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Nov 29 |
revised |
Finding an example of a discrete-time strict local martingale. deleted 47 characters in body; edited title |
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Nov 29 |
accepted | Finding an example of a discrete-time strict local martingale. |
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Jul 11 |
awarded | Yearling |
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Jun 30 |
accepted | Applying Ito to semi-group of Brownian motion in $\mathbb{R}^d$ |
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Jun 8 |
awarded | Caucus |
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May 21 |
comment |
Limit of Wiener processes Thank you, I just copy-pasted the latex. |
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May 21 |
comment |
Limit of Wiener processes My notes say that Stratonovich would be $$\displaystyle \lim_{n->\infty}\sum_{k=0}^{[2^n t]-1}\big(\frac{W_{(k+1)2^{-n}}+W_{k2^{-n}}}{2}\big)(W_{(k+1)2^{-n}}-W_{k2^{-n}}),$$ what am I missing? |
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May 21 |
comment |
Limit of Wiener processes Central limit theorem won't help, central limit theorem is for a different type of convergence. |
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May 20 |
asked | Applying Ito to semi-group of Brownian motion in $\mathbb{R}^d$ |
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May 18 |
comment |
Are hitting times of Brownian motion independent? if $c=\mathbb{P}(T_b<T_a)$, then, by OST, $0=\mathbb{E}[B_{T_a\wedge T_b}]=a(1-c)+bc$. Solve for $c$. |
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May 18 |
comment |
Hitting time of Brownian Motion with a drift @mike, What is Wald? I have managed to use OST on the exponential martingale and then just differentiate the answer with respect to $b$ to compute the expression 1). |
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May 18 |
revised |
Hitting time of Brownian Motion with a drift added 99 characters in body |
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May 18 |
comment |
Hitting time of Brownian Motion with a drift Well OK it's easy to say $T<\infty$ $\mathbb{Q}$-a.s. by the properties of BM, hence it's a.s. finite under $\mathbb{P}$ (absolute continuity). So that is not a problem anymore. |
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May 18 |
asked | Hitting time of Brownian Motion with a drift |
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May 17 |
comment |
Problem with applying Ito I have no doubt that this is correct, but could you clarify which version of Ito you are applying? My question was, essentially, how to make it rigorous. |
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May 17 |
asked | Problem with applying Ito |
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May 17 |
comment |
Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ No, I don't think you understand it properly. The first equality involves indicator function $\mathbb{1}(Z>X+Y)$, which is $1$ on the event $Z>X+Y$ and $0$ on the event $Z\leq X+Y$. Then expectation of such indicator is just the probability of the event $Z>X+Y$, by definition of expectation. |
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May 17 |
revised |
Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ deleted 2 characters in body |
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May 17 |
answered | Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ |
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May 17 |
asked | Polarisation in proving Kunita-Watanabe identinty |
