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If a stock price, $S(t)$, is governed by a geometric brownian motion. Is it possible to characterise the value of an option $V(S,t)$ as an ODE rather than a PDE (given $S$ is itself a function of $t$)? Hence is it possible to write a closed form solution as just a function of $t$?

Apologies if this is a stupid question (I'm new to QF).

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You can alway use a semidiscretization for your PDE or some kind of Galerkin method to end up with an ODE system. I suppose this would correspond to a process with either discrete timestep or discrete state space but thats a guess. – vanguard2k Oct 2 '13 at 14:53
up vote 8 down vote accepted

It is not possible for what most people think of as options, but there are classes of options for which an ODE is used.

For a nontrivial example, think of perpetual American-exercise options. Because of perpetual exercise, the option value is independent of time. In place of the Black-Scholes PDE

$$ \frac{\partial f}{\partial t} = \frac12 \sigma^2 x^2 f^{\prime \prime} + r x f^\prime-rf $$

we obtain the time-homogeneous ODE

$$ 0= \frac12 \sigma^2 x^2 f^{\prime \prime} + r x f^\prime-rf $$

Solving this ODE, one finds there is a barrier, $x^\star$, beyond which a perpetual American put should be executed. The solution is a relatively simple function

$$ K\left( \frac{K}{S} \left( 1-\frac{2r}{2r+\sigma^2} \right) \right)^{2r/\sigma^2} $$

This solution was, as far as I can tell, first derived by McKean in 1965. As you can tell, it mainly works because we were able to remove one of the (underlying price, time) variable from the ODE. Most options quite clearly have value that depends on both.

More trivial examples include bonds, which don't have any optionality to speak of but do follow the ODE

$$ \frac{dB}{dt}= -(r+h) B $$

and CDS which in the Poisson model follow the same ODE with different boundary conditions.

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Thank you for your answer. Perhaps my confusion is a calculus one rather than a finance one; you say "Most options quite clearly have a value that depends on [price and time]", but if price only depends on time could we not simply say the option value only depends on time (albeit in a more complex way)? As a trivial example I was thinking of an option such as the stock itself, i.e. $V(S,t) = S(t)e^{\rho t}$. Where the stock bleeds a continuous dividends $\rho$. – Daniel Oct 2 '13 at 14:15
Perhaps I should rephrase my question. If stock price is only a function of time, why can't we consider the option price as only a function of time (i.e. $V(t)$). – Daniel Oct 2 '13 at 14:26
I think you are asking, since $V(S,t)=V(S(t),t)$ why we cannot write that as just $V(t)$. We can, but then we're effectively expressing $V$ as a random process (like we do with $S$), which then means there's no longer any kind of PDE or ODE involved -- just the SDE. – Brian B Oct 2 '13 at 15:40
Sure you can, but then $V$ would be a stochastic process, not a deterministic function. – Eddie E. Oct 2 '13 at 15:46

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