1 vote
Accepted

Is $\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ a stopping time?

(Edited) Useful theorems Assuming nothing about the filtration: If $X=(X_t)_{t\geq 0}$ is a continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq ...
Wilfred Montoya's user avatar
1 vote
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Optional Stopping Theorem and Stopped $\sigma$-fields

For real $x$ you have $$ \{X_T\le x\}\cap\{S\le n\} = \cup_{k=0}^n\{X_k\le x, T=k, S\le n\}. $$ which is clearly $\mathcal F_n$-measurable, for each $n$. It follows that $X_T$ is $\mathcal F_S$-...
John Dawkins's user avatar
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1 vote
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Expected value of the square of a stopping time

$\def\={\mathrel{\phantom=}}$Your calculation writes\begin{gather*} E( (a^2 - S_a) I_{\{ B_{S_a} = a \}} ) + E( ((-a)^2 - S_a) I_{\{ B_{S_a} = -a \}} )\\ = a^2 - E( S_a I_{\{ B_{S_a} = a \}} ) + (-a)^...
Ѕᴀᴀᴅ's user avatar
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1 vote
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Expected value of the exponential of a stopping time

All good. Perhaps, you can add more details on the "by symmetry" Independence of $T$ and $B_T$ i.e. we have $B_t\stackrel{d}{=}-B_{t}$ and $S_{a}$ is only a function of $|B_{t}|$, which is ...
Thomas Kojar's user avatar
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