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.
