Backward Kolmogorov Equations Require Full-State Initial Payoffs
Summary
The document clarifies a common misunderstanding about the backward Kolmogorov equation for a time-homogeneous diffusion. Its initial condition is the payoff function across the relevant state space, not merely the payoff’s value at the process’s starting point. Thus, two smooth functions agreeing at the initial state can still produce different expected values at a later time if they differ elsewhere.
The explanation is conceptual rather than a worked numerical example: it identifies the error in applying uniqueness of the PDE solution using only a single matching value. It also resolves the questioner's intuition that derivatives and behavior away from the starting state matter. The result applies within the stated diffusion and PDE setup; the excerpt does not discuss technical conditions for PDE solution uniqueness or provide trading-specific applications.
Key ideas
- The backward Kolmogorov equation uses the full payoff function as its initial condition.
- Matching two payoffs at the process's starting state does not make their later expectations equal.
- The payoff values across the relevant state domain determine the PDE solution.
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# backward Kolmogorov equations - Markov properties
# backward Kolmogorov equations - Markov properties
I'm a physicist who's research has lead him into the theory of stochastic differential equations. If this question is not appropriate for this forum, please feel free to delete it.
So I've been following through Bernt Oksendal's Stochastic Differential Equations: An Introduction with Applications (5th edition). In it, he discusses a time homogeneous, one dimensional stochastic process $\{X_t\}_{t \ge 0}$, which solves the SDE $$ dX_t = a(X_t) dt + b(X_t) dW_t, \quad X_0 = x_0 \quad \mathbb{P} - a.s. $$ (This of course assumes that $a(\cdot)$ and $b(\cdot)$ are Ito integrable and I'm happy to take them as smooth infinitely differentiable bounded functions that vanish at infinity.) My question involves considering two additional (smooth) functions $f(x)$ and $g(x)$ where $f(x) \ne g(x)$ in general but do satisfy the equality, $$ f(x_0) = g(x_0). $$
My question is given this equality, must we conclude from the backward Kolomogorov equation that $$ \mathbb{E}(f(X_T) ) = \mathbb{E}(g(X_T) ) $$ for any time $T \ge 0$?
I am lead to this conclusion because the backward Kolmogorov equation says that, we can consider $$ u(t, x_0) = \mathbb{E}( g(X_t) ) $$ where $u(t, x)$ solves the PDE $$ \frac{\partial}{\partial t} u(t, x) = a(x) \frac{\partial u}{\partial x} (t, x) + \frac{1}{2}b(x) \frac{\partial^2u}{\partial x^2} (t, x) $$ with the initial condition, $u(0, x_0) = g(x_0)$.
Now instead of considering $g$ we considered the function $\mathbb{E}( f(X_t) )$. I believe I would write down the same equation subject to the same initial condition. This leads me to conclude that $ \mathbb{E}(f(X_T) ) = \mathbb{E}(g(X_T) )$, regardless of how $f(x)$ and $g(x)$ may behave away from $x_0$.
My intuition says this is not correct and that the derivative information about $f$ and $g$ must be factored into it some how. Is it that the initial condition is really a boundary condition and we are really saying that $u(0, x) = g(x)$ for any $x$?
Having typed this out I think I might have answered my own question, but any feedback would be appreciated!
thanks.
## Answer by pbr142 (score 3, accepted)
https://quant.stackexchange.com/a/10766
The initial condition for the backward Kolmogorov PDE is that $$ u(0,x) = g(x) $$ for all $x$ in the relevant domain and not just at a particular point. So if your functions $f$ and $g$ agree only at a single point the initial conditions are in fact different.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.