Brownian Motion

Asset prices, interest rates, and exchange rates move all the time, by small amounts, in response to a constant flow of news. The Poisson process is the wrong model for this kind of noise, since it stays constant most of the time and moves only by jumps. Brownian motion is the model for cumulative noise that never stops moving. It is the building block of the stochastic differential equations used to model stock prices and the source of the Black-Scholes formula.

Brownian motion plays the same role for continuous noise that the Poisson process plays for sudden events. Both arise as limits of sums of Bernoulli random variables, both have independent and stationary increments, and both generate three martingales with the same structure. The Poisson process comes from rare events, each of size one. Brownian motion comes from coin flips that occur very often, each of a tiny size. These notes construct Brownian motion as the limit of a scaled random walk, study its martingales, and then show that its paths are far rougher than the smooth functions of ordinary calculus. That roughness is what makes stochastic calculus different from ordinary calculus.

These notes draw on Shreve (2004).

Shreve, Steven E. 2004. Stochastic Calculus for Finance II: Continuous-Time Models. Springer Finance. Springer.

From Random Walks to Brownian Motion

The Symmetric Random Walk

Let X_{1}, X_{2}, \ldots be independent random variables that take the values +1 and -1 with probability 1/2 each, so that \operatorname{E}(X_{i}) = 0 and \operatorname{V}(X_{i}) = 1. The symmetric random walk S_{n} = X_{1} + \cdots + X_{n} records the cumulative gains from betting one dollar on each of n flips of a fair coin. We already know three of its properties:

• For m < n, the increment S_{n} - S_{m} = X_{m + 1} + \cdots + X_{n} depends only on the flips between m + 1 and n, so increments over disjoint periods are independent, and \operatorname{V}(S_{n} - S_{m}) = n - m.
• Both S_{n} and S_{n}^{2} - n are martingales, as shown in the notes on discrete-time martingales.
• Every step has size one, so the sum of squared increments along any path is n.

This last quantity, [S, S]_{n} = \sum_{i = 1}^{n} (S_{i} - S_{i - 1})^{2} = \sum_{i = 1}^{n} X_{i}^{2} = n, is the quadratic variation of the walk. Unlike the variance, which averages over all possible paths, quadratic variation is computed from a single path. For the random walk every path has the same quadratic variation, because every step has size one.

A Scaled Random Walk

To move to continuous time, we speed up the walk and shrink its steps. Divide a time interval into slots of length \Delta t, and at the end of each slot move up or down by \sqrt{\Delta t}. After m slots, at time t = m \Delta t, the position of the scaled random walk is B_{t}^{(n)} = \sqrt{\Delta t} \, (X_{1} + X_{2} + \cdots + X_{m}), where n = 1 / \Delta t is the number of steps per unit of time. Its mean, variance, and quadratic variation are \operatorname{E}(B_{t}^{(n)}) = 0, \qquad \operatorname{V}(B_{t}^{(n)}) = m \Delta t = t, \qquad [B^{(n)}, B^{(n)}]_{t} = \sum_{j = 1}^{m} (\sqrt{\Delta t} \, X_{j})^{2} = m \Delta t = t.

The step size \sqrt{\Delta t} is the only choice that keeps the variance at time t equal to t as the slots get shorter. With steps of size \Delta t the variance would be t \Delta t and the walk would collapse to zero, while with steps of size one it would explode. Over a short interval, the scaled walk moves by an amount of order \sqrt{\Delta t}, which is much larger than \Delta t. This simple fact is behind every result in these notes.

Figure 1: The figure plots one path of the scaled random walk over 0 \leq t \leq 1 with n = 10, n = 100, and n = 10{,}000 steps per unit of time. Each step moves up or down by \sqrt{\Delta t} = 1 / \sqrt{n}.

The Limit

As \Delta t \to 0, the number of steps up to time t grows while each step shrinks, which is the setting of the central limit theorem in the notes on generating functions. Using the characteristic function \cos(u) of a fair coin flip, also from those notes, the characteristic function of B_{t}^{(n)} is \phi_{B_{t}^{(n)}}(u) = \cos\left(u \sqrt{\Delta t}\right)^{t / \Delta t} = \left(1 - \frac{u^{2} \Delta t}{2} + o(\Delta t)\right)^{t / \Delta t} \xrightarrow[\Delta t \to 0]{} e^{-\frac{1}{2} u^{2} t}, where we used \cos(x) = 1 - \frac{1}{2} x^{2} + o(x^{2}). The limit is the characteristic function of a \mathcal{N}(0, t) random variable, so by Lévy’s continuity theorem B_{t}^{(n)} converges in distribution to \mathcal{N}(0, t).

As with the Poisson process, the construction gives us more than the distribution at a single date. For s < t, the increment B_{t}^{(n)} - B_{s}^{(n)} is built from the flips between s and t, which are independent of the flips before s. The same argument shows that it converges to a \mathcal{N}(0, t - s) random variable, whose distribution depends only on the length of the interval. In addition, the paths of the scaled walk move by at most \sqrt{\Delta t} in each step, which suggests that the limit has no jumps. These properties survive the limit and define Brownian motion.

Brownian Motion

Definition

WarningBrownian Motion

A stochastic process \{B_{t} \colon t \geq 0\} is called a Brownian motion if it has the following four properties:

1. B_{0} = 0.

2. For any finite set of times 0 \leq t_{0} < t_{1} < \cdots < t_{n}, the increments B_{t_{1}} - B_{t_{0}}, B_{t_{2}} - B_{t_{1}}, \ldots, B_{t_{n}} - B_{t_{n - 1}} are independent.

