The Basic Brownian Martingales

Brownian Motion and the Martingale Conditions

Brownian motion is a continuous-time process whose future increments have mean zero and are independent of the information available at the present time. Those two properties explain why Brownian motion itself is a martingale. The same increment calculation also produces two other important martingales: the square of Brownian motion minus elapsed time, and an exponential of Brownian motion with a deterministic correction. The filtration matters: independence of increments alone is not enough unless the process is adapted to that filtration.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a filtration $(\mathcal{F}_t)_{t\ge0}$. Let $(B_t)_{t\ge0}$ be a process with state space $\mathbb{R}$ and continuous time. Assume $B_0=0$ almost surely, paths are continuous, $B_t$ is $\mathcal{F}_t$-measurable for every $t$, and for every $0\le s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has the normal distribution with mean $0$ and variance $t-s$. These assumptions specify Brownian motion relative to the given filtration. In particular, adaptedness is an explicit assumption.

Definition: An adapted process $(M_t)_{t\ge0}$ is a martingale with respect to $(\mathcal{F}_t)$ if $\mathbb{E}|M_t|<\infty$ for every $t\ge0$, and $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely for every $0\le s\le t$. Conditional expectation equalities in this tutorial are understood almost surely.

Write $h=t-s$ and $Z=B_t-B_s$. The increment assumptions give $\mathbb{E}[Z\mid\mathcal{F}_s]=0$ and $\mathbb{E}[Z^2\mid\mathcal{F}_s]=h$. To use exponential functions, we also need the Gaussian moment-generating function. For $h>0$, if $Z$ has the normal distribution with mean $0$ and variance $h$, completing the square in its density gives, for every real $\theta$,

\begin{aligned} \mathbb{E}[e^{\theta Z}] &=\frac{1}{\sqrt{2\pi h}}\int_{-\infty}^{\infty} \exp\left(\theta z-\frac{z^2}{2h}\right)\,dz\\ &=\frac{e^{\theta^2h/2}}{\sqrt{2\pi h}}\int_{-\infty}^{\infty} \exp\left(-\frac{(z-\theta h)^2}{2h}\right)\,dz =e^{\theta^2h/2}. \end{aligned}

When $h=0$, $Z=0$ almost surely, so the same formula holds with both sides equal to $1$. Independence of $Z$ from $\mathcal{F}_s$ therefore gives $\mathbb{E}[e^{\theta Z}\mid\mathcal{F}_s]=e^{\theta^2h/2}$.

The Three Basic Brownian Martingales

Theorem: Under the Brownian and filtration assumptions above, each of the following adapted processes is a martingale with respect to $(\mathcal{F}_t)$: $$ B_t,\qquad B_t^2-t,\qquad \exp\left(\theta B_t-\frac{\theta^2t}{2}\right),\quad \theta\in\mathbb{R}. $$

Proof. Adaptedness of $B_t$ is assumed, and the other two processes are measurable functions of $B_t$ and the deterministic time $t$, so they are adapted as well. For each fixed $t$, $B_t$ is normal with mean $0$ and variance $t$; consequently, $\mathbb{E}|B_t|<\infty$, $\mathbb{E}[B_t^2]=t<\infty$, and the Gaussian moment-generating function gives $\mathbb{E}[\exp(\theta B_t-\theta^2t/2)]=1$. Thus all three processes are integrable at every time.

It remains to verify the conditional expectation identities. Fix $0\le s\le t$, put $h=t-s$, and write $B_t=B_s+Z$, where $Z=B_t-B_s$. Since $\mathbb{E}[Z\mid\mathcal{F}_s]=0$ and $B_s$ is $\mathcal{F}_s$-measurable,

\mathbb{E}[B_t\mid\mathcal{F}_s] =\mathbb{E}[B_s+Z\mid\mathcal{F}_s] =B_s.

