If $B_t$ is standard Brownian Motion, how to show that $X_t = B_t^2-t$ is a martingale?

Question

If $(B_t, \mathcal{F}_t)$ is standard Brownian Motion, I would like to show that $X_t = B_t^2-t$ is a martingale. My attempted proof works as follows:

\begin{align} E(X_{t+1}|\mathcal{F}_t) & = E(B_{t+1}^2-(t+1)|\mathcal{F}_t) \\ &= E(B_{t+1}^2|\mathcal{F}_t) -(t+1) \\ &= Var(B_{t+1}|\mathcal{F}_t) +E(B_{t+1}|\mathcal{F}_t)^2 - (t+1) \\ &= (t+1)+B_{t+1}^2 -(t+1) \end{align}

However, I believe that the decomposition into variance shouldn't work. Does anyone have any ideas where I went wrong? Thanks

Answer 1

Use the fact that $(B_t-B_s)$ is independent of $\mathcal{F}_s$ and calculate the following expectation: $$ \mathbb{E}[B_t^2\mid \mathcal{F}_s]=\mathbb{E}[(B_t-B_s+B_s)^2\mid \mathcal{F}_s]. $$

Answer 2

Define the function $f(t,x) : = x^2 - t$ and observe that $\partial_t f(t,x) = -1$ and $\partial_x^2 f(t,x) = 2$. We then see that $\partial_t f(t,x) = - \frac{1}{2} \partial_x^2 f(t,x)$ and therefore $(X_t)_{t \geq 0}$ defines a local martingale. To see that $(X_t)_{t \geq 0}$ is actually a full martingale, we need to show that for each $T > 0$, that \begin{eqnarray*} \mathbb{E} \left( \int_0^T \left| \partial_x f(t,x) \right|^2 dt \right) & < & \infty. \end{eqnarray*} To this end, we simply observe that \begin{eqnarray*} \mathbb{E} \left( \int_0^T \left| \partial_x f(t,x) \right|^2 dt \right) &=& \int_0^T \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} 2x \exp \left( - \frac{x^2}{2} \right) dx dt \\ &=& \int_0^T \frac{1}{\sqrt{2\pi t}} dt \int_{-\infty}^{\infty} 2x \exp \left( - \frac{x^2}{2} \right) dx \\ &=& 2\sqrt{T} \cdot 0 = 0 < \infty. \end{eqnarray*} So $(f(t,B_t))_{t \geq 0}$ defines a martingale.