3
votes
Accepted
Distribution of difference of first hitting times of Brownian motion
Yes, assuming that $0<b<a$, the random variables $T_a-T_b$ and $T_{a-b}$ have the same distribution. This follows from the strong Markov property of Brownian motion at time $T_b$. Note that if $...
- 12.9k
1
vote
Accepted
How can we show $\operatorname E\left[Y_\tau;t\ge\tau\mid\tau=s\right]=1_{\{\:t\:\ge\:s\:\}}\operatorname E\left[Y_s\right]$?
You are right. (1) is wrong. Here is a counter-example: let $(Y_t)$ be Brownian motion and $\tau=\inf \{t \geq 0: Y_t=1\}$. Then $Y_{\tau}=1$. If we take $s \leq t$ the LHS of (1) is $1$ and RHS is $...
- 2,186
1
vote
Accepted
martingale, stopping times
It is immediate that
$$
\{T\leqslant t\}\in \mathcal F_t
$$
for all $t$. Now, since $\sigma+\tau$ and $T$ are stopping times, we have
$$
\{(\sigma+\tau)\wedge T\leqslant t\} = \{\sigma+\tau\leqslant t\...
- 33.2k
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