The tl;dr

In this post:

  • I talk about a turning point in my life when I went from being a maths student to a researcher. It involves proving an existing theorem about Brownian motion with a more accessible and elegant approach than what’s usually in the literature.
  • I introduce the necessary maths to understand this result in an accessible way to non-experts. This post serves as a gentle introduction to the theory of random variables, stochastic processes and Brownian motion, with interactive simulations to help build intuition.
  • I derive the distribution of the first hitting time for a Brownian motion with drift without using Girsanov’s theorem, by invoking the reflection principle and properties of the Brownian bridge.


Table of contents


Maths as a formative life experience

An alternative–or perhaps even a more appropriate–title would be something like:

That one time I filled a hole in my lecturer’s course notes

Because really, when I think about the reflection principle and Brownian bridges, I don’t think of maths. Instead, I think of a period in my life at the university when I was transitioning from a student to a researcher.

Every maths researcher has a defining moment that marks this transition, and it often has to do with what they were investigating at that very moment. For me, it was the Brownian bridge that made me realise that mathematics research wasn’t another abstract thing to me like building a spaceship or doing brain surgery. That–unlike those things–I really could be a maths researcher.

To set the scene, in my third year at the University of Sydney, I took my first course in stochastic processes. If you don’t know what that is, first of all: don’t fret. My goal here is to write about maths in an educational way that hopefully non-experts can understand, at least at a high level. And secondly, stochastic processes is a subfield within probability theory, which studies how random objects behave over time.

When I took this course, I was still most definitely a maths student. It was the lecturer who offered to take me under his wing and give me an opportunity to do real mathematics research. Over the summer, my mentor had me read up on some of his papers. Any mathematician knows when you first try to sink your teeth into a new research area, you’ll need to spend at least months building up foundational theory in that domain. As such, I ended up spending a considerable amount of time going through my mentor’s full set of lecture notes, of which the stochastic processes course covered maybe only around 40%. This meant I could do two things for my mentor’s lecture notes. Firstly, I could identify typos and errors. But secondly, and more importantly…

In my mentor’s lecture notes, he wanted to include a key theorem about the behaviour of a class of stochastic processes. However, the way to prove that theorem that he was aware of was by invoking an advanced theorem that was definitely outside the scope of undergraduate mathematics. Therein lies the tension:

  • Mathematics is all about proofs.
  • Undergraduate lecture notes for maths should represent a progressive journey, where results are proven in ways accessible by undergraduates, and those results are then used to prove more important results.
  • This particular theorem that the lecturer wanted to keep was elegant and important to the subject matter.
  • There was seemingly no way to prove this theorem using undergraduate tools…

… or was there? Foreshadowing is a literary device.

My mentor left in a proof of the theorem in his lecture notes, but highlighted the step where he invoked more advanced results in red. I think he had a feeling that it could be done, but left the investigation in his backlog. And thus, when I was making my way through his lecture notes, he directed me to focus on this particular problem.

And after a couple of days, I managed to produce a proof of the theorem using only knowledge built up to that point in his lecture notes. I’d go even further and say that the proof was elegant and insightful. This was the first time I had accomplished a genuine sense of mathematical discovery outside of the didactic environment of coursework. And it was the experience that taught me I really could do research.

And thus, as my mentor added my proof into his lecture notes, they were finally complete. He had rather humorously remarked that if he were to actually give a course that included the subject matter, he would invite me as a guest lecturer. While I’m glad that this had never actualised, I’ve finally decided to write a post on it. Six years later.


The mission

An alternative title for this post that would be more appealing to goal-oriented mathematicians would be:

Deriving the distribution of the first hitting time of Brownian motion with drift without using Girsanov’s theorem

After all, mathematicians tend to be very literal and explicit.

Accessibility to non-experts

Immediately, there is a lot to unpack here. What is a “first hitting time”? A “Brownian motion”? A “drift”? A “Girsanov’s theorem”?

If you’re a non-mathematician, this is probably already enough jargon to discourage you from continuing. But actually, you are the core audience for this post. I strongly believe in the idea that maths can have wide appeal, and the major reason why it doesn’t is because currently it tends to attract a very small and esoteric audience who have seemingly given up on this idea. It’s strange to me literature is widely enjoyed and appreciated, but not maths, despite them both being context-heavy and requiring a lot of prerequisite knowledge to fully understand. I think the terrible way in which maths is taught in high school education has a lot to do with it, but alas that’s a discussion for another day.

And so the primary mission here is to talk about these maths ideas and topics, some of which are typically only encountered in postgraduate studies, in a way that’s digestible for someone who may have only studied maths at the high school level. Of course, it’ll be a challenge, and the only way we can meet it is by talking about foundational concepts at a high-level when necessary. And maybe at times such a reader may not fully understand how a proof works at a granular level, but even a high-level understanding would be enough for appreciating how maths is conducted at an advanced level.

The actual maths problem at hand

With that out of the way, what’s the exact problem that we’re trying to solve?

We’ll delve more into the details of what each of the terms mean later, but essentially a “Brownian motion with drift” is a specific kind of a stochastic process. One thing that’s commonly investigated about any stochastic process is how long it’ll take before it reaches some region in the space that it’s randomly walking in. If you think of the movement of an ant as being a stochastic process, then “how long will it take for the ant to find the exit to the room” would be such a question. we call such questions “first hitting times”. A first hitting time is a random object (“random variable”) in itself, as its dependency on the randomness of the underlying stochastic process means that the first hitting time can potentially happen at any point in time. Since a first hitting time is a random variable, it must have a probability distribution. Distributions are helpful in helping us understand the randomness of the object, for example its mean and variable are often directly calculated through its distribution.

And so, the maths problem at hand is: What’s the probability distribution of the first hitting time of a Brownian motion with drift?

