholes have shown that the option, under the 'no arbitrage' assumption, satisfies the following PDE

$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2X^2\frac{\partial^2 O}{\partial X^2} + r X \frac{\partial O}{\partial X} - rO = 0$$

in which $r$ is the risk free interest rate. If instead of the geometric brownian motion variable $X$, I reformulate this into $x=\ln X$ which is an arithmetic brownian motion variable, I can reformulate the equation as:

$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2\frac{\partial^2 O}{\partial x^2} + (r-\frac{\sigma^2}{2}) \frac{\partial O}{\partial x} - rO = 0$$

This is nothing else but a special case of the PDE's that can be solved by using the Feynman-Kac formula, which includes also the Fokker-Planck equation and the Smoluchowski equation, both related to the description of diffusion processes in physics. In the diffusion problem, O is to be interpreted as a distribution of velocities of the particle (Fokker-Planck) or of the positions of the particle (Smoluchowski). That's how we relate to what I introduced above. Also note that the Schrödinger equation in quantum mechanics is very similar in form, except you'll get complex coefficients.

The Feynman-Kac formula tells us that the solution to the PDE is:

$$O(X,t) = e^{-r(T-t)}\mathbb{E}\left[ F(X_T)|X(t)=X \right]$$

It is this expectation value that will now be represented as a pathintegral:

$$O(X,t) = e^{-r(T-t)}\int_{-\infty}^{+\infty}\left(\int_{x(t)=x}^{x(T)=x_T} F(e^{x_T}) e^{A_{BS}(x(t'))} \mathcal{D}x(t')\right) dx_T$$


$$A_{BS}(x(t'))=\int_t^{T} \frac{1}{2\sigma^2}\left(\frac{dx(t')}{dt'}-\mu\right)^2$$

is the action functional.

The reason this path integral can be built is the same explained before, here it is possible to split the conditional expectation ever further in smaller intervals:

$$\begin{array}{rcl}\mathbb{E}\left[ F(e^{x_T})|x(t)=x \right] & = & \int_{-\infty}^{+\infty} F(e^{x_T}) p(x_T|x(t)=x) dx_T \\ & = & \int_{-\infty}^{+\infty} F(e^{x_T}) \int_{\tilde{x}(t)=x}^{\tilde{x}(T)=x_T} p(x_T|\tilde{x}(\tilde{t})) p(\tilde{x}(\tilde{t})|x(t)=x) d\tilde{x}(\tilde{t}) dx_T \end{array}$$

Each of the conditional probabilities satisfying the PDE for the arithmetic brownian motion as noticed above.

I'll stop here for now, but I refer to the following article for further details.

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This doesn't talk about finance at all hardly. –  Noldorin Dec 13 '10 at 20:52
It doesn't really talk about physics, either. –  Sklivvz Dec 13 '10 at 22:13
Excellent answer @raskolnikov! I see the peanut gallery is pretty crowded today. –  user346 Dec 13 '10 at 23:29
Seems like the villagers have taken over. I vote to reopen. –  user346 Dec 14 '10 at 0:47
I voted to reopen. –  Robert Smith Dec 14 '10 at 2:18

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