Martingale Property of Brownian Motion
Let $W_t$ be a standard Brownian motion. Define the process $M_t = W_t^2 - t$.
Show that $M_t$ is a martingale with respect to the natural filtration $\mathcal{F}_s = \sigma(W_u : u \leq s)$. That is, verify that $E[M_t \mid \mathcal{F}_s] = M_s$ for all $0 \leq s < t$.
Hints
- Think about how fast $W_t^2$ grows on average. What deterministic function would you subtract to remove that drift?
- Write $W_t = W_s + (W_t - W_s)$ and expand the square. Which terms survive conditioning on $\mathcal{F}_s$?
- Use independence of increments: $E[W_t - W_s \mid \mathcal{F}_s] = 0$ and $E[(W_t - W_s)^2 \mid \mathcal{F}_s] = t - s$. The cross term vanishes and the squared increment contributes exactly $t - s$.
Worked Solution
How to Think About It: The key intuition is that Brownian motion's square $W_t^2$ is not a martingale -- it grows on average. How fast? Since $E[W_t^2] = t$, the square grows at rate 1 per unit time. So if you subtract $t$ to "detrend" it, you should kill the drift and get a martingale. The whole proof boils down to one move: decompose $W_t$ into what you knew at time $s$ and what is new, then expand the square and use independence of increments.
Approach: Write $W_t = W_s + (W_t - W_s)$, expand the square inside the conditional expectation, and use the fact that the increment $W_t - W_s$ is independent of $\mathcal{F}_s$ with mean $0$ and variance $t - s$.
Formal Solution:
We verify the martingale property $E[M_t \mid \mathcal{F}_s] = M_s$ for $s < t$.
Decompose $W_t = W_s + (W_t - W_s)$ and expand:
$$W_t^2 = W_s^2 + 2W_s(W_t - W_s) + (W_t - W_s)^2$$
Now take the conditional expectation. Since $W_t - W_s$ is independent of $\mathcal{F}_s$ and $W_s$ is $\mathcal{F}_s$-measurable:
$$E[W_t^2 \mid \mathcal{F}_s] = W_s^2 + 2W_s \cdot E[W_t - W_s] + E[(W_t - W_s)^2]$$
The increment $W_t - W_s \sim N(0, t - s)$, so:
- $E[W_t - W_s] = 0$
- $E[(W_t - W_s)^2] = t - s$
Substituting:
$$E[W_t^2 \mid \mathcal{F}_s] = W_s^2 + 0 + (t - s) = W_s^2 + t - s$$
Therefore:
$$E[M_t \mid \mathcal{F}_s] = E[W_t^2 - t \mid \mathcal{F}_s] = W_s^2 + t - s - t = W_s^2 - s = M_s$$
This confirms the martingale property.
Answer: $M_t = W_t^2 - t$ is a martingale because the conditional expectation of $W_t^2$ given $\mathcal{F}_s$ equals $W_s^2 + (t - s)$, and subtracting $t$ exactly cancels the accumulated variance, leaving $W_s^2 - s = M_s$.
Intuition
This result is really about quadratic variation. Brownian motion accumulates variance at a constant rate -- $\text{Var}(W_t) = t$ -- so $W_t^2$ is a submartingale that drifts upward at rate 1. Subtracting $t$ is the minimal correction that turns it into a fair game. The resulting martingale $M_t = W_t^2 - t$ is one of the canonical examples you should have on instant recall, because the same decomposition trick (split into known part plus independent increment, expand, use moment properties) works for proving martingale properties of exponential Brownian motion, compensated Poisson processes, and many other processes.
In practice, this fact underpins the Ito calculus machinery. When you apply Ito's lemma to $f(W_t) = W_t^2$, you get $d(W_t^2) = 2W_t \, dW_t + dt$. The $dt$ term is exactly the drift you subtract, and the $2W_t \, dW_t$ term is the martingale part. Recognizing which piece is the compensator and which is the martingale is a core skill for pricing, hedging, and building stochastic models on the desk.