A theorem is a significant mathematical result, and the theorem for this distribution already exists and is well-known. (At least by the standards of mathematicians.) However, its proof typically invokes a much more advanced result called Girsanov’s theorem. And while Girsanov’s theorem can certainly get us this distribution, we would then be troubled by the question of “but how do we prove Girsanov’s theorem?” That this question requires a much more advanced field called stochastic analysis to answer.

All of that is to say, for course on stochastic processes before stochastic analysis, we want to determine the distribution of the first hitting time without Girsanov’s theorem.

And how will we do this? By taking taking a shortcut through a bridge. A Brownian bridge, to be exact.


Building up the understanding

Alternatively, building the bridge.

Like the Lord of the Rings movies, you can’t understand the third movie if you don’t know what happens in the first two.

If you’re already well-versed with all of the concepts discussed, you can probably already skip to the proof section at the end of this post. But once again, the audience I want to cater to are non-mathematicians, so in this section we’ll build up the necessary knowledge to mathematically understand the theorem and its proof. To such a reader, this section will likely be the most valuable. It serves as a gentle introduction to random variables, stochastic processes, and the reflection principle.

I sprinkle in several interactive simulations to help with visualising and building intuition.

Random variables

A random variable is simply any object such that its value isn’t deterministic. That is, it’s a variable that’s random. (See, I told you mathematicians tend to be very literal.) Coin flips, dice rolls, stock prices, the position of a drop of dye in a beaker, are all examples of random variables. But how do we actually mathematically formulate a random variable?

Let $(\Omega, \mathcal F,\mathbb P)$ be a probability space, where:

  • $\Omega$ is the set of possible outcomes. I like to think of this is a set of parallel universes, where if somehow you know exactly which universe you’re in, then you can theoretically determine the outcome of every random variable.
  • $\mathcal F$ is a sigma-algebra. This is the set of all possible “events” that we can calculate probabilities for.
    • For example, the statement “getting heads on the first flip and tails on the second and third flips” is an event.
  • $\mathbb P$ is a probability measure. This is a function that assigns probabilities to each event in $\mathcal F$. In the parallel universe interpretation of $\Omega$, given an event, the measure $\mathbb P$ tells the portion of parallel universes for which that event occurs.
    • For example, $\mathbb P(\text{rolling 6 on a dice}) = 1/6$ is equivalent to saying that “of all possible parallel universes, getting a 6 by rolling a particular dice happens in exactly one in six of those universes.”

This is the formulation of probability using measure theory, which is a branch of maths that studies measures (e.g. lengths, areas, volumes). Why measure theory? Two reasons. Firstly, probabilities are just measures. Secondly, due to complicated reasons about maths itself, the popular mainstream formulation of maths allows for the existence of non-measurable sets, which are things that you can’t measure because attempting to measure them breaks the definiton of what a measure is. By formulating probability theory using measure theory (“measure-theoretic probability theory”), we mitigate those issues.

A random variable $X(\omega)$ is a function that takes an outcome $\omega\in\Omega$ as input and outputs some value (1 to 6 for a dice roll, “head” or “tail” for a coin flip, a positive real number for a stock price). This is where the parallel universes interpretation of $\Omega$ comes handy: where $\omega\in\Omega$ is a single parallel universe, $X(\omega)$ has a single deterministic value–it’s no longer random! But if we don’t know which parallel universe we’re in, then $X$ might be any value–it’s random. Randomness comes from not knowing exactly which universe (or history) we exist in. Whether you believe in the existence of parallel universes is a separate question–this is just how probabilities are formulated in maths.

There is also another condition required for $X$ to be a random variable: it needs to be measurable, but we can safely skip this as the nuances of measure theory aren’t required for our discussion.

If we don’t know which parallel universe we’re in, then how do we gain information about the value of the random variable $X$?

That’s where its probability distribution comes in. In this post, we’re going to focus on one-dimensional continuous-valued random variables only. That is, where $X(\omega)\in \mathbb R$ (where $\mathbb R$ is the set of all real values, e.g. 0, -1, 0.12, $\pi$). For example, the stock price at a future point in time can be modelled as a continuous-valued random variable that takes positive real values only.

The probability distribution of a continuous random variable $X$ is given by its probability density function, which can be written as $f_X(x)$ or $\mathbb P(X \in dx)$. The probability of the random variable taking a value within some interval $[a, b]$ is given by $\mathbb P(X\in [a, b]) = \int_a^b f_X(x)\,dx = \int_a^b \mathbb P(X \in dx)$. So calculating probabilities, or statistical quantities that involve probabilities (e.g. mean and variance), is just a matter of integration. For better or for worse.

For a probability density to be valid, the random variable must “take a real value with probability one”, that is, it must be true that

\[\mathbb P(X\text{ takes a real value}) = \mathbb P(-\infty < X < +\infty) = \int_{-\infty}^{+\infty} f_X(x)\, dx = \int_{-\infty}^{+\infty} \mathbb P(X \in dx) = 1.\]

I want to highlight that $\mathbb P(X \in dx)$ is a particularly useful way of representing a density. Think of $dx$ as “an infinitesimally small region around $x$”, or “the interval $(x-\varepsilon, x+\varepsilon)$ for an infinitesimally small $\varepsilon$” (remember high school calculus?). That way, we can make sense of a probability density as a probability of an event (but a really specific one) that we can play around with when it’s convenient (and we will do so later). This is a lot more useful than writing $\mathbb P(X = x)$, which is always zero due to the way integration works: $\mathbb P(X = x) = \int_x^x f_X(x)\,dx = 0$.

Note that to describe the distribution of a random variable, it’s simply enough to give its density or its cumulative distribution function $F(x) = \mathbb P(X \le x)$, as integrating the former gives you the latter while differentiating the latter gives you the former. So when we talk about the “distribution” of a random variable, you can think of either way of representing it.

