The distribution of Brownian motion enjoys many interesting symmetries. The reflection of a Brownian motion about any time \(s\) is also a Brownian motion.
Lemma 1. (Reflection at time \(s\)) Let \(B_t\) be a standard Brownian motion. Then, the process \((-B_t,t \geq 0)\) is a Brownian motion. More generally, for any \(s \geq 0\), the process \((\tilde{B_t},t\geq 0)\) defined by:
\[\begin{align*} \tilde{B}_t = \begin{cases} B_t & \text{ if } t\leq s\\ B_s - (B_t - B_s) & \text{ if }t > s \end{cases} \end{align*}\]
is a Brownian motion.
Claim. \((-B_t,t\geq 0)\) is a Brownian motion.
Proof.
We have, \(-B(0) = 0\).
For any increment \(s < t\), the increment \((-B_t) - (-B_s) = B_s - B_t\) is a Gaussian random variable with mean \(0\) and variance \(t - s\).
For any choice of \(n\) times, \(0 \leq t_1 \leq t_2 \leq \ldots \leq t_n\), the increments:
\[\begin{align*} (B_{0} - B_{t_1}), (B_{t_1} - B_{t_2}), \ldots, (B_{t_n} - B_{t_{n-1}}) \end{align*}\]
are independent
The paths \(-B_t(\omega)\) are continuous.
Thus, \((-B_t,t\geq 0)\) is a standard Brownian motion.
Claim. \((\tilde{B_t},t\geq 0)\) is a Brownian motion.
Proof.
Let \(s \geq 0\) be any arbitrary time.
We have, \(\tilde{B}(0) = 0\).
Consider any increment \(\tilde{B}(t_2) - \tilde{B}(t_1)\), \(t_2 < t_1\).
Case I. \(s \leq t_1 < t_2\)
Here
\[\begin{align*} \tilde{B}(t_2) - \tilde{B}(t_1) &= B(s) - (B(t_2) - B(s)) - (B(s) - (B(t_1) - B(s))) \\ &= -(B(t_2) - B(t_1)) \end{align*}\]
Hence, \(\tilde{B}(t_2) - \tilde{B}(t_1) \sim \mathcal{N}(0,t_2 - t_1)\).
Case II. \(t_1 < s < t_2\)
Here
\[\begin{align*} \tilde{B}(t_2) - \tilde{B}(t_1) &= B(s) - (B(t_2) - B(s)) - B(t_1)\\ &= (B(s) - B(t_1)) - (B(t_2) - B(s)) \end{align*}\]
\(B(s) - B(t_1)\) and \(B(t_2) - B(s)\) are independent random variables. Moreover, \(B(s) - B(t_1) \sim \mathcal{N}(0,s - t_1)\) and \(B(t_2) - B(s) \sim \mathcal{N}(0,t_2 - s)\). Consequently, \(\tilde{B}(t_2) - \tilde{B}(t_1) \sim \mathcal{N}(0,t_2 - t_1)\).
Case III. \(t_1 < t_2 \leq s\)
Here
\[\begin{align*} \tilde{B}(t_2) - \tilde{B}(t_1) &= B(t_2) - B(t_1) \end{align*}\]
so \(\tilde{B}(t_2) - \tilde{B}(t_1) \sim \mathcal{N}(0,t_2 - t_1)\).
Finally, the paths \(\tilde{B}(t,\omega)\) are continuous. Hence, \((\tilde{B}(t),t\geq 0)\) is a standard brownian motion.
Reflection Principle
It turns out that the above reflection property holds even if \(s\) is replaced by a stopping time. I prove this here.
Lemma 2. (Reflection Principle) Let \((B_t,t \geq 0)\) be a standard Brownian motion and \(\tau\) be a stopping time for its filtration. Then, the process \((\tilde{B}(t),t\geq 0)\) defined by the reflection at time \(\tau\):
\[\begin{align*} \tilde{B}(t) &= \begin{cases} B_t & \text{ if } t\leq \tau\\ B_\tau - (B_t - B_\tau) & \text{ if }t > \tau \end{cases} \end{align*}\]
is also a standard Brownian motion.
Bachelier’s formula
It is an amazing fact, that some simple manipulations using stopping time yield the complete distribution of the first passage time \(\tau_a\) of a Brownian motion as well as the distribution of the running maximum of the Brownian path on an interval of time \([0,T]\). This is surprising since the maximum of the Brownian path on \([0,T]\), denoted by \(\sup_{0\leq t \leq T} B_t\) is a random variable that depends on the whole path on \([0,T]\). This beautiful result is due to Bachelier.
Proposition 3. (Bachelier’s formula) Let \((B_t,t\leq T)\) be a standard Brownian motion on \([0,T]\). Then, the CDF of the random variable \(\sup_{0 \leq t\leq T} B_t\) is:
\[\begin{align*} \mathbb{P}\left(\sup_{0\leq t \leq T} B_t \leq a\right) = \mathbb{P}(|B_T| \leq a) \quad \text{ for any }a\geq 0 \end{align*}\]
In particular, its PDF is:
\[\begin{align*} f_{max}(a) = \frac{2}{\sqrt{2\pi T}} e^{-a^2/2T} \end{align*}\]
In other words, the random variable \(\sup_{0 \leq t \leq T} B_t\) (the maximum of the brownian motion at any time \(t\)) has the same distribution as \(|B_T|\) (the terminal distribution of the absolute value of the brownian motion).
This equality holds in distribution for a fixed \(t\). As a bonus corollary, we get the distribution of the first passage time at \(a\).