3. For any 0 \leq s < t, the increment B_{t} - B_{s} has a \mathcal{N}(0, t - s) distribution.

4. With probability one, the path t \mapsto B_{t} is continuous.

The first two properties are the same as for the Poisson process. The third replaces a Poisson increment with mean \lambda (t - s) by a normal increment with variance t - s, and the fourth rules out the jumps that are the defining feature of the Poisson process. Brownian motion is sometimes called the Wiener process, and some texts denote it by W_{t}.

Since B_{t} = B_{t} - B_{0} is normal with mean zero and variance t, we have \operatorname{E}(B_{t}) = 0 and \operatorname{V}(B_{t}) = t. For s \leq t, writing B_{t} = B_{s} + (B_{t} - B_{s}) and using the independence of the increment from B_{s}, \operatorname{Cov}(B_{s}, B_{t}) = \operatorname{Cov}(B_{s}, B_{s}) + \operatorname{Cov}(B_{s}, B_{t} - B_{s}) = s + 0 = s. Therefore, \operatorname{Cov}(B_{s}, B_{t}) = \operatorname{E}(B_{s} B_{t}) = \min(s, t) for all s, t \geq 0. The value of Brownian motion at a later date is its current value plus independent noise, so for s \leq t the correlation between B_{s} and B_{t} is \sqrt{s / t}. It depends on the ratio of the two dates rather than on the gap between them, and fades as t grows relative to s.

Figure 2: The figure plots three simulated paths of a Brownian motion over 0 \leq t \leq 10. The dashed curves mark \pm 2 \sqrt{t}, which contain about 95% of the distribution of B_{t} at each date.

Compare Figure 2 with the paths of the Poisson process. Both processes have independent and stationary increments, but a Brownian path is never flat. Over a short interval of length \mathop{}\!\mathrm{d}t, the increment \mathop{}\!\mathrm{d}B_{t} = B_{t + \mathop{}\!\mathrm{d}t} - B_{t} has mean zero and standard deviation \sqrt{\mathop{}\!\mathrm{d}t}, so the path moves by an amount far larger than \mathop{}\!\mathrm{d}t in every short interval, in one direction or the other.

Martingales of Brownian Motion

Let \mathcal{F}_{t} denote the information generated by observing \{B_{s} \colon s \leq t\} up to time t. As for the Poisson process, increments of Brownian motion are independent of the past, so conditional expectations of functions of B_{t} - B_{s} given \mathcal{F}_{s} reduce to ordinary expectations.

WarningMartingales of Brownian Motion

Let \{B_{t}\} be a Brownian motion. The following processes are martingales with respect to \{\mathcal{F}_{t}\}:

1. The Brownian motion B_{t} itself.

2. The process B_{t}^{2} - t.

3. For any real \theta, the exponential martingale Z_{t} = \exp\left(\theta B_{t} - \frac{1}{2} \theta^{2} t\right).

Proof

Fix s < t and write \Delta B = B_{t} - B_{s}, which is independent of \mathcal{F}_{s} and has a \mathcal{N}(0, t - s) distribution, so that \operatorname{E}(\Delta B) = 0 and \operatorname{E}(\Delta B^{2}) = t - s.

For the first process, \operatorname{E}(B_{t} \mid \mathcal{F}_{s}) = B_{s} + \operatorname{E}(\Delta B) = B_{s}. For the second, expanding B_{t}^{2} = B_{s}^{2} + 2 B_{s} \Delta B + \Delta B^{2} gives \operatorname{E}(B_{t}^{2} - t \mid \mathcal{F}_{s}) = B_{s}^{2} + 2 B_{s} \operatorname{E}(\Delta B) + \operatorname{E}(\Delta B^{2}) - t = B_{s}^{2} + (t - s) - t = B_{s}^{2} - s. For the third, the MGF of a \mathcal{N}(0, t - s) random variable is \operatorname{E}(e^{\theta \Delta B}) = e^{\frac{1}{2} \theta^{2} (t - s)}, so that \operatorname{E}(Z_{t} \mid \mathcal{F}_{s}) = Z_{s} \, \operatorname{E}\left(e^{\theta \Delta B}\right) e^{-\frac{1}{2} \theta^{2} (t - s)} = Z_{s}.

These are the same three martingales we found for the Poisson process, and the proofs are identical line by line. The first two are the continuous-time versions of the random walk martingales S_{n} and S_{n}^{2} - n. The exponential martingale has the form e^{\theta B_{t} - t K(\theta)} found in the notes on generating functions, where K(\theta) = \frac{1}{2} \theta^{2} is the cumulant generating function of B_{1}. The exponential martingale is the tool we will later use to change probability measures, and the next section uses it to compute the probability that a process ever reaches a barrier.

Application: Hitting a Barrier

Many questions in finance ask whether a process ever reaches a given level: whether the value of a firm’s assets falls far enough to trigger default, whether a stock reaches the barrier of a knock-out option, or whether a fund loses half of its value. A simple model for the logarithm of such a value is a Brownian motion with drift, X_{t} = \mu t + \sigma B_{t}, which grows on average at the rate \mu and has volatility \sigma > 0. Let \tau be the first time X_{t} reaches either an upper barrier a > 0 or a lower barrier -b < 0. This is the gambler’s ruin problem from the notes on ruin probabilities, in continuous time and with a drift.