We don’t need to go much further into the theory of random variables for the objectives of this post, but I’ll mention two key quantities, the mean and the variance of a random variable:

\[\begin{align} \mathbb E[X] &= \int_{-\infty}^{+\infty} x f_X(x)\, dx\\ \text{Var}(X) &= \int_{-\infty}^{+\infty} \left(x - \mathbb E[X]\right)^2 f_X(x)\, dx \end{align}\]

The most commonly used continuous random variable is one that you’re probably familiar with yourself: a normally distributed random variable, for which it has the density $f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e ^ {-(x - \mu)^2 / 2\sigma^2}$, where $\mu$ and $\sigma^2$ are its mean and variance respectively. We write “$X \sim N(\mu, \sigma^2)$” to mean that the random variable $X$ is of this normal distribution.

Let’s look at some random samples from a $N(\mu, \sigma^2)$ random variable.

$\mu =$   $\sigma =$  



Example: uniform distribution Let $U$ be a uniformly distributed random variable over $[a, b]$. This means it has a probability density given by $f_U(x) = 1/(b - a)$ when $x\in[a,b]$ and zero everywhere else. Then, it's mean and variance are given by $$\begin{align} \mathbb E[X] &= \int_{a}^{b} \frac{x}{b - a}\, dx = \frac12\frac{b^2 - a^2}{b - a} = \frac12\frac{(b-a)(b+a)}{b - a} = \frac{a + b}{2},\\ \text{Var}(X) &= \int_{a}^{b} \left(x - \mathbb E[X]\right)^2 \frac{1}{b - a}\, dx \\ &= \int_{a}^{b} \frac{x^2}{b - a}\, dx - \mathbb E[X]^2 \\ &= \frac{b^3 - a^3}{3(b-a)} - \left(\frac{a + b}{2}\right)^2 \\ &= \frac{a^2 + ab + b^2}{3} - \frac{a^2 + 2ab - b^2}{4} \\ &= \frac{(b-a)^2}{12}, \end{align}$$ where we applied the difference of two squares in the calculation of the expectation, and in the calculation of the variance, we applied the formula that $\text{Var}(X) = \mathbb E[X^2] - \mathbb E[X]^2$ (easy to verify yourself) and the difference of two cubes.


I want to highlight a useful property on how normally distributed random variables scale. Let $N\sim N(0,1)$ and $X = \sigma N$. Then, $X \sim N(0,\sigma^2)$. That is, every normally distributed random variable with zero mean can be expressed the scaled version of a $N(0,1)$-distributed random variable.

Proof The probability of $X$ taking values below $a$ is given by $$\mathbb P(X \le a) = \mathbb P(\sigma N \le a) = \mathbb P(N \le a/\sigma) = \int_{-\infty}^{a/\sigma} f_N(y)\,dy = \int_{-\infty}^{a/\sigma} \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\,dy.$$ Let's do the change of variables $y = x/\sigma$, so $dy = (1/\sigma)dx$ and hence $$\mathbb P(X \le a) = \int_{-\infty}^{a} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}\,dx,$$ therefore we see that the density of $X$ is the same as that for $N(0,\sigma^2)$, so $X\sim N(0,\sigma^2)$.


Finally, I want to quickly mention independence. Two random variables $X$ and $Y$ are independent if knowing the value of one random variable doesn’t give me any information about the other. For example, flipping a coin multiple times is independent, as knowing how one toss landed doesn’t give me any information about the next toss. However, whether it’s currently raining and whether the sky is cloudy are dependent random variables; if I know the former is true then the latter is probably also true.

And with that, we’re now ready to talk about stochastic processes.

Stochastic processes

When my lecturer had described stochastic processes as “randomness indexed by time”, he was being literal. They are random variables indexed by time. We’re only going to be talking about continuous-time stochastic processes in this post. This is as opposed to discrete-time stochastic processes, such as my money during the course of 100 rounds on a slot machine.

A (continuous-time) stochastic process is given by $\{X_t : t\in[0,T]\}$, where $X_t$’s are random variables and $T$ is some end time (possibly infinite).

Technically stochastic processes are also “adapted to filtrations”, but we don’t need to go into that much nuance here.

One common question that’s asked about stochastic processes is: what are the properties of the first hitting times for that particularly stochastic process? Before we formulate what a first hitting time is, we need to first formulate the concept of a “first”.

Let $S = (0,1)$ be a set of real numbers between zero and one, but excluding the end points. That is, 0.5, 0.001 are in $S$, but zero and one are not. What’s the minimum element of this set $\min S$? You might be tempted to say zero, but that would be incorrect because zero isn’t in this set! What about 0.001? But that can’t be the smallest number in $S$, because 0.0001 would be smaller. And so we can keep going; for every element in $S$ you provide, I can provide a smaller number in $S$, and so $\min S$ doesn’t actually exist–it’s undefined. And same with the maximum element $\max S$.

The lack of a guarantee of the existence of $\min$ and $\max$ is exactly why mathematicians invented the concepts of the infimum and supremum. The infimum of a set is the largest lower bound–a lower bound is a number such that every element in a set is greater than that. So for example, -1 is a lower bound for $S$, but -0.5 would be a larger lower bound. Zero is a lower bound for $S$, and it’s easy to see that any number higher than that will cease to be a lower bound. Therefore, $\inf S = 0$. Similarly, the supremum is the smallest upper bound, so $\sup S = 1$. Infima and suprema are guaranteed to always exist, and in cases where minima and maxima exist within the given set , they coincide. For this reason, for the remainder of this post, I’ll be using $\inf$ and $\sup$, but in your head you can effectively read them as $\min$ and $\max$.

