# Given that $B(t)$ is standard Brownian motion. Is $\overline{B}(t) = B(t+s)-B(s)$ a standard Brownian motion?

1) $$B(0) = 0$$ is satisfied, because $$\overline{B}(0) = B(0+s) - B(s) = B(s) - B(s) = 0$$.

3) Assumption that $$\bar{B}(t)-\bar{B}(s) \sim N(0,t-s)$$ is not satisfied, because:

$$\overline{B}(t)-\overline{B}(s) = B(t+s)-B(s) - ((B(2s)-B(s)) = B(t+s) - 2B(s)$$

From definition $$B(s)$$ is normally distributed, but how to show that $$B(t+s)$$ is normally distributed? If they are normally distributed, then $$E[B(t+s)-B(s)] = E[B(t+s)] - 2E[B(s)] = 0$$. But here again, it only is so, if $$B(t+s)$$ is Brownian motion. Do I need to separately prove that $$B(t+s)$$ is Brownian motion or is there a shortcut?

$$D[B(t+s)-B(2s)] = D[B(t+s)]-D[B(2s)] + 2Cov[B(t+s), B(2s)] = t+s - 2s + 2Cov[B(t+s), B(2s)]$$

1. Case when $$t\le s$$: $$D[B(t+s)-B(2s) = t -s + 2(t+s) = 3t +s \neq t-s$$

2. Case when $$s \le t$$: $$D[B(t+s)-B(2s)] = t -s + 4s = t+3s \neq t-s$$

$$\overline{B}(t) - \overline{B}(t') = B(t+s)-B(s) - B(t'+s)+B(s) = B(t+s)-B(t'+ s) \\ E[B(t+s)-B(t'+s)] = B(s) - B(s) = 0 \\ D[B(t+s)-B(t'+s)] = D[B(t+s)] +D[B(t'+s)] + 2Cov[B(t+s), B(t'+s)] \\ \text{ 1) Case when } t \le t' \\ t+t'+2s + 2Cov[B(t+s),B(t'+s -(t+s) + t+s)] = t+t'+2s+2(t+s) = 3t+4s+t'$$

So, here the assumption is not satisfied. But I probably have made an error somewhere. Also, in order to conclude that $$D[B(t+s)]$$ is equal to $$t+s$$, $$B(t+s)$$ should be a Brownian motion, but the only given is that B(t) is Brownian motion.

2) Assumption that increments of $$\overline{B}(t)$$ are independent. I don't know how to prove or disprove this. $$\overline{B}(t) = \overline{B}(t)-\overline{B}(0) = B(t+s) - B(s) - B(s)+ B(s) = B(t+s)-B(s) \\ \overline{B}(t+h) - \overline{B}(h) = B(t+h+s) - B(s) - B(h+s) + B(s) = B(t+h+s) - B(h+s)$$ From there it looks like increments are not independent. Not sure how to formally show it apart from $$B(t+s) - B(s) \neq B(t+h+s) - B(h+s)$$

Let $$t_1 < t_2 < ... < t_{i-1} < t_{i} < ... < t_n \ \ \forall n$$

Let $$z_i = \overline{B}(t_i+s)$$ and $$z_{i-1} = \overline{B}(t_{i-1}+s)$$ Then from definition of standard Brownian motion $$z_i - z_{i-1}$$ are independet $$\forall i\in 1,...,n$$ And $$\overline{B}(t_{i-1}) - \overline{B}(t_i) = B(t_{i-1}+s)-B(s)-B(t_i+s)+B(s) = B(t_{i-1}+s)-B(t_i+s) \Rightarrow \text{ increments } \overline{B}(t_{i-1})-\overline{B}(t_i) \text{ are independet } \forall i = 1,...,n$$

• You have a typo in verifying assumption 3, in your first line. Also, you need to consider a general increment say $\bar B(t) - \bar B(t')$ not just the case $t' = s$. Hint for Assumption 2, increments of a process are invariant to a (constant) shift of the value of the process. Jan 3, 2021 at 14:48
• @passerby51 Do you mean that $2B(s)$ should be written as $B(s)$? I just thought that the constant can be taken out of functions' argument. Jan 3, 2021 at 15:04
• $2B(s)$ should be $B(2s)$. Jan 3, 2021 at 15:08
• @passerby51 Now I tried to test the third assumption using general increment. Ant it looks bad :| . Jan 3, 2021 at 15:18
• @passerby51 I also tried to test the third assumption. Jan 3, 2021 at 15:31

First, check out what independent increments means. You need to consider a general sequence of points say $$t_0 < t_1 < \cdots < t_n$$.
• $$\bar B(t_1) - \bar B(t_0) = B(t_1 +s) - B(t_0+s)$$. You know something about the distribution of an increment of $$B(\cdot)$$, that is, you know the distribution of the right-hand side of the equation just mentioned, without any calculations. What can you conclude about the distribution of $$\bar B(t_1) - \bar B(t_0)$$?
• Consider the random variables $$\bar B(t_{i}) - \bar B(t_{i-1})$$ for $$i=1,\dots,n$$ and show that they can related to random variables $$B(z_i) - B(z_{i-1})$$, $$i=1,\dots,n$$ for some points $$\{z_i\}$$.
Does $$B(\cdot)$$ have independent increments? What can you conclude about $$\bar B(\cdot)$$ then?
• I showed that increments for random variables are independent using the second hint. From the first hint I get that the increments for $\overline{B}(t)$ are normally distributed and that they are independent. What about the third assumption for general case? Jan 3, 2021 at 21:06
• You should not conclude that $\bar B(t)$ are independent for different $t$ (they are not). From the first hint you can directly conclude that $\bar B(t_1) - \bar B(t_0)$ has a normal distribution with mean 0 and variance .... . Just review the definition of the Brownian motion. Jan 3, 2021 at 21:27