WarningHitting Probabilities

Let X_{t} = \mu t + \sigma B_{t} with \mu \neq 0, and let \gamma = 2 \mu / \sigma^{2}. The probability that X_{t} reaches a before -b is \operatorname{P}(X_{\tau} = a) = \frac{1 - e^{-\gamma b}}{1 - e^{-\gamma (a + b)}}. \tag{1} If \mu > 0, the probability that X_{t} ever reaches -b is e^{-\gamma b}.

Proof

Applying the exponential martingale with \theta \sigma in place of \theta shows that \exp\left(\theta \sigma B_{t} - \frac{1}{2} \theta^{2} \sigma^{2} t\right) is a martingale for every \theta. Since \theta \sigma B_{t} = \theta X_{t} - \theta \mu t, this martingale equals \exp\left(\theta X_{t} - \left(\theta \mu + \tfrac{1}{2} \theta^{2} \sigma^{2}\right) t\right). Choosing \theta = -\gamma = -2 \mu / \sigma^{2} makes the coefficient of t zero, so Y_{t} = e^{-\gamma X_{t}} is a martingale.

The stopped process Y_{t \land \tau} is also a martingale, so \operatorname{E}(Y_{t \land \tau}) = Y_{0} = 1 for every t. Since X_{t \land \tau} stays between -b and a, the stopped process is bounded. The stopping time is finite with probability one, because \{\tau > t\} \subset \{-b < X_{t} < a\} and X_{t} is normal with standard deviation \sigma \sqrt{t}, so \operatorname{P}(\tau > t) \leq \operatorname{P}(-b < X_{t} < a) \to 0 as t \to \infty. Letting t \to \infty gives \operatorname{E}(Y_{\tau}) = 1, the continuous-time version of the optional stopping theorem in the notes on discrete-time martingales. Because the paths are continuous, X_{\tau} equals -b or a exactly. Writing p = \operatorname{P}(X_{\tau} = a), p \, e^{-\gamma a} + (1 - p) \, e^{\gamma b} = 1, and solving for p and multiplying its numerator and denominator by e^{-\gamma b} gives (1). If \mu > 0, then \gamma > 0 and e^{-\gamma (a + b)} \to 0 as a \to \infty, so p \to 1 - e^{-\gamma b} and the probability of ever reaching -b is e^{-\gamma b}.

The argument is the one used for the gambler and for the insurer: find a martingale, stop it at the hitting time, and compare its value at the start with its value at the stopping time. The quantity e^{-\gamma} plays the role of q / p for the biased random walk. There is an important difference with Lundberg’s inequality in the notes on Poisson processes. The surplus of the insurer jumps below zero by a random amount, so its value at the ruin time is unknown and the argument gives only a bound. Brownian paths are continuous and hit the barrier exactly, so here the argument gives the exact probability. As \mu \to 0, (1) converges to b / (a + b), the probability that the fair gambler wins.

Example 1 The logarithm of the value of an endowment fund, relative to its initial value, follows a Brownian motion with drift \mu = 0.04 and volatility \sigma = 0.15 per year. The fund loses half of its value when X_{t} falls by b = \ln 2 = 0.6931. With \gamma = 2 \times 0.04 / 0.15^{2} = 3.5556, the probability that the fund ever loses half of its value is e^{-\gamma b} = e^{-3.5556 \times 0.6931} = 0.085. The probability that it doubles before it halves follows from (1) with a = b = \ln 2 and equals 0.9216.

The drift \mu in this example is the growth rate of the logarithm of the fund’s value, which is lower than the expected return on the fund. The notes on modeling stock prices derive the relation between the two.

Total and Quadratic Variation

Variation of Smooth Functions

To measure how much a function moves over [0, T], take a partition \Pi = \{t_{0}, t_{1}, \ldots, t_{n}\} with 0 = t_{0} < t_{1} < \cdots < t_{n} = T, and let \lVert{\Pi}\rVert = \max_{j} (t_{j + 1} - t_{j}) be its mesh, the length of its longest subinterval. The total variation and the quadratic variation of a function f are V_{T}(f) = \lim_{\lVert{\Pi}\rVert \to 0} \sum_{j = 0}^{n - 1} |f(t_{j + 1}) - f(t_{j})|, \qquad [f, f]_{T} = \lim_{\lVert{\Pi}\rVert \to 0} \sum_{j = 0}^{n - 1} (f(t_{j + 1}) - f(t_{j}))^{2}. Total variation is the distance traveled by the function, up and down. Quadratic variation adds up squared moves instead.

Suppose f has a continuous derivative. By the mean-value theorem, each increment can be written as f(t_{j + 1}) - f(t_{j}) = f'(t_{j}^{*}) (t_{j + 1} - t_{j}) for some t_{j}^{*} in [t_{j}, t_{j + 1}]. The sum of absolute increments is then a Riemann sum, \sum_{j = 0}^{n - 1} |f'(t_{j}^{*})| (t_{j + 1} - t_{j}) \xrightarrow[\lVert{\Pi}\rVert \to 0]{} \int_{0}^{T} |f'(t)| \mathop{}\!\mathrm{d}t, so the total variation is finite. For the sum of squared increments, one factor of t_{j + 1} - t_{j} is at most \lVert{\Pi}\rVert, so \sum_{j = 0}^{n - 1} f'(t_{j}^{*})^{2} (t_{j + 1} - t_{j})^{2} \leq \lVert{\Pi}\rVert \sum_{j = 0}^{n - 1} f'(t_{j}^{*})^{2} (t_{j + 1} - t_{j}) \xrightarrow[\lVert{\Pi}\rVert \to 0]{} 0 \times \int_{0}^{T} f'(t)^{2} \mathop{}\!\mathrm{d}t= 0. A smooth function has finite total variation and zero quadratic variation. Over a short interval it moves by an amount proportional to the length of the interval, and squaring such small moves makes them negligible.

