Quadratic Variation of a square-integrable Lévy process

Question

I am having a problem with the following question. I have tried using the definition of square integrable martingales and quadratic variation, but just can't seem to get anywhere. Can anybody offer me any assistance?

Let $(X_t)_{t \geq 0}$ be a square-integrable Lévy process. Show that $\langle X \rangle_t = t E(X_1^2)$ if $(X_t)_{t \geq 0}$ is a martingale.

I'm not sure how to proceed. If anyone can start me off, that'll be great.

Answer 1

Hints:

1. Since $(X_t)_{t \geq 0}$ is a martingale and $X_0=0$, we have $\mathbb{E}(X_t)=0$ for all $t \geq 0$.
2. If a random variable $Y$ has finite second moment, then $$\mathbb{E}(Y) = \frac{1}{\imath} \frac{d}{d\xi} \chi(\xi) \bigg|_{\xi=0} \qquad \quad \mathbb{E}(Y^2) = - \frac{d^2}{d\xi^2} \chi(\xi) \bigg|_{\xi=0}$$ where $\chi$ denotes the characteristic function of $Y$. Set $$\chi_t(\xi) := \mathbb{E}e^{\imath \, \xi X_t}.$$ Conclude from the Lévy-Khintchine formula that $$\chi_t(\xi) = (\chi_1(\xi))^t$$ and, by differentiating this expression, that $$\mathbb{E}(X_t^2) = t \mathbb{E}(X_1^2).$$
3. By the definition of $\langle X \rangle_t$, it suffices to show that $$M_t := X_t^2 - t \mathbb{E}(X_1^2)$$ is a martingale. To this end, fix $s \leq t$ and write $$\begin{align*} \mathbb{E}(M_t \mid \mathcal{F}_s) &= \mathbb{E}\big( ((X_t-X_s)+X_s)^2 \mid \mathcal{F}_s \big) - t \mathbb{E}(X_1^2) \\ &= \mathbb{E}((X_t-X_s)^2 \mid \mathcal{F}_s) + 2 \mathbb{E}(X_s (X_t-X_s) \mid \mathcal{F}_s) + \mathbb{E}(X_s^2 \mid \mathcal{F}_s)- t \mathbb{E}(X_1^2). \end{align*}$$ Now use that $(X_t)_{t \geq 0}$ has independent increments (i.e. $X_t-X_s$ and $\mathcal{F}_s$ are independent), the stationarity (i.e. $X_t-X_s \sim X_{t-s}$) and step 2.