For the squared process, expand the square and use the conditional first and second moments of $Z$:

\begin{aligned} \mathbb{E}[B_t^2-t\mid\mathcal{F}_s] &=\mathbb{E}[(B_s+Z)^2-t\mid\mathcal{F}_s]\\ &=B_s^2+2B_s\mathbb{E}[Z\mid\mathcal{F}_s] +\mathbb{E}[Z^2\mid\mathcal{F}_s]-t\\ &=B_s^2+h-t =B_s^2-s. \end{aligned}

For the exponential process, factor out the $\mathcal{F}_s$-measurable term and apply the Gaussian moment-generating function:

\begin{aligned} \mathbb{E}\left[\exp\left(\theta B_t-\frac{\theta^2t}{2}\right) \middle|\mathcal{F}_s\right] &=\exp\left(\theta B_s-\frac{\theta^2t}{2}\right) \mathbb{E}[e^{\theta Z}\mid\mathcal{F}_s]\\ &=\exp\left(\theta B_s-\frac{\theta^2t}{2}\right) e^{\theta^2(t-s)/2}\\ &=\exp\left(\theta B_s-\frac{\theta^2s}{2}\right). \end{aligned}

Each conditional expectation is the value of the corresponding process at time $s$. Along with adaptedness and integrability, this proves all three martingale claims. The calculations include $s=t$, when $Z=0$ and the identities reduce to equality of a quantity with itself. $\square$

The deterministic correction terms are essential. The squared process gains conditional mean $t-s$ over an interval, so subtracting time removes that increase. The exponential process gains a conditional multiplicative factor $e^{\theta^2(t-s)/2}$, so subtracting $\theta^2t/2$ in its exponent cancels that factor.

Worked Example: Conditional Mean and Variance of a Future Value

Let $B_t$ satisfy the stated assumptions. Compute $\mathbb{E}[B_3\mid\mathcal{F}_1]$ and $\mathbb{E}[(B_3-B_1)^2\mid\mathcal{F}_1]$. Here $s=1$, $t=3$, so the increment variance is $h=3-1=2$.

Write $B_3=B_1+Z$, where $Z=B_3-B_1$ is independent of $\mathcal{F}_1$, has mean $0$, and has variance $2$. Therefore,

\mathbb{E}[B_3\mid\mathcal{F}_1] =B_1+\mathbb{E}[Z\mid\mathcal{F}_1] =B_1+0=B_1.

For the increment's squared value, independence and its variance give

\mathbb{E}[(B_3-B_1)^2\mid\mathcal{F}_1] =\mathbb{E}[Z^2\mid\mathcal{F}_1] =\operatorname{Var}(Z)+(\mathbb{E}Z)^2 =2+0^2=2.

Check: the increment has the normal distribution with mean $0$ and variance $3-1=2$, so its conditional mean is $0$ and its conditional second moment is $2$. The future-value calculation consequently gives $B_1$, as required by the martingale identity.

Worked Example: Checking the Squared Martingale

Take $s=\tfrac12$ and $t=2$. Then $h=2-\tfrac12=\tfrac32$. Compute $\mathbb{E}[B_2^2-2\mid\mathcal{F}_{1/2}]$ directly and compare it with the squared-martingale value at time $\tfrac12$.

Set $Z=B_2-B_{1/2}$. Its conditional mean is $0$, and its conditional second moment is $\tfrac32$. Expanding the square gives

\begin{aligned} \mathbb{E}[B_2^2-2\mid\mathcal{F}_{1/2}] &=\mathbb{E}\left[\left(B_{1/2}+Z\right)^2-2 \middle|\mathcal{F}_{1/2}\right]\\ &=B_{1/2}^2+2B_{1/2}(0)+\frac32-2\\ &=B_{1/2}^2-\frac12. \end{aligned}

At time $s=\tfrac12$, the process $B_s^2-s$ equals $B_{1/2}^2-\tfrac12$. Thus the computed conditional expectation agrees with the claimed martingale value.

Check: the increase in conditional second moment from time $\tfrac12$ to time $2$ is $\tfrac32$, exactly the elapsed time $2-\tfrac12$. Subtracting $2$ at the later time leaves the correction $-\tfrac12$, matching the earlier time correction.

