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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 |