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I`m having problems understanding this (in the paper https://rucore.libraries.rutgers.edu/rutgers-lib/30482/PDF/1/play/ page 23 at the top) sufficient condition of a weak solution of an SDE. Is it trivial (and why) if:

$\frac{\partial}{\partial T} \int_{\mathbb{R}^n}f(x) p_X(x,T) dx = \int_{\mathbb{R}^n}f(x)(A^* p_X(x,T))$

holds for every $f \in C_{0}^2$, $p_X(x,t)$ probability density function of a stochastic process $X(T)$ and

$A^* p_X(x,t) := - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} (b_i(x,t)p_X(x,t)) + \frac{1}{2} \sum_{i,j = 1}^n \frac{\partial^2}{\partial x_i \partial x_j} (a_{ij}(x,t)p_X(x,t))$

that then $p_X(x,t)$ is a weak solution (Difference between weak ( or martingale ) and strong solutions to SDEs) of

$\frac{\partial p_X}{\partial t} = A^* p_X(x,t)$.

I hope someone can enlighten me, thank you very much in advance!

Tom
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I think you are confused by two notions of weak solution.

  1. Weak solution of an SDE : there exists a filtered probability space supporting a brownian motion and an adapted continuous process $X$ satisfying the SDE.
  2. Weak solution of a PDE : https://en.wikipedia.org/wiki/Weak_solution

By the very definition of a weak solution for a PDE, it is clear $p_X(x,t)$ is a weak solution.

Bad-Zol
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  • Hi, thanks for your fast answer, I thought about it for a while and would check if I`m correct... (even though there are multiple definitions for weak solutions of PDEs, and for me the "general case" is not a real definition of a weak solution in my opinion): Multiplying my PDE and integrating by parts of $\frac{\partial p_X){\partial t}$ = $A^* p_X(x,t)$ should yield the first equation and therefore is equivalent if f is at least $C_0$? – Tom Jan 20 '22 at 15:30