Variation of the Poisson Process

A Poisson path is flat except at its jumps, and it has finitely many jumps in [0, T], no two at the same time. As the mesh goes to zero, each subinterval eventually contains at most one jump, so each increment N_{t_{j + 1}} - N_{t_{j}} is either zero or one. Its absolute value and its square both equal the increment itself, so both sums count the jumps, V_{T}(N) = [N, N]_{T} = N_{T}. The Poisson process has finite total variation, like a smooth function, but its quadratic variation is not zero: each jump contributes its square however fine the partition. The compensated process M_{t} = N_{t} - \lambda t subtracts a smooth function, whose squared increments vanish in the limit, and the cross terms are at most 2 \lambda \lVert{\Pi}\rVert N_{T}, so [M, M]_{T} = N_{T} as well.

The quadratic variation of M is random, and its expectation \operatorname{E}(N_{t}) = \lambda t is the compensator in the martingale M_{t}^{2} - \lambda t of the Poisson process. The random version is a martingale too: M_{t}^{2} - [M, M]_{t} = (M_{t}^{2} - \lambda t) - (N_{t} - \lambda t) is a difference of two martingales.

Quadratic Variation of Brownian Motion

Brownian motion moves by about \sqrt{t_{j + 1} - t_{j}} over each subinterval, so its squared moves are of the same order as t_{j + 1} - t_{j} and do not become negligible. They add up to the length of the interval.

WarningQuadratic Variation of Brownian Motion

For a partition \Pi of [0, T], let Q_{\Pi} = \sum_{j = 0}^{n - 1} (B_{t_{j + 1}} - B_{t_{j}})^{2}. Then \operatorname{E}(Q_{\Pi}) = T and \operatorname{V}(Q_{\Pi}) \leq 2 \lVert{\Pi}\rVert T. Therefore, Q_{\Pi} converges to T in mean square as \lVert{\Pi}\rVert \to 0, and [B, B]_{T} = T.

Proof

Write \Delta B_{j} = B_{t_{j + 1}} - B_{t_{j}} and \Delta t_{j} = t_{j + 1} - t_{j}. Each increment is \mathcal{N}(0, \Delta t_{j}), so \operatorname{E}(\Delta B_{j}^{2}) = \Delta t_{j} and \operatorname{E}(Q_{\Pi}) = \sum_{j = 0}^{n - 1} \Delta t_{j} = T. A standard normal random variable Z has \operatorname{E}(Z^{4}) = 3, and \Delta B_{j} has the same distribution as \sqrt{\Delta t_{j}} \, Z, so \operatorname{E}(\Delta B_{j}^{4}) = 3 \Delta t_{j}^{2} and \operatorname{V}(\Delta B_{j}^{2}) = \operatorname{E}(\Delta B_{j}^{4}) - \operatorname{E}(\Delta B_{j}^{2})^{2} = 3 \Delta t_{j}^{2} - \Delta t_{j}^{2} = 2 \Delta t_{j}^{2}. The increments are independent, so the variance of their sum is the sum of their variances, \operatorname{V}(Q_{\Pi}) = 2 \sum_{j = 0}^{n - 1} \Delta t_{j}^{2} \leq 2 \lVert{\Pi}\rVert \sum_{j = 0}^{n - 1} \Delta t_{j} = 2 \lVert{\Pi}\rVert T. Therefore, \operatorname{E}\left((Q_{\Pi} - T)^{2}\right) = \operatorname{V}(Q_{\Pi}) \to 0 as \lVert{\Pi}\rVert \to 0. By Chebyshev’s inequality, \operatorname{P}(|Q_{\Pi} - T| > \varepsilon) \leq \operatorname{V}(Q_{\Pi}) / \varepsilon^{2} \to 0 for every \varepsilon > 0, so Q_{\Pi} also converges to T in probability.

The scaled random walk has quadratic variation t on every path, and Brownian motion inherits this property in the limit. Each squared increment (B_{t_{j + 1}} - B_{t_{j}})^{2} is random, but on a fine partition there are many of them, they are independent, and their sum has a vanishing variance. Quadratic variation is computed from a single path, and the result is essentially the same on every path. This is the main contrast with the Poisson process, whose quadratic variation [N, N]_{t} = N_{t} is random.

Figure 3: The figure computes the sums of squared increments (left) and of absolute increments (right) along one simulated Brownian path on [0, 1], using partitions with n = 2, 4, \ldots, 2^{16} equal subintervals. The dashed lines mark the quadratic variation T = 1 and the expected sum of absolute increments \sqrt{2 n / \pi} for each partition.

