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Let $X_t$ be a Geometric Brownian Motion (GBM) with initial state $x_0$ at $t=0$ and dynamics:

$$ dX_t = \mu X_tdt + \sigma X_tdW_t $$

Let us define the GBM hitting time of level $x$:

$$ \begin{align} \tau & = \min\,\,\{t \geq 0: X_t=x\} \\[6pt] & = \min\left\{ t \geq 0:W_t+\left(\mu-\frac{\sigma^2}{2}\right)\frac{t}{\sigma}=\frac{1}{\sigma}\log\frac{x}{x_0}\right\} \end{align}$$

Under the following conditions:

$$ \begin{align} & (\text{A}) \quad \sigma > 0 \\[12pt] & (\text{B}) \quad \lambda > 0 \\[12pt] & (\text{C}) \quad x > x_0 \\[6pt] & (\text{D}) \quad \lambda\left(\theta+\frac{\lambda}{2}\right) >0 \end{align} $$

Do we have:

$$ \mathbb{P}\left(\tau < \infty\right)=1 \text{ ?}$$

By letting $\theta=(\mu-\sigma^2/2)/\sigma$ and $\hat{W}_t=W_t+\theta t$, I have defined the martingale $M_t$:

$$ M_t=e^{\lambda\hat{W}_t-\lambda\theta t-\frac{\lambda^2}{2}t} $$

Then by studying the limiting behaviour of the stopped (martingale) process $M_{\min(t,\tau)}$ for each case $\{\tau<\infty\}$ and $\{\tau=\infty\}$, noticing that $\mathbb{E}[M_{\min(t,\tau)}]=\mathbb{E}[M_0]=1$, and applying the dominated convergence theorem, I have concluded that:

$$ \mathbb{E}\underbrace{\left[\mathbf{1}_{\{\tau \,<\, \infty\}}e^{\lambda\hat{W}_{\tau}-\lambda\theta\tau-\frac{\lambda^2}{2}\tau}\right]}_{M_{\infty}}=1$$

Now, from $(\text{A})$, $(\text{C})$ and $(\text{D})$, I would be tempted to make $M_{\infty}$ dependent on $\lambda$, then showing that $M_{\infty}(\lambda)$ is bounded from above by $e^{(\lambda/\sigma) \log(x/x_0)}>1$ and converges to $\mathbf{1}_{\{\tau \, < \, \infty\}}$ to apply the dominated convergence theorem again to conclude that:

$$ \lim_{\lambda \rightarrow \infty}\mathbb{E}[M_{\infty}(\lambda)] = \mathbb{E}\left[\lim_{\lambda \rightarrow \infty}M_{\infty}(\lambda)\right] = \mathbb{P}(\tau < \infty) =1 $$

Would this be correct?

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  • $\begingroup$ How is $\lambda$ to be chosen? $\endgroup$ – John Dawkins Jul 2 '17 at 22:06
  • $\begingroup$ @JohnDawkins just a positive real number. $\endgroup$ – Morris Fletcher Jul 4 '17 at 15:17
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If you should take $\lambda\to0$ instead of $\lambda\to\infty$, your method works for $\theta\ge0$ without need for the LIL. Indeed, if $\theta\ge0$, then $M_\infty(\lambda)\le\exp(\lambda\hat W_\tau)\le\exp(\hat W_\tau)=(x/x_0)^{1/\sigma}$ for all $0<\lambda<1$, so the result follows by letting $\lambda\to0$ by the bounded convergence theorem. To prove that $\mathbb P(\tau<\infty)<1$ for $\theta<0$, you probably need to appeal to the LIL.

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  • $\begingroup$ Thank you for the answer @Janson. Just a couple of questions: do we need $\theta \geq 0$, or is condition $(\text{D})$ enough? The 2nd one might be silly, but I thought the dominated convergence theorem worked when the convergence is towards $\infty$; how do we justify the applicability of the theorem when the convergence is towards $0$? $\endgroup$ – Morris Fletcher Jul 8 '17 at 10:55
  • $\begingroup$ You need $\theta\ge0$ - John Dawkins' answer demonstrates necessity. Condition (D) on its own doesn't make a lot of sense. Do you mean for all $\lambda>0$? Then it is equivalent to $\theta\ge0$. For some $\lambda>0$? Then this is true for any $\theta$. Taking $\lambda\to\infty$ doesn't work. Your dominating function depends on $\lambda$, and actually tends to $+\infty$ as $\lambda\to\infty$. The dominated convergence theorem works just fine if you let $\lambda\to0$ instead, as my answer shows. $\endgroup$ – Jason Jul 8 '17 at 16:45
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Keep in mind that for large $t$, $W_t$ will frequently get as large as $\sqrt{bt\log\log t}$ (for each $0<b<2$) but not as large as $\sqrt{bt\log\log t}$, infinitely often, if $b>2$, by the law of the iterated logarithm. Therefore under your conditions (A) and (C), $W_t$ will hit a line with slope $(\mu-\sigma^2/2)/\sigma$ and intercept $\log(x/x_0)/\sigma$ (i) with probability 1 if $\mu-\sigma^2/2\le 0$, but (ii) with probability less than 1 if $\mu-\sigma^2/2>0$.

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  • $\begingroup$ Thank you @JohnDawkins. However I am not sure I follow you, I am not familiar with the LIL. Could you be a bit more explicit please? $\endgroup$ – Morris Fletcher Jul 3 '17 at 18:48
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    $\begingroup$ The LIL states that, with probability 1, $\limsup_{t\to\infty}{W_t\over\sqrt{2t\log\log t}}=1$. For your purposes, the takeaway is that the Brownian motion will hit a line of positive slope. with intercept greater than $W_0$, only with probability less than $1$. $\endgroup$ – John Dawkins Jul 4 '17 at 16:29
  • $\begingroup$ I see thank you @JohnDawkins. Thus do you see where does my original argument go wrong? $\endgroup$ – Morris Fletcher Jul 5 '17 at 9:38
  • $\begingroup$ Your evaluation of the limit $\lim_{\lambda\to\infty}M_\infty(\lambda)$ is in doubt; should it not br $0$? $\endgroup$ – John Dawkins Jul 5 '17 at 16:24
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    $\begingroup$ You have your signs the wrong way around - $\tau$ is the first hitting time of the line $-\frac1\sigma(\mu-\sigma^2/2)t+\log(x/x_0)/\sigma$. $\endgroup$ – Jason Jul 7 '17 at 21:41

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