I find the below problem very strange since the derivative in the SDE is squared. Can I even use Feynman-Kac here because the derivative is squared? An in that case how would that formula look like?
Am I missing something here?
2026-02-23 15:15:33.1771859733
Can I use Feyman-Kac here when the derivative is squared?
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Look at the Feynmac-Kac formula here: https://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula
It does not matter that the terminal condition be zero, because you have a term $q(t,x)$ (which corresponds to $f(x,t)$ in their article)