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 Yearling
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Dec
4
awarded  Yearling
Dec
3
answered using Feynman-Kac formula
Nov
12
comment Expectation, exponential of an additive functional of Brownian motion
Is this related to Novikov condition? If yes there is other criteria implying the martingality of the exponential of local martingales.
Nov
3
comment Extension of Dynkin's formula, conclude that process is a martingale
This question was already answered here mathoverflow.net/questions/221585/…
Nov
2
awarded  Critic
Nov
2
comment Question on viscosity solution through stochastic differential equations
Is the function $a$ dependent on $u$ or on $x$ only. In the former case it is not obvious to link it to stochastic differential equations easily as the PDE is not linear and your process $X_t$ will depend on $u$ which is unknown here. If $a$ depends only on $x$ you can represent it via Feyman-Kac and all what you need for the existence of the solution to the SDE equation on $X_t$ is that $a$ is Lipschitzian.
Oct
22
comment Expectation equation, harmonic functions, do not understand why equation is true
I was correcting the formula in the same time you posted your comment :)
Oct
22
revised Expectation equation, harmonic functions, do not understand why equation is true
added 15 characters in body
Oct
22
answered Expectation equation, harmonic functions, do not understand why equation is true
Oct
5
comment What function is a Gaussian integral
You're welcome.
Oct
3
awarded  Yearling
Oct
3
revised What function is a Gaussian integral
added 55 characters in body
Oct
3
answered What function is a Gaussian integral
Oct
2
comment A Question on Chinese Remainder Theorem
@Igor, The other solutions differ by multiple of $p_1\cdots p_n$ so necessarily it is the smallest positive one. We can see it directly as follows, if we choose $y$ between $0$ and $p_1\cdots p_n-1$ as in the Chinese reminder theorem, we have on the one hand $2y=-1 \mod p_i$ so $p_i$ divides $2y+1$ for each $i$ and then $p_1\cdots p_n$ divides $2y+1$. On the other hand we have $2y+1\leq 2p_1\cdots p_n-1$ so necessarily $2y+1 = p_1\cdots p_n$.
Oct
1
comment A Question on Chinese Remainder Theorem
$y$ is simple to express in your case, it is just $$y=\frac{p_1\cdots p_n-1}{2}$$. indeed $2y = -1 \mod p_1 = p_1-1 \mod p_1$. then if $p$ is prime such that $y=\frac{p-1}{2} \mod p$ then $p$ divides $2y+1$ and hence is one of the $p_i$
Jan
28
revised Equality of two conditional expectations
added 99 characters in body
Jan
28
comment Equality of two conditional expectations
You're right, I see that the purpose of the first part of imateapot answer is to justify that this conditional expectation is $g(X)$ measurable when $X$ and $g(X)$ are independent. I'll leave the answer like that so the mistake can be clear to the reader
Jan
28
answered Equality of two conditional expectations
Dec
18
revised Do we need Feller condition if the process jumps?
deleted 6 characters in body
Dec
17
revised Do we need Feller condition if the process jumps?
deleted 4 characters in body