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May 25, 2019 at 13:24 vote accept user78370
May 4, 2019 at 11:49 comment added Nawaf Bou-Rabee "Change of variables" is an elegant way to describe it, and indeed, this stochastic representation of the solution to (3) crucially depends on $\tilde V \ge 0$.
May 3, 2019 at 21:13 comment added user78370 Interesting, I didn't think of this very simple "change of variables". I suppose in that case that the "niceness" of the Feynman-Kac formula depends somehow on how nice $V$ is, since we can obtain some rather unwieldy Markov process as a result.
May 3, 2019 at 11:22 history answered Nawaf Bou-Rabee CC BY-SA 4.0