An (a.s.) continuous process $(X_t)_{t\geq 0}$ is a Brownian motion if $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale

Question

Problem

Let $X=(X_t)_{t\geq0}$ be an (a.s.) continuous $\mathbb{R}$-valued process with $X_0=0$ such that $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a $\mathbb{C}$-valued local martingale for all $\lambda\in\mathbb{R}$. Show that $X$ is a standard $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-Brownian motion.

Opening remark

The standard definition of a Brownian motion goes as follows:

A $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process $B=(B_t)_{t\geq 0}$ is called a standard $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-Brownian motion if:

1. $B_0=0$ (a.s.)
2. $B$ is (a.s.) continuous.
3. $\forall s<t: B_t-B_s\sim\mathcal{N}(0,t-s)$
4. $B_t$ has independent increments.

This and this solution for related problems make use of the so-called Lévy characterisation:

A $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process $B=(B_t)_{t\geq 0}$ is called a standard $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-Brownian motion if:

1. $B_0=0$ (a.s.)
2. $B_t$ is an (a.s.) continuous martingale.
3. $B$ has the quadratic variation $\left\langle B\right\rangle_t = t$.

This equivalent definition looks very useful, but I wish to check all the conditions of the standard definition. (This is mainly because the Lévy characterisation is not mentioned in my lecture notes.) Finally, note that I have already solved the following similar problem:

Let $B$ be a standard $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-Brownian motion. Show that $(e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0}$ is a $\mathbb{C}$-valued martingale for all $\lambda\in\mathbb{R}$.

Trivial outline for the proof

First, note that $(X_t)_{t\geq 0}$ is $(\mathcal{F}_t)_{t\geq 0}$-adapted because the local martingale $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is. $(X_t)_{t\geq 0}$ is therefore a $\mathbb{R}$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process by assumption.

1. $X_0=0$ (a.s.) by assumption.
2. $X$ is (a.s.) continuous by assumption.
3. ???
4. ???

Some ideas

If I knew (3.), I could try to show (4.) by calculating $\text{Cov}\left(X_t - X_s,X_v - X_u\right)=\mathbb{E}\left[\left(X_t - X_s\right)\left(X_v - X_u\right)\right] = 0$ for all $s<t\leq u<v$, but I feel like the information I have about $X$ is too limited to do that. Also, I have no idea how to make use of the fact that $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale. Any help is appreciated here.

Questions

1. Is checking all defining properties of a Brownian motion a reasonable approach here? If yes, how can we check (3.) and (4.)? Mere hints would be appreciated.
2. Are there any interesting theorems (other than the Lévy characterisation) which show that a given process is a Brownian motion?

Answer 1

The main idea is to use the following fact:

The conditional characteristic function $\mathbb{E}\left[e^{i\lambda X_s}\vert\mathcal{F}_s\right]$ of $X_s$ is equal to $e^{i\lambda\mu - \frac{1}{2}\lambda^2\sigma^2}$ (= the characteristic function of a $\mathcal{N}\left(\mu,\sigma^2\right)$-random variable) if and only if $X_s$ is independent of $\mathcal{F}_s$ and $\mathcal{N}\left(\mu,\sigma^2\right)$-distributed.

In this case:

The conditional characteristic function $\mathbb{E}\left[e^{i\lambda \left(X_t - X_s\right)}\vert\mathcal{F}_s\right]$ of $X_t-X_s$ is equal to $e^{- \frac{1}{2}\lambda^2\left(t-s\right)}$ (= the characteristic function of a $\mathcal{N}\left(0,t-s\right)$-random variable) if and only if $X_t - X_s$ is independent of $\mathcal{F}_s$ and $\mathcal{N}\left(0,t-s\right)$-distributed.

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First, we show that $Z:=(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is not only a local martingale but a martingale as well. Note that it is an adapted process because it is a local martingale.