This brings us back to first hitting times. Let $a\in \mathbb R$ be a threshold value. The first hitting time for when the stochastic process crosses the threshold $a$ is given by $\tau_a = \inf\{t \ge 0 : X_t = a\}$. As $\tau_a$ depends on the stochastic process, first hitting times are thus a random variable. And of course, first hitting times aren’t guaranteed to happen. For example, if a stochastic process starts at $X_0 = 0$ and we’re looking for the first time it hits $a = 100$, then it’s entirely possible that the process never reaches it; it may simply just randomly walk underneath it for all of time. In such cases, $\tau_a = +\infty$ by the definition of infimum (the infimum of an empty set is infinity).

One trivial fact to note is that $X_{\tau_a} = a$, because obviously at the time that the first hitting time happens, the stochastic process is at the threshold value.

Let’s now study our first stochastic process.

Brownian motion

Let $\{B_t : t \ge 0\}$ be a stochastic process that takes values in $\mathbb R$. It’s a Brownian motion if:

  • The difference between its position at two different points in time (“increment”) is normally distributed. That is, for every $t \ge 0$ and $\Delta \ge 0$, $B_{t+\Delta} - B_t \sim N(0, \Delta)$.
  • Increments are independent of the past (“Markov property”). That is, for every $t \ge 0$ and $\Delta \ge 0$, $B_{t+\Delta} - B_t$ is independent of all $B_s$ for $s \le t$.
  • It has continuous paths. That is, $B_t(\omega)$ is continuous in $t$ for every $\omega\in\Omega$. “Continuous” just means that there are no sudden jumps, i.e. you can trace the entire path with a pen without needing to lift it up.
  • It starts from the origin, i.e. $B_0 = 0$.

Let’s look at simulations of the Brownian motion.


Brownian motion is a model for a lot of things. It was named after the botanist Robert Brown when he described how a single pollen of a plant moved in water (of course, that would be a Brownian motion in $\mathbb R^3$). It’s also commonly used for modelling quantities in finance, and even the motions of stars and black holes. A major reason for why Brownian motion is used so often is because of its mathematical elegance.

Brownian motions are an example of Gaussian processes. A stochastic process is a Gaussian process if any vector of its marginals $(X_{t_1}, X_{t_2},\dots, X_{t_n})$ follows a multivariate normal distribution. A neat property about a Gaussian process is that if two stochastic processes are Gaussian and they have the same expectation and covariance structure $\text{Cov}(X_s, X_t) = \mathbb E[(X_s - \mathbb E[X_s])(X_t - \mathbb E[X_t])]$, then the stochastic processes have the same distribution. This means we can’t really tell them apart–that they’re the same for all practical intents and purposes.

Why is this true? The distributions of random variables can be studied by looking at their characteristic functions $\varphi_X(\theta) = \mathbb E[e^{i\theta X}]$. If you know a little bit about differential equations or engineering, you might recognise that this is actually just the Fourier transform. So once again, another example of the same concept appearing in totally different contexts! Characteristic functions are useful because if two random variables have the same characteristic function, then they have the same distribution. The characteristic function of a normal distribution can be straightforwardly computed as $\varphi(\theta) = \exp(-i\mu \theta - (\sigma\theta)^2/2)$. Similarly, a multivariate normal characteristic function looks like $\varphi(\boldsymbol{\theta}) = \exp(i\boldsymbol{\theta}^\intercal\boldsymbol{\mu} - \boldsymbol{\theta}^\intercal\boldsymbol{\Sigma}\boldsymbol{\theta}/2)$, where $\boldsymbol{\Sigma}$ is the covariance matrix. As only the mean vector and covariance matrix are the only parameters in the multivariate normal characteristic function, two multinormal distributions are the same if they have the same mean and covariances.


Using the Gaussian property along with the previous result on the scaling of normal random variables, it’s straightforward to show that Brownian motions are self-similar. That is, for every $c > 0$, it holds that $\{B_{ct} : t \ge 0\}$ has the same distribution as $\{\sqrt c B_t : t \ge 0\}$. This is a really neat fractal property of the Brownian motion where zooming into (the time axis of) a Brownian motion ($c < 1$) will look like a Brownian motion.

Let’s zoom into some Brownian motion sample paths below to examine this fractal-like nature. The plot on the right looks like a Brownian motion, doesn’t it? Just different axis scales, as expected.


The reflection principle

The reflection principle is a useful way to formalise the idea that at any point of a Brownian motion’s path, if we flip the direction of the Brownian motion moving forward after that point, the entire path of the resulting stochastic process is still a Brownian motion.

Let’s look at simulations of reflected Brownian motions at fixed reflection times. The reflected sample path looks just like a standard Brownian motion. After all, do you even know which path is the reflected one?

Reflection time:    


Let $a > 0$ be a threshold value and $\tau_a$ be the first hitting time of this threshold, i.e. $\tau_a = \inf\{t \ge 0: B_t = a\}$. Our motivating question is: what’s the distribution of the first hitting time $\tau_a$?

We derive the reflection principle as follows:

\[\begin{align} \mathbb P\left(\tau_a \le t\right) &= \mathbb P\left(\left\{\tau_a \le t\right\}\cap\left\{B_t \ge a\right\}\right) + \mathbb P\left(\left\{\tau_a \le t\right\}\cap\left\{B_t < a\right\}\right) \\ &= \mathbb P\left(B_t \ge a\right) + \mathbb P\left(\left\{\tau_a \le t\right\}\cap\left\{B_t < a\right\}\right) \\ &= \mathbb P\left(B_t \ge a\right) + \mathbb P\left(\tau_a \le t\right)\mathbb P\left(B_t < a | \tau_a \le t\right)\\ &= \mathbb P\left(B_t \ge a\right) + \mathbb P\left(\tau_a \le t\right)\mathbb P\left(B_t - B_{\tau_a} < a - B_{\tau_a}| \tau_a \le t\right)\\ &= \mathbb P\left(B_t \ge a\right) + \mathbb P\left(\tau_a \le t\right)\mathbb P\left(B_t - B_{\tau_a} < 0| \tau_a \le t\right)\\ &= \mathbb P\left(B_t \ge a\right) + \mathbb P\left(\tau_a \le t\right)\mathbb P\left(B_t - B_{\tau_a} < 0\right)\\ &= \mathbb P\left(B_t \ge a\right) + \frac12\mathbb P\left(\tau_a \le t\right), \end{align}\]

where our equalities hold because:

