Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{X_t, t\ge 0\}$ be a standard Brownian motion process. For a fixed positive number s and all $t\ge 0$, we define $Y_t = X_{t+s} - X_s$. Is $\{Y_t, t\ge0\}$ a standard Brownian motion?

Attempt: I know that I have to use the method of verifying a standard Brownian motion: i.e.
1. $Y_0 = 0$
2. $Y_t$, $t \ge 0$ possesses independent increments
3. For every $t \ge 0$, $Y_t$ follows $N(0, \sigma^2t)$ where $\sigma = 1$ since we are verifying if $Y_t$ is a standard Brownian motion.

But I am stuck on the necessary steps to prove/verify $Y_t$

Any help would be greatly appreciated!

share|cite|improve this question
I don't understand... All you have to do is to prive 1, 2 and 3? – GHL Dec 8 '13 at 17:50
i think so, like i understand i have to use methods 1,2 and 3, but am not sure about the detailed steps necessary for calculation – Jack Wills Dec 8 '13 at 17:58
Well, can you prove 1? To prove 2, you will need to use the fact that $X_t$ satisfies 2. To prove 3, you will need to use the fact that $X_t$ satisfies 3. – GHL Dec 8 '13 at 18:01
@JackWills : Could you tell me if you're using some outside software to do your MathJax code? It was written in the same bizarre way I've seen several times in the last few days. Please look at my edits. You shouldn't be going in and out of MathJax six times in one short mathematical expression. – Michael Hardy Dec 8 '13 at 19:29
So... You have $Y_0 = X_{0+s} - X_s = 0$. Then for 2 and 3 you indeed have to take $0 < t_1 < t_2 < t_3 < t_4$, express $Y_{t_i}$ in term of $X_{t_i}$ and you're pretty much done! – GHL Dec 9 '13 at 21:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.