First Passage Times¶

Using the CDF formula \(\mathbb{P}(\tau_a \leq t) = 2\Phi(-a/\sqrt{t})\):

\(\mathbb{P}(\tau_1 \leq 1)\): With \(a = 1\), \(t = 1\):

\(\mathbb{P}(\tau_1 \leq 4)\): With \(a = 1\), \(t = 4\):

\(\mathbb{P}(\tau_2 \leq 4)\): With \(a = 2\), \(t = 4\):

Note that \(\mathbb{P}(\tau_2 \leq 4) = \mathbb{P}(\tau_1 \leq 1)\), which is consistent with the scaling \(\tau_{ca} \overset{d}{=} c^2 \tau_a\) (here \(c = 2\), so \(\tau_2 \overset{d}{=} 4\tau_1\)).

We verify \(\int_0^\infty f_{\tau_a}(t)\,dt = 1\) using the substitution \(u = a/\sqrt{t}\).

Then \(t = a^2/u^2\) and \(dt = -2a^2/u^3\,du\). When \(t \to 0^+\), \(u \to \infty\); when \(t \to \infty\), \(u \to 0^+\):

Substituting \(t = a^2/u^2\), so \(t^{-3/2} = u^3/a^3\):

Since \(\int_0^\infty e^{-u^2/2}\,du = \sqrt{\pi/2}\):

We compute \(\mathbb{E}[\tau_a^{1/2}] = \int_0^\infty t^{1/2} f_{\tau_a}(t)\,dt\) using the Lévy density:

Substitute \(u = a^2/(2t)\), so \(t = a^2/(2u)\) and \(dt = -a^2/(2u^2)\,du\):

This integral diverges logarithmically! Let us redo this more carefully. We have \(r = 1/2\), so the integrand at \(t \to \infty\) behaves as \(t^{1/2} \cdot t^{-3/2} = t^{-1}\), which is not integrable. However, the Gaussian factor \(e^{-a^2/(2t)}\) decays slowly (approaching 1) for large \(t\).

Instead, use the Laplace transform approach. From \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), we can use the identity:

Substitute \(\beta = a\sqrt{2\alpha}\), so \(\alpha = \beta^2/(2a^2)\) and \(d\alpha = \beta/(a^2)\,d\beta\):

This shows \(\mathbb{E}[\tau_a^{-1/2}] < \infty\). For \(\mathbb{E}[\tau_a^{1/2}]\), the tail of \(f_{\tau_a}(t)\) is \(\sim \frac{a}{\sqrt{2\pi}} t^{-3/2}\), and \(t^{1/2} \cdot t^{-3/2} = t^{-1}\), which is not integrable at infinity. But the Gaussian factor provides just enough decay: using \(u = a^2/(2t)\), the integral becomes \(\frac{a}{\sqrt{2\pi}} \int_0^\infty u^{-1} e^{-u}\,du\), which is \(\frac{a}{\sqrt{2\pi}} \cdot \Gamma(0)\) — this diverges. So actually \(\mathbb{E}[\tau_a^{1/2}]\) is finite only because the condition \(r < 1/2\) is strict. In fact, \(\mathbb{E}[\tau_a^r] < \infty\) iff \(r < 1/2\), so \(r = 1/2\) is the borderline case. To show finiteness for \(r < 1/2\), take any such \(r\). The integrand for large \(t\) behaves as \(t^r \cdot t^{-3/2} = t^{r - 3/2}\), which is integrable at \(\infty\) iff \(r - 3/2 < -1\), i.e., \(r < 1/2\). The integral near \(t = 0\) converges due to the factor \(e^{-a^2/(2t)}\) which decays faster than any power. Hence \(\mathbb{E}[\tau_a^r] < \infty\) for all \(r < 1/2\).

From the Laplace transform \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), moments are obtained by differentiation:

For \(n = 1\): \(\frac{d}{d\alpha} e^{-a\sqrt{2\alpha}} = -\frac{a}{\sqrt{2\alpha}} e^{-a\sqrt{2\alpha}}\), and as \(\alpha \to 0^+\), \(\frac{a}{\sqrt{2\alpha}} \to \infty\), so \(\mathbb{E}[\tau_a] = \infty\).

For the variance: \(\text{Var}(\tau_a) = \mathbb{E}[\tau_a^2] - (\mathbb{E}[\tau_a])^2\). Since \(\mathbb{E}[\tau_a] = \infty\), the variance is automatically \(\infty\).

Alternatively, even if we consider \(\mathbb{E}[\tau_a^2]\) directly, differentiating twice gives terms involving \(\alpha^{-3/2}\) which diverge as \(\alpha \to 0^+\). Therefore \(\text{Var}(\tau_a) = \infty\).

By the scaling property of Brownian motion, \(\{W_{c^2 t}\}_{t \geq 0} \overset{d}{=} \{c\,W_t\}_{t \geq 0}\) as processes. Define \(\widetilde{W}_t = W_{c^2 t}/c\), which is a standard Brownian motion.

Then:

Substitute \(t = c^2 s\), so we want the first time \(W_{c^2 s} = ca\), i.e., \(W_{c^2 s}/c = a\), i.e., \(\widetilde{W}_s = a\):

since \(\widetilde{W}\) is a standard Brownian motion and \(\tau_a\) under \(\widetilde{W}\) has the same distribution as \(\tau_a\) under \(W\).

We verify the identity \(\frac{\partial}{\partial a}\mathbb{E}[e^{-\alpha\tau_a}] = -\sqrt{2\alpha}\,\mathbb{E}[e^{-\alpha\tau_a}]\).

Since \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), differentiate with respect to \(a\):

This confirms the identity. The interpretation is that increasing the target level \(a\) by a small amount \(da\) reduces the Laplace transform by a factor proportional to \(\sqrt{2\alpha}\), reflecting the additional time needed to travel the extra distance \(da\).

For Brownian motion with drift \(\mu\), \(X_t = W_t + \mu t\), the Laplace transform of the first passage time to \(a > 0\) is:

Setting \(\mu = 0\):

This matches the formula for standard Brownian motion. The drift term \(\mu\) modifies the exponent: when \(\mu > 0\) (positive drift toward \(a\)), the factor \(\sqrt{2\alpha + \mu^2} - \mu < \sqrt{2\alpha}\), so the Laplace transform is larger (closer to 1), reflecting that the hitting time is stochastically smaller. When \(\mu < 0\) (drift away from \(a\)), the factor increases, reflecting longer expected hitting times.