  1. Holds due to the law of total probability, i.e. “event $A$” is the same as “(event $A$ and event $B$) or (event $A$ and not event $B$)”. E.g. the statements “I’m in class” and “I’m either in class and paying attention, or in class and not paying attention” are equivalent.
  2. Holds due to the fact that for $B_t \ge a$ to happen, the Brownian motion must have hit $a$ before or at time $t$, so $\tau_a \le t$ must also happen, hence $\{\tau_a \le t\}\cap\{B_t \ge a\} = \{B_t \ge a\}$. E.g. “I’m in class” and “I’m in class and this class is taking place” are equivalent, because I can’t be in class if the class isn’t happening in the first place.
  3. Holds due to the definition of conditional probability, i.e. $\mathbb P(A|B) = \mathbb P(A\cap B) / \mathbb P(B)$.
  4. Holds due to subtracting both sides of an inequality by the same quantity.
  5. Holds due to $B_{\tau_a} = a$ by the definition of the hitting time $\tau_a$.
  6. Holds due to the Markov property that the increment over $[\tau_a, t]$ is independent of the information up to time $\tau_a$. Since the event $\{\tau_a \le t\}$ is contained in the formation up to time $\tau_a$ (obviously, because at the moment that the first hit happens you’ll know that it happened), the increment is thus independent of this event.
  7. Holds due to the fact that $B_t - B_{\tau_a}$ is distributed as $N(0, t - \tau_a)$, which is a symmetric probability distribution around zero so $\mathbb P\left(B_t - B_{\tau_a} < 0\right) = 1/2$.

Rearranging what we have, it holds that $\mathbb P\left(\tau_a \le t\right) = 2\mathbb P\left(B_t \ge a\right)$. Now, note that $\{\tau_a \le t\} = \{\sup_{0\le s\le t} B_s \ge a\}$, as obviously the first hitting time for reaching $a$ happening before time $t$ is equivalent to saying that the Brownian motion has a peak at or above $a$ before time $t$. Thus, we obtain the reflection principle.

Theorem (Reflection principle, Part 1). Let ${B_t : t \ge 0}$ be a Brownian motion, $a > 0$ and $\tau_a = \inf\{t \ge 0: B_t = a\}$. Then, it holds that

\[\mathbb P(\sup\nolimits_{0\le s\le t} B_s \ge a) = \mathbb P\left(\tau_a \le t\right) = 2\mathbb P\left(B_t \ge a\right).\]

Technically, in our proof of the reflection principle we didn’t fully use the fact that the stochastic process was a Brownian motion. In fact, our proof will still work for any stochastic process that:

  • is continuous,
  • has the Markov property where increments are independent of the past, and
  • has increments such that $\mathbb P(X_t - X_s < 0) = 1/2$ (e.g. has a symmetric distribution centred at zero),

thus the reflection principle holds for such stochastic processes.

How do we make sense of this result, and what we did in the proof? I find it easier to interpret when arranged as $\mathbb P\left(B_t \ge a\right) = \mathbb P(\tau_a \le t)/2$. Here’s how to think about it:

  • If the Brownian motion does meet or exceed $a$ in the time $[0,t]$, then after such point two things are possible:
    • either it’ll finish at time $t$ in a position that exceeds $a$, i.e. $B_t \ge a$, or
    • it’ll finish at time $t$ in a position lower than $a$, i.e. $B_t \le a$.
  • Hence, $\mathbb P\left(B_t \ge a\right) + \mathbb P\left(B_t \le a\right) = \mathbb P(\tau_a \le t)$ (which we arrived at more formally above).
  • But due to the symmetry of the Brownian motion, the probability of either happening is exactly the same, i.e. $\mathbb P\left(B_t \ge a\right) = \mathbb P\left(B_t \le a\right)$.
  • Therefore, the probability that $B_t \ge a$ is exactly half of the probability of the Brownian motion exceeding $a$ before $t$ in the first place, i.e. $\mathbb P\left(B_t \ge a\right) = \mathbb P(\tau_a \le t)/2$.

Let’s simulate this exact setup: given a threshold value, we reflect the path of the Brownian motion after the first hitting time. Pay attention to the symmetry: if a reflection happens, then for every path that ends above the threshold, there is also another path that ends below it.

Threshold value:    


The formulation of the reflection principle as $\mathbb P(\sup\nolimits_{0\le s\le t} B_s \ge a) = 2\mathbb P\left(B_t \ge a\right)$ is incredibly useful because it directly gives us a way to compute the maximum position of a Brownian motion up until time $t$, which is exactly what $\sup_{0\le s\le t}B_s$ is. Specifically, by noting that due to the symmetry of the normal distribution, $2\mathbb P(B_t \ge a) = \mathbb P(\{B_t \ge a\}\cup\{-B_t \ge a\}) = \mathbb P(|B_t| \ge a)$, we thus have

\[\mathbb P(\sup\nolimits_{0\le s\le t} B_s \ge a) = \mathbb P(|B_t| \ge a),\]

which says that the maximum of a Brownian motion until time $t$ has the same distribution as $|B_t|$, which is called a folded Brownian motion.

The formulation of the reflection principle as $\mathbb P\left(\tau_a \le t\right) = 2\mathbb P\left(B_t \ge a\right) = 2(1 - \mathbb P\left(B_t < a\right))$ immediately answers the question of what the distribution of the first hitting time is for a Brownian motion, as $\mathbb P\left(B_t < a\right)$ is simply the cumulative distribution function for a $N(0,t)$ random variable. We can differentiate the expression with respect to $t$ to get the probability density for $\tau_a$ if we wanted to.

