Proving a Wiener Process is a Martingale

Question

Let $W_t$ be a Wiener process. Let $X_t = W_t^2 − t$. Show that {${X_t ;t ≥ 0}$} is a martingale with respect to the natural filtration.

To prove that it is in fact a Martingale I must prove 2 properties:

1) $\mathbb E [|X_t|] < ∞ $

2) $\mathbb E [X_t | F_s] = X_s$ for $ 0 ≤ s ≤ t$

For the first part I have: $$\mathbb E [|X_t|] = \mathbb E [|W_t^2 -t|] ≥ t \mathbb E [W_t^2] +t =2t $$

Although I am not sure if this is correct?

For the second part I am unsure what the answer is, if anyone can help, I'd be very grateful.

Answer 1

For the first part, $X_t$ is just a sum of a constant and an integrable random variable, so it is integrable. For the second:

$E[ W_t ^2 - W_s ^2 | F_s ] = E[(W_t - W_s)(W_t + W_s) | F_s]= E[(W_t - W_s)W_t | F_s] + E[(W_t - W_s)W_s | F_s] $

The right summand equals zero, since $W_s$ is a martingale, and since $W_s$ goes outside of the expectation, because it is $F_s$-measurable. For the other summand: $=E[(W_t - W_s)(W_t - W_s + W_s) | F_s]= E[(W_t - W_s)^2 | F_s] +0= E[(W_t - W_s)^2]= \text{Var}(W_t - W_s)= t -s $, where we passed from conditional to simple expectation, because $W_t - W_s$ is independent from $F_s$.

Hence $E[ W_t ^2 - W_s ^2 | F_s] = t-s$ , as desired.

Answer 2

For the first part you showed that $E|X_t|>\infty$ not $E|X_t|<\infty$. To show the later just use the triangle inequality and the fact that $W_t \sim n(0,t)$ since $W_t$ is a Wiener Process. $$ E|W_t^2-t|<E|W_t|^2+t=2t < \infty $$

For the second part we need to use that property of Wiener Processes that the increments are independent. In particular, $W_t-W_u$ is independent of $W_u$. $$\begin{align} E(W_t^2-t| \mathcal F_u)&=E\big((W_t-W_u+W_u)^2-t|\mathcal F_u\big) \\&= E\big((W_t-W_u)^2+W_u^2+2W_u(W_t-W_u)-t|\mathcal F_u\big) \\&=t-u+W_u^2 +0-t \\&=W_u^2-u \end{align}$$