1. Recall that $\left\lvert e^{ix}\right\rvert=1$ for all $x\in\mathbb{R}$. $Z_t$ is integrable: $$\mathbb{E}\left[\left\lvert Z_t\right\rvert\right] = \mathbb{E}\left[\left\lvert e^{i\lambda X_t + \frac{1}{2}\lambda^2 t}\right\rvert\right] = \mathbb{E}\left[\left\lvert e^{i\lambda X_t}\right\rvert \cdot\left\lvert e^{\frac{1}{2}\lambda^2 t}\right\rvert\right] = e^{\frac{1}{2}\lambda^2 t}<\infty\ \text{(a.s.)}$$
2. Let $(\tau_n)_{n\in\mathbb{N}}$ be a reducing sequence of $Z$. We then have that $Z^{\tau_n}$ is a martingale and $\lim_{n\to\infty}Z^{\tau_n}_t=Z_t$ (a.s.). Note that $\left\lvert Z^{\tau_n}_t\right\rvert$ is dominated by $\left\lvert Z_t\right\rvert$: $$\left\lvert Z^{\tau_n}_t\right\rvert = e^{\frac{1}{2}\lambda^2 \left(t\land\tau\right)} \leq e^{\frac{1}{2}\lambda^2 t} = \left\lvert Z_t\right\rvert\ \text{(a.s.)}$$ The martingale condition for $Z$ now follows: \begin{align*} &\mathbb{E}\left[Z_t\vert \mathcal{F}_s\right] &&\vert\ \text{Lebesgue’s dominated convergence theorem}\\ =\ &\lim_{n\to\infty}\mathbb{E}\left[Z^{\tau_n}_t\vert\mathcal{F}_s\right] &&\vert\ \text{martingale condition for}\ Z^{\tau_n}\\ =\ &\lim_{n\to\infty} Z^{\tau_n}_s = Z_s &&\text{(a.s.)} \end{align*}

We now show that $X$ is a standard $\mathbb{R}$-valued Brownian motion. Note that $(X_t)_{t\geq 0}$ is adapted because the martingale $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is. $(X_t)_{t\geq 0}$ is therefore a $\mathbb{R}$-valued adapted process by assumption.

1. $X_0 = 0$ (a.s.) by assumption.
2. $X$ is (a.s.) continuous by assumption.
3. The martingale condition for $Z=(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ yields: \begin{align*} &\mathbb{E}\left[e^{i\lambda X_t + \frac{1}{2}\lambda^2 t}\vert\mathcal{F}_s\right] = e^{i\lambda X_s + \frac{1}{2}\lambda^2 s} &&\vert\ X_s\ \text{is}\ \mathcal{F}_s\text{-measurable}\\ \Leftrightarrow\ &\mathbb{E}\left[e^{i\lambda X_t - i\lambda X_s}\vert\mathcal{F}_s\right] = e^{-\frac{1}{2}\lambda^2 t + \frac{1}{2}\lambda^2 s}\\ \Leftrightarrow\ &\mathbb{E}\left[e^{i\lambda \left(X_t-X_s\right)}\vert\mathcal{F}_s\right] = e^{-\frac{1}{2}\lambda^2 \left(t-s\right)} \end{align*} Since the conditional characteristic function of $X_t-X_s$ is equal to the characteristic function of a $\mathcal{N}\left(0,t-s\right)$-random variable, it follows that $X_t-X_s$ is independent of $\mathcal{F}_s$ and $\mathcal{N}\left(0,t-s\right)$-distributed.

Answer 2

By using martingale-property and taking derivative at $\lambda=0$, we get martingale property for $X$

$$E[X_{t}|\mathcal{F}_{s}]=X_{s},$$

where we can take derivative due to dominated convergence theorem since $|e^{iX}|=1$. Similarly, using martingale again and taking derivative twice in $E[e^{i\lambda(X_{t}-X_{s})+\lambda^2(t-s)/2}]=1$, we get

$$E[(X_{t}-X_{s})^{2}]=t-s.$$

Using the martingale for $X$ we also get $$E[(X_{t_{4}}-X_{t_{3}})(X_{t_{2}}-X_{t_{1}})]=0$$

when $$t_{1}<...<t_{4}$. As mentioned in https://almostsuremath.com/2010/01/18/quadratic-variations-and-integration-by-parts/#scn_ibp_thm1, semimartingales are guaranteed to have quadratic variations

Theorem 1 (Quadratic Variations and Covariations) Let $X,Y$ be semimartingales. Then, there exist cadlag adapted processes $[X]$ and $[X,Y]$ satisfying the following. For any sequence $P_n$ of stochastic partitions of $\mathbb R_+$ such that, for each $t\ge 0$, the mesh $\vert P_n^t\vert$ tends to zero in probability as $n\rightarrow\infty$, the following limits hold \begin{array}{rl} \displaystyle [X]^{P_n}&\displaystyle\xrightarrow{\rm ucp}[X],\\ \displaystyle [X,Y]^{P_n}&\displaystyle\xrightarrow{\rm ucp}[X,Y], \end{array}
as $n\rightarrow\infty$. Furthermore, convergence also holds in the semimartingale topology.

Therefore, it suffices to show that the variance of the quadratic variation goes to zero

$$E[[X]_{t}^{2}]-(E[[X]_{t}])^{2}\to 0.$$

This follows from the previous computation

\begin{array}{rl} \displaystyle {\rm Var}[V^n] &\displaystyle = \sum_{k=1}^n {\rm Var}\left[(X_{kt/n}-X_{(k-1)t/n})^2\right]=\sum_{k=1}^n2t^2/n^2=2t^2/n\to 0 \end{array}