There is yet more to the reflection principle. If we go back to the proof at the beginning of this section and replace $B_t$ with $2B_{\tau_a} - B_t$ after the first equality, the rest of the proof still holds one-to-one, and so we arrive at a similar result that for every $a \ge 0$, $\mathbb P(\tau_a \le t) = 2\mathbb P(2B_{\tau_a} - B_t \ge a)$. Rearranging this to the cumulative probability function for $2B_{\tau_a} - B_t$, we see that

\[\mathbb P(2B_{\tau_a} - B_t \le a) = 1 - \mathbb P(2B_{\tau_a} - B_t > a) = 1 - \frac12\mathbb P(\tau_a \le t) = 1 - \mathbb P(B_t > a) = \mathbb P(B_t \le a),\]

that is, $2B_{\tau_a} - B_t$ and $B_t$ have the same distribution. Technically we also need to show that the above holds for $a \le 0$, but once again it’s easy to check that this is true by defining $\tau_a = \inf\{0\le s\le t : B_t \le a\}$ for $a < 0$ and repeating our proof again.

This leads to our second formulation of the reflection principle.

Theorem (Reflection principle, Part 2). Let $\{B_t : t \ge 0\}$ be a Brownian motion, $a > 0$ and $\tau_a = \inf\{t \ge 0: B_t = a\}$. Let

\[\tilde B_t = \begin{cases} B_t\quad&\text{for } t \le \tau_a,\\ 2B_{\tau_a} - B_t \quad& \text{for } t > \tau_a. \end{cases}\]

Then, the process $\{\tilde B_t : t\ge 0\}$ is also a Brownian motion.

Technically, we need to verify that $\{\tilde B_t : t\ge 0\}$ is indeed a Brownian motion by showing that it meets the definition of a Brownian motion, but this is very easy to do by exploiting the fact that $\{B_t : t\ge 0\}$ is a Brownian motion.

This version of the reflection principle formalises very explicitly the idea that if a Brownian motion gets “reflected” at the first hitting time for touching some threshold, it’s sample path still just looks like a typical Brownian motion. Or in other words, that act of reflection had zero effect from a probabilistic view. If somebody made the Brownian motion reflect after the first hitting time and and didn’t tell us, we wouldn’t be able to tell!

To prove this point, I’ve simulated a Brownian motion that gets reflected at the first hitting time of some positive threshold. I’m not going to tell you what this threshold is. Can you tell?


The second version of the reflection principle is the one that we’ll use to prove the main theorem later.


Brownian bridge

So far, we’ve managed to obtain the distribution of the first hitting time of a standard Brownian motion in the form of the reflection principle. We want to investigate this in the case of a more general Brownian motion, but before we do so we must introduce the Brownian bridge, which will be central in our pursuit.

Let $\{B_t : t \in [0,T]\}$ be a Brownian motion over a fixed time interval $[0,T]$. Let’s say that you know what the position of $B_T$ ahead of time. How might this arise? Maybe there’s a particle that moves like a Brownian motion, and you know where the particle is currently. Then, you have an “information gap” where you don’t know the exact path of how the particle got to where it is currently. For the sake of simplicity, we’ll express this as $\{(B_t | B_T) : t \in [0,T]\}$, where the “$| B_T$” means “given the fact that we know what $B_T$ is” (i.e. we’re “conditioning on $B_T$” in the language of conditional probability). Then this new stochastic process is called a Brownian bridge.

Let’s talk a bit more about how this conditioning works. Since we know where $B_T$ is, $B_T | B_T$ is deterministic. For example, if $B_T = a$, then $(B_T | (B_T = a)) = a$. However, knowing $B_T$ is not enough to completely determine where $B_t$ is for $t < T$, so $B_t | B_T$ is still random. That said, of course we can expect that knowing $B_T$ gives some information about $B_t$. For example, due to the continuity of the Brownian motion, the closer $t$ is to $T$, the more we would expect $B_t$ to be closer to $B_T$ on average.

Why is this stochastic process called a “Brownian bridge”? Obviously it’s “Brownian” because it’s derived from a Brownian motion. But why “bridge”? Well, we know that $B_0 = 0$ and we know that $B_T = a$, so really this stochastic process is simply tracing out a continuous path from $(t, x) = (0, 0)$ to $(t, x) = (T, a)$. That is, it’s creating a bridge between those two points, albeit a rough and jagged one.

Let’s look at simulations of the Brownian bridge.

Conditioned on $B_{10} =$  


The “marginal” distribution of a Brownian bridge is given by $B_t | B_T \sim N((t/T)B_T, t(T-t) / T)$. It’s not too hard to show this. Hint: show that $B_t - (t/T)B_T$ is independent of $B_T$ and also make use of the law of total expectation in the calculations. The distribution makes a lot of sense, right? Naturally as $t$ approaches $T$, the Brownian bridge gets closer to $B_T$ on average, and also the variance decreases (as we already know what $B_T$ is). From this, it’s easy to see that the Brownian bridge is a Gaussian process.

Another result that is fairly straightforward to compute is its covariance structure, namely $\text{Cov}(B_s B_t | B_T) = s(T-t)/T$ for $s < t < T$. This is where we will terminate our short-lived discussion of the Brownian bridge, as we now have everything we need to describe and prove the main theorem of interest.

But before we move on, you may wonder “if this post is about Brownian motions, then why did we briefly talk about Brownian bridges?” The answer is simple: conditioning. In probability theory, a common strategy for deriving probabilistic quantities (probabilities, expectations, etc) is by dealing with conditional versions first. E.g. if I want to compute $\mathbb E[X]$, I might first start looking at the simpler problem of $\mathbb E[X | \{\text{some informative event that simplifies the expression}\}]$. In fact, we’ve already used this strategy in our derivation of the reflection principle, and if you wanted to derive the mean, variance and covariance results of Brownian bridge in this section yourself, you’d want to apply the strategy of conditioning also. As we will shortly see, the Brownian bridge naturally pops out when we apply the strategy of conditioning to a Brownian motion.