Worked Example: An Exponential Martingale

For $\theta=2$, compute the conditional expectation of the exponential process at $t=\tfrac32$, given $\mathcal{F}_1$. The process at time $u$ is $\exp(2B_u-2u)$, since $\theta^2u/2=4u/2=2u$.

Here $Z=B_{3/2}-B_1$ has variance $h=\tfrac32-1=\tfrac12$. The Gaussian moment-generating function gives $\mathbb{E}[e^{2Z}\mid\mathcal{F}_1]=e^{2^2h/2}=e^{4(1/2)/2}=e$. Hence

\begin{aligned} \mathbb{E}\left[e^{2B_{3/2}-3}\mid\mathcal{F}_1\right] &=e^{2B_1-3}\mathbb{E}[e^{2Z}\mid\mathcal{F}_1]\\ &=e^{2B_1-3}e =e^{2B_1-2}. \end{aligned}

At time $1$, the process is $\exp(2B_1-2\cdot1)=e^{2B_1-2}$, so the answer is exactly its time-$1$ value.

Check: the conditional Gaussian factor is $e$ because the increment variance is $\tfrac12$ and $\theta=2$. Multiplying the time-$\tfrac32$ correction $e^{-3}$ by this factor gives $e^{-2}$, which is the correction at time $1$.

Common Mistakes and Why the Assumptions Matter

A frequent computational error is to omit the conditional second moment of the increment when expanding a square. For example, at $s=0$ and $t=1$, $B_0=0$ and $B_1$ has variance $1$. The incorrect calculation $\mathbb{E}[B_1^2-1\mid\mathcal{F}_0]=0-1=-1$ has treated $\mathbb{E}[B_1^2]$ as zero. The correction is $\mathbb{E}[B_1^2]=\operatorname{Var}(B_1)+(\mathbb{E}B_1)^2=1+0=1$, so $\mathbb{E}[B_1^2-1\mid\mathcal{F}_0]=1-1=0=B_0^2-0$.

There is also a filtration pitfall. Suppose the filtration is trivial, $\mathcal{F}_t=\{\varnothing,\Omega\}$ for every $t$, while $(B_t)$ has its usual Brownian distributions and independent increments. Its increments are independent of each $\mathcal{F}_s$, but $B_t$ is not $\mathcal{F}_t$-measurable for $t>0$. Thus the process is not adapted and does not meet the definition of a martingale with respect to this filtration. Increment independence cannot replace the adaptedness requirement.

Finally, the exponential correction depends on the chosen parameter: for $\exp(\theta B_t)$ without the term $-\theta^2t/2$, the conditional expectation from $s$ to $t$ is $\exp(\theta B_s)e^{\theta^2(t-s)/2}$, not $\exp(\theta B_s)$ in general. Only when $\theta=0$ or $s=t$ does that extra factor equal $1$. Checking the factor using the increment variance $t-s$ prevents a common sign or factor-of-two error.

These martingales provide tools for organizing conditional calculations and for later arguments involving stopping times. At this stage, the claims concern deterministic times $s\le t$; no conclusion about evaluating a martingale at an unbounded random time follows from these computations alone.

Check Your Understanding

For each question, identify the increment interval and state which martingale identity or Gaussian moment is being used.

For $0\le s<t$, calculate $\mathbb{E}[B_t-B_s\mid\mathcal{F}_s]$ and $\mathbb{E}[(B_t-B_s)^2\mid\mathcal{F}_s]$.
Expand $B_t^2=(B_s+(B_t-B_s))^2$ and use the conditional increment moments to verify the martingale identity for $B_t^2-t$.
For a real $\theta$, state the Gaussian moment-generating function needed to check the exponential process and identify its deterministic correction.
In the exponential example, replace $\theta=2$ by $\theta=-1$. What factor is contributed by the increment from time $s$ to time $t$?
Explain why a process with increments independent of $\mathcal{F}_s$ may still fail to be a martingale with respect to $(\mathcal{F}_t)$.

What You'll Learn