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Let $X_t$ be a solution to the SDE, $dX_t=X_t \,dt+X_t\,dW_t $, $X_0=x>0$ where $W_t$ is brownian motion, then the solution to this SDE is $X_t=xe^{\frac{t}{2}+W_t}$.

Let $\tau=\inf_{t>0} \{t:X_t\ge R\}$. I am not sure how to calculate the expectation of the stopping time $\mathbb{E}_x[\tau]$.

Thank you

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    Do you know what the expected value of the hitting time of a given level $a$ is for standard Brownian motion (with or without drift)?2011-11-13
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    Can you solve similar hitting times problems? Which ones?2011-11-13
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    I am not very comfortable with hitting time actually :(2011-11-13
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    I know how to do the basic ones, like first passage time2011-11-13
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    @cardinal, I know how to calculate the problem you mentioned2011-11-13
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    This *is* a first-passage time problem.2011-11-13
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    Both with and without drift (the latter is especially trivial, the former is more relevant to this problem)?2011-11-13
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    so we calculate $P(\tau\le t)$?2011-11-13
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    Here is a **hint** that may help: Consider $Y_t = \log X_t$. What kind of process is this? Can you redefine $\tau$ in terms of $Y_t$? Now, proceed...2011-11-13
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    so we calculate $P(\tau\le t)$= $2P\{X_t>R \}$?? $Y_t$ is just normal2011-11-13
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    how do you redefine $\tau$?2011-11-13
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    @TashaChen: if $X_t \geq R$ then $Y_t = \log X_t\geq ...$ (or maybe $Y_t\leq \dots$)2011-11-14

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Observe that $(W_t)_{t\geqslant0}$ is a martingale, hence the optional stopping theorem yields $\mathbb{E}[W_\tau]=\mathbb{E}[W_0]$. Thus, $\mathbb{E}[W_\tau]=\mathbb{E}_x[\log(X_\tau)-\log(X_0)-\frac{1}{2}\tau]=0$ and $\mathbb{E}_x[\tau]=2\log(R/x)$.