Verifying a Brownian motion through the Laplace transform

Question

Let $X(t)$ be a continuous stochastic process and $\mathcal G(t)$ be the $\sigma$-algebra generated by $\{X(\tau) : \tau\leq t \}$. Suppose that for any $0\leq s\leq t$ and $\lambda\in\mathbb C$ $$\mathbb{E}\left[\exp(\lambda X(t))\, |\, \mathcal G(s) \right] = \exp\left(\frac{1}{2}|\lambda|^2(t-s) + \lambda X(s) \right). $$ Prove, that $X(t)$ is a Brownian motion.

My attempt. From the equation above it is easy to conclude, that for any $0\leq t_1\leq t_2$ $$\mathbb{E}\left[\exp(\lambda ( X(t_2) - X(t_1)) \right] = \exp\left(\frac{1}{2}|\lambda|^2(t_2-t_1)\right), $$ and, as the Laplace transform determines the distribution completely, $$ X(t_2) - X(t_1) \sim N(0, t_2-t_1). $$

Now, my question is how to prove the independence of increments ?

Answer 1

To prove that the process $X(t)$ is a Brownian motion one has to show that increments are normally distributed and are independent.

1. Here I use the fact that the Laplace transform (characteristic functions) determines the distribution of a random variable. For any $0\leq s\leq t$, $\lambda\in\mathbb C$ have $$\mathbb{E}\left[\exp(\lambda ( X(t) - X(s)) \right] =\\ \mathbb{E}\left[\mathbb{E}\left(\exp(\lambda ( X(t) - X(s)) \,|\, \mathcal G(s)\right)\right]=\\ \mathbb{E} \left[\exp(-\lambda X(s))\mathbb{E}\left(\exp(\lambda ( X(t)) \,|\, \mathcal G(s)\right)\right] = \\ \mathbb{E}\left[\exp(-\lambda X(s)) \exp\left(\frac{1}{2}|\lambda|^2(t-s) + \lambda X(s)\right)\right]=\\ \exp\left(\frac{1}{2}|\lambda|^2(t-s)\right).$$ Thus we conclude that $X(t_{t})-X(t_{s}) \sim N(0, t-s)$.

2. The independence of increments follows from Kac's theorem for characteristic functions.

   Theorem (Kac's theorem) The random variables $X, Y$ are independent if and only if $$\mathbb E\left[\exp(i ( \alpha X + \beta Y) \right] =\mathbb E\left[\exp(i\alpha X)\right]\cdot\mathbb E\left[\exp(i\beta Y)\right]$$ for all $\alpha, \beta \in \mathbb R$.

   (The result of this kind discussed a lot on MSE: here, and here. ) Using the same argument as in 1. obtain $$\Bbb E[\exp(i(\alpha X(s)+\beta(X(t)-X(s))))]=\\ \Bbb E[\exp(i\alpha X(s))]\cdot\Bbb E[\exp(i\beta(X(t)-X(s)))].$$ This means that increments $X(s), X(t)-X(s)$ are independent. In the same way one can show independence of any number of increments.