Proof. Consider \(\mathbb{P}(\sup_{t \leq T} B_t \geq a)\). By splitting this probability over the event of the endpoint, we have:
\[\begin{align*} \mathbb{P}(\sup_{t \leq T} B_t \geq a) &= \mathbb{P}(\sup_{t \leq T} B_t \geq a, B_T > a) + \mathbb{P}(\sup_{t \leq T} B_t \geq a, B_T \leq a) \end{align*}\]
Note also that \(\mathbb{P}(B_T = a) = 0\). Hence, the first probability equals \(\mathbb{P}(B_T \geq a)\). As for the second, consider the time \(\tau_a\). On the event considered, we have \(\tau_a \leq T\) and using the reflection principle (lemma 2), we get:
\[\begin{align*} \mathbb{P}(\sup_{t \leq T} B_t \geq a, B_T \leq a) &= \mathbb{P}(\sup_{t \leq T} B_t \geq a, \tilde{B}_T \geq a) \end{align*}\]
Observe that, since \(\tau_a \leq T\), the event \(\{\sup_{t \leq T} B_t \geq a\}\) is the same as \(\{\sup_{t\leq T} \tilde{B}(t) \geq a\}\). Therefore the above probability is:
\[\begin{align*} \mathbb{P}(\sup_{t \leq T} B_t \geq a) &= \mathbb{P}(B_T > a) + \mathbb{P}(\sup_{t \leq T} \tilde{B}_t \geq a, \tilde{B}_T \geq a) \end{align*}\]
But, \(\tilde{B}_t\) is also a standard brownian motion and has the same distribution as \(B_t\). \(\mathbb{P}(B_t \in S) = \mathbb{P}(\tilde{B}_t \in S)\) by the reflection principle. So, we can simply drop the tilde signs and write:
\[\begin{align*} \mathbb{P}(\sup_{t \leq T} B_t \geq a) &= \mathbb{P}(B_T > a) + \mathbb{P}(\sup_{t \leq T} {B}_t \geq a, {B}_T \geq a)\\ &=\mathbb{P}(B_T > a) + \mathbb{P}({B}_T \geq a)\\ &= \mathbb{P}(B_T \leq -a) + \mathbb{P}(B_T \geq a) \\ & \quad \{\text{ By symmetry of the Gaussian distribution }\}\\ &= \mathbb{P}(B_T \leq -a \cup B_T \geq a) \\ &= \mathbb{P}(|B_T| \geq a) \end{align*}\]
We conclude that:
\[\begin{align*} \mathbb{P}(\sup B_t \leq a) = \mathbb{P}(|B_T| \leq a) \end{align*}\]
as claimed.
To derive the PDF, we can always write:
\[\begin{align*} \mathbb{P}(\sup_{t \leq T} B_t \geq a) &= \mathbb{P}(B_T > a) + \mathbb{P}({B}_T \geq a)\\ &= 2\mathbb{P}(B_T \geq a)\\ &= 2(1 - F_{B_T}(a)) \end{align*}\]
So:
\[\begin{align*} F_{\sup B_t}(a) &= 1 - 2(1 - F_{B_T}(a))\\ \frac{d}{da}(F_{\sup B_t}(a)) &= 2 \frac{d}{da}(F_{B_T}(a))\\ f_{\sup B_t}(a) &= 2 f_{B_T}(a)\\ &= \frac{2}{\sqrt{2\pi T}}\exp\left[-\frac{a^2}{2T}\right] \end{align*}\]
Distribution of the first passage time \(\tau_a\)
Corollary 4. Let \(a \geq 0\) and let \(\tau_a = \min \{t \geq 0: B_t \geq a\}\). Then:
\[\begin{align*} \mathbb{P}(\tau_a \leq T) = \mathbb{P}\left(\sup_{0 \leq t \leq T} B_t \geq a\right) = \int_{a}^{\infty} \frac{2}{\sqrt{2\pi T}}e^{-x^2/2T} dx \end{align*}\]
In particular, the random variable \(\tau_a\) has PDF:
\[\begin{align*} f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi}} \frac{e^{-a^2/2t}}{t^{3/2}} \end{align*}\]
This implies that it is heavy-tailed and \(\mathbb{E}[\tau_a] = \infty\).
Proof.
The maximum on \([0,T]\) is larger than or equal to \(a\), if and only if, \(\tau_a \leq T\). Therefore, the events \(\{\sup_{0 \leq t \leq T} B_t \geq a\}\) and \(\{\tau_a \leq T\}\) are the same. So, the CDF \(\mathbb{P}(\tau_a \leq t)\) of \(\tau_a\) is
\[\begin{align*} F_{\tau_a}(t) = \int_{a}^{\infty} \frac{2}{\sqrt{2\pi t}} e^{-\frac{x^2}{2t}} dx \end{align*}\]
To get the PDF, it remains to differentiate the integral with respect to \(t\). This is easy to do, once we realise a change of variable \(u = x/\sqrt{t}\), \(du = dx/\sqrt{t}\) that:
\[\begin{align*} F_{\tau_a}(t) &= \int_{a/\sqrt{t}}^{\infty} \frac{2}{\sqrt{2\pi}} e^{-\frac{u^2}{2}}du\\ &= 2(1 - \Phi\left(\frac{a}{\sqrt{t}}\right)) \end{align*}\]
Differentiating on both sides with respect to \(t\), we get:
\[\begin{align*} f_{\tau_a}(t) &= - 2\phi\left(\frac{a}{\sqrt{t}}\right) \left(-\frac{1}{2}\right) \frac{a}{t^{3/2}}\\ &= \frac{a}{t^{3/2}} \frac{e^{-a^2/2t}}{\sqrt{2\pi}} \end{align*}\]
This closes the proof.