The Multiplication Table

In differential notation, [B, B]_{t} = t reads (\mathop{}\!\mathrm{d}B)^{2} = \mathop{}\!\mathrm{d}t. Two more limits complete the picture. Let M_{\Pi} = \max_{j} |B_{t_{j + 1}} - B_{t_{j}}| be the largest move of the path over the partition. Brownian paths are continuous, so M_{\Pi} \to 0 as \lVert{\Pi}\rVert \to 0, and \left|\sum_{j = 0}^{n - 1} (B_{t_{j + 1}} - B_{t_{j}}) (t_{j + 1} - t_{j})\right| \leq M_{\Pi} T \to 0, \qquad \sum_{j = 0}^{n - 1} (t_{j + 1} - t_{j})^{2} \leq \lVert{\Pi}\rVert T \to 0. Together these results give the multiplication table of Brownian motion, (\mathop{}\!\mathrm{d}B)^{2} = \mathop{}\!\mathrm{d}t, \qquad (\mathop{}\!\mathrm{d}B)(\mathop{}\!\mathrm{d}t) = 0, \qquad (\mathop{}\!\mathrm{d}t)^{2} = 0, sometimes called box calculus. The table does not say that each squared increment equals the length of its interval. It says that sums of squared increments behave like sums of time increments, and the notes on stochastic calculus use it in exactly that way to derive Itô’s formula.

For the Poisson process, [N, N]_{t} = N_{t} reads (\mathop{}\!\mathrm{d}N)^{2} = \mathop{}\!\mathrm{d}N. The sum of the products (N_{t_{j + 1}} - N_{t_{j}}) (t_{j + 1} - t_{j}) is at most \lVert{\Pi}\rVert N_{T}, so (\mathop{}\!\mathrm{d}N)(\mathop{}\!\mathrm{d}t) = 0, and the multiplication table of the Poisson process is (\mathop{}\!\mathrm{d}N)^{2} = \mathop{}\!\mathrm{d}N, \qquad (\mathop{}\!\mathrm{d}N)(\mathop{}\!\mathrm{d}t) = 0, \qquad (\mathop{}\!\mathrm{d}t)^{2} = 0. Comparing the two tables:

• The quadratic variation of Brownian motion is deterministic. The quadratic variation of the compensated Poisson process is random, and only its expectation, \operatorname{E}(N_{t}) = \lambda t, is deterministic.
• In both cases the expected quadratic variation equals the variance of the process, \operatorname{V}(B_{t}) = t and \operatorname{V}(M_{t}) = \lambda t.
• Brownian paths accumulate quadratic variation continuously through infinitely many tiny moves. Poisson paths accumulate it only at the jumps.

Infinite Total Variation

The quadratic variation result implies that Brownian paths travel an infinite distance over any interval. For each subinterval, (B_{t_{j + 1}} - B_{t_{j}})^{2} \leq M_{\Pi} \, |B_{t_{j + 1}} - B_{t_{j}}|, and summing over j gives \sum_{j = 0}^{n - 1} |B_{t_{j + 1}} - B_{t_{j}}| \geq \frac{Q_{\Pi}}{M_{\Pi}}. As the mesh goes to zero, the numerator approaches T > 0 while the denominator goes to zero, so the sum of absolute increments grows without bound. The total variation of Brownian motion over [0, T] is infinite. The right panel of Figure 3 shows the sum growing like \sqrt{n}, since each of the n increments has an expected absolute value proportional to \sqrt{T / n}.

A path with a continuous derivative on some interval would have finite total variation there, so Brownian paths are not smooth on any interval, however short. In fact a stronger result holds: with probability one, a Brownian path does not have a derivative at any point. The proof of this result requires a separate argument, but Figure 1 and Figure 2 already show why: a Brownian path wiggles at every scale, and zooming in never makes it look straight.

Infinite total variation is the reason we need a new calculus. An integral \int_{0}^{T} f(t) \mathop{}\!\mathrm{d}g(t) can be defined path by path as a limit of sums \sum f(t_{j}) (g(t_{j + 1}) - g(t_{j})) whenever f is continuous and g has finite total variation, and the point in each subinterval at which f is evaluated does not matter in the limit. With g = B it does matter. For f = B, evaluating at the right endpoint instead of the left changes the sum by \sum (B_{t_{j + 1}} - B_{t_{j}})^{2} = Q_{\Pi}, which converges to T rather than to zero. The Itô integral in the notes on stochastic calculus evaluates the integrand at the left endpoint and is defined as a limit in mean square, relying on the same independence of increments that made \operatorname{V}(Q_{\Pi}) vanish. The nonzero quadratic variation then adds the extra term in Itô’s formula that has no counterpart in ordinary calculus.

Summary

The table below compares Brownian motion with the Poisson process.

Property	Poisson process N_{t}	Brownian motion B_{t}
Increment over (s, t]	Poisson, mean \lambda (t - s)	Normal, variance t - s
Discrete-time origin	Rare Bernoulli events	Scaled coin flips
Paths	Step functions, jumps of size one	Continuous, never flat
Martingale	N_{t} - \lambda t	B_{t}
Second martingale	(N_{t} - \lambda t)^{2} - \lambda t	B_{t}^{2} - t
Exponential martingale	e^{\theta N_{t} - \lambda t (e^{\theta} - 1)}	e^{\theta B_{t} - \frac{1}{2} \theta^{2} t}
Total variation	N_{t} (finite)	Infinite
Quadratic variation	N_{t} (random)	t (deterministic)

Practice Problems

Problem 1 Let \{B_{t}\} be a Brownian motion.

1. Compute \operatorname{V}(B_{1} + B_{2}), \operatorname{V}(B_{2} - 2 B_{1}), and \operatorname{Cov}(B_{1}, B_{3} - B_{1}).
2. Find the distribution of B_{1} + B_{2} + B_{3}.
3. Compute \operatorname{P}(B_{1} > 0, B_{2} > B_{1}).

