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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!

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closed as off-topic by Did, USER91500, Soke, Jonas, Claude Leibovici May 28 at 7:29

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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
1  
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

Collecting all mentioned arguments.

0*. The function $t\mapsto Y_t = X_{t+s}-X_s$ is continuous a.s. (I believe that this is also necessarily).

  1. $Y_0 = X_{0+s}-X_s = 0$.

  2. Observe that $Y_{t_2} - Y_{t_1} = X_{t_2+s} - X_{t_1+s}$ holds for any $0\leq r_1\leq r_2$. Then for all $0\leq r_1\leq r_2\leq \dots\leq r_n$ $Y_{t_2} - Y_{t_1}, Y_{t_3} - Y_{t_2}, \dots Y_{t_n} - Y_{t_{n-1}}$ are independent random variables because $X_{t_2+s} - X_{t_1+s}, X_{t_3+s} - X_{t_2+s}, \dots X_{t_n+s} - X_{t_{n-1}+s}$ are independent. So $y_t$ possesses independent increments.

  3. Note that $X_{t_2+s} - X_{t_1+s} \sim N(0, t_2-t_1)$ ant it implies $Y_{t_2} - Y_{t_1+s} \sim N(0, t_2-t_1)$. In particular $Y_{t} \sim N(0, t)$.

That is why $Y_t$ is a standard Brownian motion.

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