The theorem

We’ve seen previously how we can apply the reflection principle to derive the distribution of the first hitting time for a standard Brownian motion. Let’s now consider a more general problem. Let $\mu\in\mathbb R$ and $\sigma > 0$, then $\{\mu t + \sigma B_t : t \ge 0\}$ is a scaled Brownian motion with drift (or a generalised Brownian motion). From our previous discussion of normal random variables, we can see that $\mu t + \sigma B_t \sim N(\mu t, \sigma^2 t)$. The drift parameter $\mu$ determines the direction and speed of the linear drift, while the scale parameter $\sigma$ determines the variance of the Brownian randomness.

Let’s look at simulations of the generalised Brownian motion (with drift).

$\mu =$   $\sigma =$  


Let $a > 0$ be a threshold value and $\tau_a = \inf\nolimits_{0 \le s \le t}\{\mu t + \sigma B_t \ge a\}$ be the first hitting time of when the generalised Brownian motion hits the threshold $a$. Our question is the same as before: what’s the distribution of $\tau_a$?

Theorem (first hitting time of generalised Brownian motion). Let $a > 0$ and $t > 0$. Then,

\[\mathbb P(\tau_a \le t | \mu t + \sigma B_t \in dx) = \begin{cases} &\exp\left(-\frac{2a(a - x)}{\sigma^2 t}\right)\quad&\text{if } x \le a\\ &1 &\text{if } x > a, \end{cases}\]

and

\[\mathbb P(\tau_a \le t) = 1 - \Phi\left(\frac{a - \mu t}{\sqrt{\sigma^2 t}}\right) + e^{2\mu a / \sigma^2} \Phi\left(\frac{-a -\mu t}{\sqrt{\sigma^2 t}}\right),\]

and

\[\mathbb E[\tau_a] \begin{cases} <+\infty\quad&\text{if } \mu > 0,\\ =+\infty\quad&\text{if } \mu \le 0, \end{cases}\]

where $\Phi$ is the cumulative distribution function of a standard normal $N(0,1)$ random variable.

Isn’t it strange that the first formula, which describes the probability distribution of the first hitting time of a Brownian motion with drift, doesn’t actually depend on the drift parameter $\mu$ when conditioned by the process at time $t$? I wonder why! Hint: keep reading.

The second formula is the cumulative distribution function for $\tau_a$, and it’s straightforwardly obtained by applying the law of total probability by integrating the first formula across all values of $x$.

The third result tells us that if the drift is in the positive direction, then the average hitting time is finite. But if the drift is in the negative direction or zero, then the mean time is infinite. This is a straightforward consequence of the second formula. In particular, if $\mu < 0$, then $\mathbb P(\tau_a = +\infty) > 0$. The interesting case is when $\mu = 0$, for which $\mathbb P(\tau_a < +\infty) = 1$ (i.e. the first hitting time is almost surely finite) and yet $\mathbb E[\tau_a] = +\infty$.

How do we show this? The results about the probabilities are easy to show, using the property that the cumulative distribution function $\Phi(x)$ goes to zero when $x\to-\infty$ and goes to one when $x\to+\infty$. The result on the finiteness of the mean requires some additional knowledge. Firstly, it's a well-known but perhaps counterintuitive fact that $$\mathbb E[X] = \int_0^{+\infty}(1 - F_X(x))\,dx$$ if the random variable $X$ doesn't take any negative values. Hence, when computing $\mathbb E[\tau_a]$, the inside of the integral is $$f_\mu(t) = \Phi\left(\frac{a - \mu t}{\sqrt{\sigma^2 t}}\right) - e^{2\mu a / \sigma^2} \Phi\left(\frac{-a -\mu t}{\sqrt{\sigma^2 t}}\right).$$ To check whether the integral is finite, we need to understand the behaviour of $f_\mu(t)$ as $t\to+\infty$. The simplest way to do this is by looking at the asymptotic behaviour of $\Phi(x) = (1+\text{erf}(x))/2$. The asymptotic behaviour of $\text{erf}(x)$ for large $x$ is $C\exp(-x^2)/x$ for some constant $C>0$.


Therefore, the second and third results in the theorem are merely consequences of the first result. That is, all we need to do is to prove the first one. But therein lies the problem. The usual derivation of this result in the literature is to invoke Girsanov’s theorem, which allows us to “transform away” the drift component of the problem by changing the underlying probability measure $\mathbb P$. By doing this, we can straightforwardly apply our first hitting time result for a non-drifting Brownian motion. Unfortunately, Girsanov’s theorem is rather advanced, and requires background knowledge of the more advanced field of stochastic calculus to prove.

Our challenge is to prove the first result in the above theorem using only the knowledge we have built up until now in this post.


The proof

Alternatively, walking the bridge.