Solution

1. Using \operatorname{Cov}(B_{s}, B_{t}) = \min(s, t), \operatorname{V}(B_{1} + B_{2}) = 1 + 2 + 2 \times 1 = 5 and \operatorname{V}(B_{2} - 2 B_{1}) = 2 + 4 - 4 \times 1 = 2. The increment B_{3} - B_{1} is independent of B_{1}, so \operatorname{Cov}(B_{1}, B_{3} - B_{1}) = 0, which also follows from \operatorname{Cov}(B_{1}, B_{3}) - \operatorname{V}(B_{1}) = 1 - 1.
2. A linear combination of the values of a Brownian motion is a linear combination of independent normal increments, so it is normal. Its mean is zero and its variance is 1 + 2 + 3 + 2 (1 + 1 + 2) = 14, so B_{1} + B_{2} + B_{3} \sim \mathcal{N}(0, 14). Writing it as 3 B_{1} + 2 (B_{2} - B_{1}) + (B_{3} - B_{2}) gives the same variance, 9 + 4 + 1.
3. B_{1} and B_{2} - B_{1} are independent standard normals, so the probability is \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Problem 2 Let \{B_{t}\} be a Brownian motion. Check the four properties of the definition to decide whether each process is a Brownian motion.

1. -B_{t}.
2. c B_{t / c^{2}} for a constant c > 0.
3. B_{s + t} - B_{s} for a fixed s > 0.
4. \sqrt{t} \, Z, where Z \sim \mathcal{N}(0, 1).

Solution

1. Yes. Negating every increment keeps them independent, and -\mathcal{N}(0, t - s) has the same distribution as \mathcal{N}(0, t - s). The paths are still continuous and start at zero.
2. Yes. The increments of c B_{t / c^{2}} are c times increments of B over disjoint intervals, so they are independent, and the increment from s to t is normal with variance c^{2} (t - s) / c^{2} = t - s. Speeding up time by 1 / c^{2} and scaling space by c gives back a Brownian motion, which is why zooming in on a Brownian path never makes it look smoother.
3. Yes. It starts at zero, its increments are increments of B after time s, which are independent and have the right normal distributions, and its paths are continuous. Brownian motion starts afresh at every fixed time.
4. No. \sqrt{t} \, Z \sim \mathcal{N}(0, t) for every t, and the paths are continuous, but the increments are not independent: the whole path is determined by Z. For example, \operatorname{Cov}(\sqrt{1} \, Z, (\sqrt{2} - 1) Z) = \sqrt{2} - 1 \neq 0, and \operatorname{V}(\sqrt{2} \, Z - Z) = (\sqrt{2} - 1)^{2} = 0.17 instead of 1. Having the right distribution at each date is not enough.

Problem 3 The log of a stock price, relative to its current value, is X_{t} = \mu t + \sigma B_{t}, with \mu = 0.06 and \sigma = 0.20 per year.

1. Compute the probability that the stock is below its current price in 1, 10, and 30 years.
2. Show that the probability approaches zero as the horizon grows, but only slowly. What makes the drift dominate in the long run?

Solution

1. X_{t} \sim \mathcal{N}(\mu t, \sigma^{2} t), so \operatorname{P}(X_{t} < 0) = \mathop{\Phi}(-\mu \sqrt{t} / \sigma). This gives \mathop{\Phi}(-0.30) = 0.382 for one year, \mathop{\Phi}(-0.95) = 0.171 for ten years, and \mathop{\Phi}(-1.64) = 0.050 for thirty years.
2. The mean grows like t and the standard deviation like \sqrt{t}, so the ratio \mu \sqrt{t} / \sigma grows without bound and the probability goes to zero. It grows only like \sqrt{t}, though, so even after thirty years there is a 5% chance of a loss. This calculation does not mean that stocks are safer in the long run: the dollar amount at risk also grows with the horizon.

Problem 4 (★) Let \{B_{t}\} be a Brownian motion and 0 < s < t.

1. Use the properties of the bivariate normal distribution to show that \operatorname{E}(B_{s} \mid B_{t}) = \frac{s}{t} B_{t}, \qquad \operatorname{V}(B_{s} \mid B_{t}) = \frac{s (t - s)}{t}.
2. Compare \operatorname{E}(B_{s} \mid B_{t}) with \operatorname{E}(B_{t} \mid \mathcal{F}_{s}).
3. A log price follows a Brownian motion with volatility 0.20 per year, and over a year it rose by 0.30. What are the mean and standard deviation of its change over the first six months?

Solution

1. B_{s} and B_{t} are bivariate normal with means zero, variances s and t, and covariance s. The conditional mean is \frac{\operatorname{Cov}(B_{s}, B_{t})}{\operatorname{V}(B_{t})} B_{t} = \frac{s}{t} B_{t}, and the conditional variance is \operatorname{V}(B_{s}) (1 - \rho^{2}) = s \left(1 - \frac{s}{t}\right) = \frac{s (t - s)}{t}.
2. Looking forward, \operatorname{E}(B_{t} \mid \mathcal{F}_{s}) = B_{s}: the best forecast is today’s value. Looking backward, the best estimate of the intermediate value is a fraction s / t of the final value, on the straight line from 0 to B_{t}. The conditional variance vanishes at both ends and is largest in the middle. A Brownian path with both ends fixed is called a Brownian bridge.
3. Write the log price as 0.2 B_{t}, so B_{1} = 1.5. Then \operatorname{E}(0.2 B_{1/2} \mid B_{1} = 1.5) = 0.2 \times 0.75 = 0.15 and \operatorname{SD}= 0.2 \sqrt{1/2 \times 1/2} = 0.10. Half of the annual change is expected in each half of the year, with considerable uncertainty around it.

