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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 ?

Nikita Evseev
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    Take the next step and compte $\Bbb E[\exp(\alpha X(s)+\beta(X(t)-X(s)))]$ for $0\le s\le t$. – John Dawkins May 28 '16 at 17:58
  • Computing $\Bbb E[\exp(\alpha X(s)+\beta(X(t)-X(s)))] = \exp(\frac{1}{2}(\alpha^2s + \beta^2(t-s)))$. Does it imply independence, and why? – Nikita Evseev May 29 '16 at 07:09
  • It shows $\Bbb E[\exp(\alpha X(s)+\beta(X(t)-X(s)))]=\Bbb E[\exp(\alpha X(s))]\cdot\Bbb E[\exp(\beta(X(t)-X(s)))]$ for all real (or complex, for that matter) $\alpha$ and $\beta$. – John Dawkins May 29 '16 at 16:17
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    @NikitaEvseev If you don't know what to make of JohnDawkin's comment, then have a look at this answer: http://math.stackexchange.com/a/287321/36150 – saz May 29 '16 at 19:19
  • @JohnDawkins and saz thanks for your help. Below I post the solution. – Nikita Evseev May 30 '16 at 05:48
  • Apparently, here $X(t)$ is $\mathbb R^2$ valued (random) and $\lambda X(t)$ should be understood as the inner product in $\mathbb R^2$. Otherwise, for a 1-dimensional Brownian motion you don't get $|\lambda|^2$ in the exponent of the right hand side. – Sergio A. Yuhjtman Oct 10 '18 at 20:29

1 Answers1

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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.

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Nikita Evseev
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