# Distribution of hitting time of line by Brownian motion

I came across the following question:

Let $T_{a,b}$ denote the first hitting time of the line $a + bs$ by a standard Brownian motion, where $a > 0$ and $−\infty < b < \infty$ and let $T_a = T_{a,0}$ represent the first hitting time of the level $a$.

1) For $\theta > 0$, by using the fact that $\mathbb{E}e^{-\theta T_a}=e^{-a\sqrt{2\theta}}$, or otherwise, derive an expression for $Ee^{-\theta T_{a,b}}$, for each $b$, $−\infty < b < \infty$.

2) Hence, or otherwise, show that, for $t > 0$, $$\mathbb{P}[T_{a,b}\leq t] = e^{-2ab}\phi\left(\frac{bt-a}{\sqrt{t}}\right)+1-\phi\left(\frac{a+bt}{\sqrt{t}}\right).$$

For the first part, I ended up, by changing measure, with the (unverified) expression

$$\mathbb{E}e^{-\theta T_{a,b}}=\exp\left(-a\left[b+\sqrt{2\left(\theta+\frac{b^2}{2}\right)}\right]\right).$$

What's the cleanest way to do the second part? It seems I could either do some kind of inverse transform on the moment generating function, or calculate the moment generating function of the given distribution. Both of these seem difficult. Am I missing something, or do I just need to persevere?

Thank you.

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See first few slides of these notes. –  Sasha Apr 19 '12 at 5:56
@Sasha Thanks. The question suggests there should be some (not too horrible) way to use the mgf in the second part. –  Ben Derrett Apr 19 '12 at 8:51

First part

The probability density of $T_{a,0}$ is well-known: $$f_{T_{a,0}}(t) = \frac{a}{\sqrt{2 \pi}} t^{-3/2} \exp\left( -\frac{a^2}{2t} \right)$$ From here, for $\theta >0$,
$$\mathbb{E}\left( \mathrm{e}^{-\theta T_{a,0}} \right) = \int_0^\infty \frac{a}{\sqrt{2 \pi t}} \exp\left( -\theta t -\frac{a^2}{2t} \right) \frac{\mathrm{d} t}{t} \stackrel{t = a^2 u}{=} \int_0^\infty \frac{1}{\sqrt{2 \pi u}} \exp\left( -\theta a^2 u -\frac{1}{2 u} \right) \frac{\mathrm{d} u}{u}$$ According to Grandstein and Ryzhyk, formula 3.471.9, see also this math.SE question, we have: $$\mathbb{E}\left( \mathrm{e}^{-\theta T_{a,0}} \right) = \frac{1}{\sqrt{2 \pi}} \cdot \left. 2 \left(2 \theta a^2\right)^{\nu/2} K_{\nu}\left( 2 \sqrt{\frac{\theta a^2}{2}} \right) \right|_{\nu = \frac{1}{2}} = \sqrt{\frac{2}{\pi}} \sqrt{2\theta} a K_{1/2}(a \sqrt{2 \theta} ) = \mathrm{e}^{-a \sqrt{2 \theta}}$$

The time $T_{a,b}$ for standard Brownian motion $B(t)$ to hit slope $a+ b t$, is equal in distribution to the time for Wiener process $W_{-b, 1}(t)$ to hit level $a$. Thus we can use Girsanov theorem, with $M_t = \exp(-b B(t) - b^2 t/2)$: $$\mathbb{E}_P\left( \mathrm{e}^{-\theta T_{a,b}} \right) = \mathbb{E}_Q\left( \mathrm{e}^{-\theta T_{a,0}} M_{T_{a,0}} \right) = \mathbb{E}_Q\left( \mathrm{e}^{-\theta T_{a,0}} \mathrm{e}^{-b a - b^2 T_{a,0}/2} \right) = \exp(-b a - a \sqrt{b^2 + 2\theta})$$

Second part

In order to arrive at $\mathbb{P}(T_{a,b} \leqslant t)$ notice that $$\mathbb{P}(T_{a,b} \leqslant t) = \mathbb{E}_Q\left( [T_{a,0} \leqslant t] \mathrm{e}^{-b a - b^2 T_{a,0}/2} \right) = \int_0^t \frac{a}{\sqrt{2 \pi s}} \exp\left( -b a - \frac{b^2 s}{2} -\frac{a^2}{2s} \right) \frac{\mathrm{d} s}{s}$$ The integral is doable by noticing that $$-b a - \frac{b^2 s}{2} -\frac{a^2}{2s} = -\frac{(a+b s)^2}{2s} = -2a b -\frac{(a-b s)^2}{2s}$$ and $$\frac{a}{s^{3/2}} = \frac{\mathrm{d}}{\mathrm{d} s} \frac{-2a}{\sqrt{s}} = \frac{\mathrm{d}}{\mathrm{d} s} \left( \frac{b s - a}{\sqrt{s}} - \frac{b s + a}{\sqrt{s}}\right)$$ Hence $$\begin{eqnarray} \mathbb{P}(T_{a,b} \leqslant t) &=& \int_0^t \frac{1}{\sqrt{2\pi}} \exp\left(- \frac{(a+bs)^2}{2 s}\right) \mathrm{d} \left( - \frac{b s + a}{\sqrt{s}} \right) + \\ &\phantom{+}& \int_0^t \frac{1}{\sqrt{2\pi}} \exp(-2ab) \exp\left(- \frac{(b s-a)^2}{2 s}\right) \mathrm{d} \left( \frac{b s - a}{\sqrt{s}} \right) \\ &=& -\Phi\left( \frac{b t + a}{\sqrt{t}} \right) + \lim_{t \searrow 0} \Phi\left( \frac{b t + a}{\sqrt{t}} \right) + \\ &\phantom{=}& \mathrm{e}^{-2 a b} \Phi\left(\frac{b t - a}{\sqrt{t}} \right) - \mathrm{e}^{-2 a b} \lim_{t \searrow 0} \Phi\left(\frac{b t - a}{\sqrt{t}} \right) \end{eqnarray}$$ where $\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-z^2/2} \mathrm{d} z$ is the cumulative distribution function of the standard normal variable. Since we assumed $a > 0$, $$\lim_{t \searrow 0} \Phi\left( \frac{b t + a}{\sqrt{t}} \right) = \Phi(+\infty) = 1 \qquad \lim_{t \searrow 0} \Phi\left( \frac{b t - a}{\sqrt{t}} \right) = \Phi(-\infty) = 0$$ and we arrive at c.d.f of the inverse Gaussian random variable: $$\mathbb{P}(T_{a,b} \leqslant t) = 1 - \Phi\left( \frac{b t + a}{\sqrt{t}} \right) + \mathrm{e}^{-2 a b} \Phi\left( \frac{b t - a}{\sqrt{t}} \right)$$

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@BenDerrett Evidently my answer is not helpful. Did you manage to solve it to your satisfaction. If so, I would appreciate it, if you dropped a hint, or better yet, posted it as a solution. Thanks. –  Sasha Apr 20 '12 at 14:42
Thanks for taking the time to write this. This is probably the best way to show 2). The question suggests there's an easier way to show this, assuming 1). –  Ben Derrett Apr 20 '12 at 16:03
@BenDerrett I suspect that the problem intends to prove 2) by differentiating to find pdf, computing the Laplace transform of the pdf. I would very much like to know if there is a more elegant way. So if you find out of one, please take a moment to share. –  Sasha Apr 20 '12 at 17:18