Problem 5 (★) A trader buys a stock and sells it once the price has risen by 20% or fallen by 10%. The log price relative to the purchase price is X_{t} = \mu t + \sigma B_{t} with \mu = 0.05 and \sigma = 0.25 per year.

1. Find the barriers a and b for the log price, and compute the probability that the trade ends with the gain.
2. Compare with the probability for \mu = 0.
3. Without the take-profit order, what is the probability that the trade ever loses 10%?

Solution

1. The log barriers are a = \ln 1.2 = 0.182 and b = -\ln 0.9 = 0.105. With \gamma = 2 \times 0.05 / 0.25^{2} = 1.6, (1) gives \operatorname{P}(X_{\tau} = a) = \frac{1 - e^{-1.6 \times 0.105}}{1 - e^{-1.6 \times 0.287}} = 0.421.
2. With \mu = 0 the probability of the gain is b / (a + b) = 0.366. The drift raises the probability of the gain, but by less than one might expect, because over the few months the trade lasts the volatility dominates the drift.
3. e^{-\gamma b} = e^{-1.6 \times 0.105} = 0.845. A stock with a positive drift is still very likely to fall by 10% at some point.

Problem 6 (★) Let \{B_{t}\} be a Brownian motion.

1. Show that B_{t}^{3} - 3 t B_{t} is a martingale.
2. Expand the exponential martingale Z_{t} = \exp\left(\theta B_{t} - \frac{1}{2} \theta^{2} t\right) as a power series in \theta, and show that the coefficients of \theta, \theta^{2}, and \theta^{3} are proportional to B_{t}, B_{t}^{2} - t, and B_{t}^{3} - 3 t B_{t}.
3. Explain why every coefficient of the series is a martingale, and find the coefficient of \theta^{4}.

Solution

1. Write \Delta B = B_{t} - B_{s}, independent of \mathcal{F}_{s} with \operatorname{E}(\Delta B) = \operatorname{E}(\Delta B^{3}) = 0 and \operatorname{E}(\Delta B^{2}) = t - s. Then \operatorname{E}(B_{t}^{3} \mid \mathcal{F}_{s}) = B_{s}^{3} + 3 B_{s}^{2} \operatorname{E}(\Delta B) + 3 B_{s} \operatorname{E}(\Delta B^{2}) + \operatorname{E}(\Delta B^{3}) = B_{s}^{3} + 3 (t - s) B_{s}, and \operatorname{E}(3 t B_{t} \mid \mathcal{F}_{s}) = 3 t B_{s}, so \operatorname{E}(B_{t}^{3} - 3 t B_{t} \mid \mathcal{F}_{s}) = B_{s}^{3} - 3 s B_{s}. This is the continuous-time version of the martingale S_{n}^{3} - 3 n S_{n} in the practice problems on discrete-time martingales.
2. Multiplying the series e^{\theta B_{t}} = 1 + \theta B_{t} + \frac{1}{2} \theta^{2} B_{t}^{2} + \frac{1}{6} \theta^{3} B_{t}^{3} + \cdots and e^{-\frac{1}{2} \theta^{2} t} = 1 - \frac{1}{2} \theta^{2} t + \cdots, Z_{t} = 1 + \theta B_{t} + \frac{\theta^{2}}{2} (B_{t}^{2} - t) + \frac{\theta^{3}}{6} (B_{t}^{3} - 3 t B_{t}) + \cdots
3. \operatorname{E}(Z_{t} \mid \mathcal{F}_{s}) = Z_{s} holds for every \theta, so the coefficients of each power of \theta on both sides must agree, and each coefficient is a martingale. Collecting the \theta^{4} terms, \frac{1}{24} B_{t}^{4} - \frac{1}{4} t B_{t}^{2} + \frac{1}{8} t^{2} = \frac{1}{24} (B_{t}^{4} - 6 t B_{t}^{2} + 3 t^{2}), so B_{t}^{4} - 6 t B_{t}^{2} + 3 t^{2} is a martingale. The exponential martingale generates a whole family of polynomial martingales.

Problem 7 (★★) Let X_{t} = \mu t + \sigma B_{t}, and let \tau be the first time X_{t} reaches a or -b, where a, b > 0.

1. Let \mu = 0. Use the martingale X_{t}^{2} - \sigma^{2} t, stopped at \tau, to show that \operatorname{E}(\tau) = a b / \sigma^{2}. Justify the limit t \to \infty.
2. Let \mu \neq 0. Use the martingale X_{t} - \mu t to show that \operatorname{E}(\tau) = \frac{(a + b) \operatorname{P}(X_{\tau} = a) - b}{\mu}.
3. Compute the expected duration of the trade in the previous problem, and compare it with the case \mu = 0.