The maximum of a scaled Brownian bridge

We first begin by considering the problem without the drift component. Let $\{B_s : s\in[0,t]\}$ be a standard Brownian motion over $[0,t]$. Let $\tau_a^0 = \inf\{\sigma B_s: s\in[0,t]\}$ be the first passage time for the scaled Brownian motion $\{\sigma B_s : s\in[0,t]\}$. Then, for $x \le a$, we have

\[\begin{align} \mathbb P(\{\tau_a^0 \le t\}\cap\{\sigma B_t \in dx\}) &= \mathbb P(\{\tau_a^0 \le \sigma^2 t\}\cap\{B_{\sigma^2 t} \in dx\})\\ &= \mathbb P(2a - B_{\sigma^2 t} \in dx)\\ &= \mathbb P(N(2a, \sigma^2 t) \in dx)\\ &= \frac{1}{\sqrt{2\pi\sigma^2 t}} \exp\left(-\frac{(x - 2a)^2}{2\sigma^2 t}\right)\,dx, \end{align}\]

where the equalities hold because:

  1. Holds due to the self-similarity of Brownian motion.
  2. Holds due to the second version of the reflection principle and applying the fact that $\sigma B_{\tau_a^0} = B_{\sigma^2 \tau_a^0} = a$ when $\tau_a^0 \le t$ is true.
  3. Holds due to the fact that $2a - B_{\sigma^2 t}\sim N(2a, \sigma^2 t)$. Note that $2a - B_{\sigma^2 t}$ and $2a + B_{\sigma^2 t}$ has the same distribution due to the fact that the distribution of $B_{\sigma^2 t}$ is symmetric around zero.

We now apply the definition of conditional probability (in terms of probability densities) to get

\[\begin{align} \mathbb P(\tau_a^0 \le t | \sigma B_t \in dx) &= \frac{\mathbb P(\{\tau_a^0 \le t\}\cap\{\sigma B_t\in dx\})}{\sigma B_t\in dx}\\ &= \frac{1}{\sqrt{2\pi\sigma^2 t}} \exp\left(-\frac{(x - 2a)^2}{2\sigma^2 t}\right)\,dx \bigg/ \frac{1}{\sqrt{2\pi\sigma^2 t}} \exp\left(-\frac{x^2}{2\sigma^2 t}\right)\,dx\\ &= \exp\left(-\frac{2a(a-x)}{\sigma^2 t}\right). \end{align}\]

This is all we need from this calculation. But what’s the significance of this result? Well, notice that $\{\tau_a^0 \le t\} = \{\sup\nolimits_{0\le s\le t}\sigma B_s \ge a\}$, so

\[\mathbb P (\sup\nolimits_{0\le s\le t}\sigma B_s \ge a | \sigma B_t \in dz) = \exp\left(-\frac{2a(a-x)}{\sigma^2 t}\right).\]

This is an interesting result in and of itself: since we’ve conditioned on $\sigma B_t$, this means we’ve actually derived the distribution for the maximum of a (scaled) Brownian bridge between times $[0,t]$. How neat!

A Brownian bridge embeds drift

If you compare what we have just derived to what we’re trying to prove, they’re exactly the same, except for the fact that we’ve left out the drift. So what we’re going to do next is to consider the more general case with the drift in play and set up for an application of what we’ve just derived. That is to say, we’re going to extend the reflection principle via the Brownian bridge.

We propose that conditioning upon $\{\mu t + \sigma B_t \in dx\}$, the generalised Brownian motion $\{\mu s + \sigma B_s : s\in[0,t]\}$ becomes the scaled Brownian bridge $\{(\sigma B_s | \sigma B_t) : s\in[0,t]\}$. To show this, we can take advantage of the fact that it’s a Gaussian process and thus we only need to show that the expectation and covariance structures match that of a scaled Brownian bridge.

The expectation is easily computed as

\[\mathbb E[\mu s + \sigma B_s | \mu t + \sigma B_t \in dx] = \mu s + \sigma\mathbb E[B_s | B_t \in (dx-\mu t)/\sigma] = \mu s + \sigma\frac{s}{t}\frac{dx - \mu t}{\sigma} = \frac{s}{t}dx,\]

where the second equality holds due to the marginal expectation of Brownian bridges we highlighted earlier. We see that this expectation indeed aligns with the expectation of a Brownian bridge.

We now consider the covariance structure. Let $s_1 < s_2 < t$, then

\[\begin{align} \text{Cov}(\mu s_1 + \sigma B_{s_1}, \mu s_2 + \sigma B_{s_2} |\mu t + \sigma B_t \in dx) &= \text{Cov}(\sigma B_{s_1}, \sigma B_{s_2} | B_t \in (dx-\mu t)/\sigma)\\ &= \sigma^2 \text{Cov}(B_{s_1}, B_{s_2} | B_t \in (dx-\mu t)/\sigma)\\ &= \sigma^2 \frac{s_1(t-s_2)}{t}, \end{align}\]

where the equalities hold because:

  1. Holds due to the covariance of linear combinations.
  2. ^^^
  3. Holds due to the covariance of standard Brownian bridges we highlighted earlier.

We see that this covariance structure indeed matches with that of a scaled Brownian bridge. Therefore, indeed the generalised Brownian motion $\{\mu s + \sigma B_s : s\in[0,t]\}$ becomes a scaled Brownian bridge $\{(\sigma B_s | \sigma B_t \in dx) : s\in[0,t]\}$ when conditioned on $\{\mu t + \sigma B_t \in dz\}$. And so we see that upon conditioning, the drift parameter vanishes completely! This is why the drift parameter doesn’t appear in the formula that we’re trying to derive. But how should we make sense of this intuitively? Well, recall that the expectation of a Brownian bridge $\mathbb E[B_s | B_t] = (s/t) B_t$ is a linear drift in and of itself. So what’s actually happening is that the drift term in the generalised Brownian motion becomes absorbed by the expectation structure of the Brownian bridge, as conditioning on any future point causes the Brownian bridge to drift towards that point on average anyway.

Therefore, applying everything we’ve derived in this section, it holds that for $x \le a$,

\[\mathbb P(\tau_a \le t | \mu t + \sigma B_t \in dx) = \mathbb P(\tau_a^0 \le t | \sigma B_t \in dx) = \exp\left(-\frac{2a(a-x)}{\sigma^2 t}\right).\]

For $x > a$, $\mathbb P(\tau_a \le t | \mu t + \sigma B_t \in dx) = 1$ as the generalised Brownian motion being at $x$ at time $t$ can only happen if $\tau_a \le t$ is also true.

That’s it!