Solution

1. X_{t}^{2} - \sigma^{2} t = \sigma^{2} (B_{t}^{2} - t) is a martingale, and so is its stopped version, so \operatorname{E}(X_{t \land \tau}^{2}) = \sigma^{2} \operatorname{E}(t \land \tau) for every t. As t \to \infty, t \land \tau increases to \tau, so the right side converges to \sigma^{2} \operatorname{E}(\tau) by monotone convergence. On the left, X_{t \land \tau}^{2} \leq \max(a, b)^{2} and X_{t \land \tau} \to X_{\tau}, so the left side converges to \operatorname{E}(X_{\tau}^{2}) by dominated convergence. With \operatorname{P}(X_{\tau} = a) = b / (a + b), \sigma^{2} \operatorname{E}(\tau) = a^{2} \frac{b}{a + b} + b^{2} \frac{a}{a + b} = a b. This is the limit of the random walk with small ticks in the practice problems on ruin probabilities.
2. X_{t} - \mu t = \sigma B_{t} is a martingale, so \operatorname{E}(X_{t \land \tau}) = \mu \operatorname{E}(t \land \tau). The same two convergence arguments give \operatorname{E}(X_{\tau}) = \mu \operatorname{E}(\tau). Writing p = \operatorname{P}(X_{\tau} = a), \operatorname{E}(X_{\tau}) = a p - b (1 - p) = (a + b) p - b, and dividing by \mu gives the result.
3. With a = 0.1823, b = 0.1054, and p = 0.4205, \operatorname{E}(\tau) = (0.2877 \times 0.4205 - 0.1054) / 0.05 = 0.312 years, about 16 weeks. With \mu = 0, \operatorname{E}(\tau) = a b / \sigma^{2} = 0.307 years. The drift barely changes the duration, which is set by the volatility and the width of the band.

Problem 8 (★★) The log price of a stock is X_{t} = \mu t + \sigma B_{t}, observed at n equally spaced dates over [0, T], with \Delta = T / n. The realized variance is RV = \sum_{j = 1}^{n} (X_{j \Delta} - X_{(j - 1) \Delta})^{2}.

1. Show that \operatorname{E}(RV) = \sigma^{2} T + \mu^{2} T^{2} / n. Hint: each increment is \mathcal{N}(\mu \Delta, \sigma^{2} \Delta).
2. Use \operatorname{V}(Y^{2}) = 2 v^{2} + 4 m^{2} v for Y \sim \mathcal{N}(m, v) to show that \operatorname{V}(RV) \to 0 as n \to \infty, so that RV converges to \sigma^{2} T in mean square.
3. The natural estimator of the drift is \hat{\mu} = X_{T} / T. Show that \operatorname{SD}(\hat{\mu}) = \sigma / \sqrt{T}, whatever the value of n.
4. With \sigma = 0.20, \mu = 0.08, and daily observations over one year, n = 252, compute \operatorname{SD}(RV) relative to \sigma^{2} T and \operatorname{SD}(\hat{\mu}) relative to \mu. How many years of data would make \operatorname{SD}(\hat{\mu}) = 0.04?

Solution

1. Each increment has \operatorname{E}(\Delta X^{2}) = (\mu \Delta)^{2} + \sigma^{2} \Delta, and adding up the n terms gives \operatorname{E}(RV) = \mu^{2} T^{2} / n + \sigma^{2} T.
2. The increments are independent, so \operatorname{V}(RV) = n \left(2 \sigma^{4} \Delta^{2} + 4 \mu^{2} \Delta^{2} \sigma^{2} \Delta\right) = \frac{2 \sigma^{4} T^{2}}{n} + \frac{4 \mu^{2} \sigma^{2} T^{3}}{n^{2}} \xrightarrow[n \to \infty]{} 0. Together with the first part, \operatorname{E}\left((RV - \sigma^{2} T)^{2}\right) = \operatorname{V}(RV) + (\mu^{2} T^{2} / n)^{2} \to 0. This is the quadratic variation result: the drift does not contribute to the quadratic variation.
3. \hat{\mu} = \mu + \sigma B_{T} / T, so \operatorname{SD}(\hat{\mu}) = \sigma \sqrt{T} / T = \sigma / \sqrt{T}. The estimator uses only the endpoints, and observing the path more often within [0, T] does not help.
4. Ignoring the small drift term, \operatorname{SD}(RV) \approx \sqrt{2 / 252} \, \sigma^{2} = 0.0036, about 9% of \sigma^{2} = 0.04. By contrast, \operatorname{SD}(\hat{\mu}) = 0.20, two and a half times the drift itself. To get \operatorname{SD}(\hat{\mu}) = 0.04 we need \sqrt{T} = 0.20 / 0.04 = 5, or 25 years of data. Volatility can be measured precisely from a short sample of frequent data, while expected returns require a long history. This is why models in finance often treat volatility as known and the expected return as uncertain.

Problem 9 (★) A stock price moves by one tick at the events of a Poisson process \{N_{t}\} with rate \lambda, up or down with probability 1/2 each, independently of the event times, as in the practice problems on Poisson processes. Its change since time zero is X_{t} = \sum_{i = 1}^{N_{t}} Y_{i}, with Y_{i} = \pm 1.

1. Show that the total variation and the quadratic variation of X over [0, t] both equal N_{t}.
2. Given that X_{t}^{2} - \lambda t is a martingale, show that X_{t}^{2} - [X, X]_{t} is also a martingale. Explain why Brownian motion has only one martingale of this kind, B_{t}^{2} - t.

Solution

1. As the mesh shrinks, each subinterval contains at most one tick, and a tick moves X by \pm 1, whose absolute value and square are both 1. Both sums count the ticks, so V_{t}(X) = [X, X]_{t} = N_{t}.
2. X_{t}^{2} - [X, X]_{t} = X_{t}^{2} - N_{t} = (X_{t}^{2} - \lambda t) - (N_{t} - \lambda t) is a difference of two martingales. Jumps make the quadratic variation random, so X has two compensators for X_{t}^{2}: the random quadratic variation N_{t} and its expectation \lambda t. The quadratic variation of Brownian motion is the deterministic t, which equals its own expectation, so the two compensators coincide and B_{t}^{2} - [B, B]_{t} = B_{t}^{2} - t.