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 $$\overline{B}(t_{i-1}) - \overline{B}(t_i) = B(t_{i-1}+s)-B(s)-B(t_i+s)+B(s) = B(t_{i-1}+s)-B(t_i+s) \Rightarrow \text{ increments } \overline{B}(t_{i-1})-\overline{B}(t_i) \text{ are independet } \forall i = 1,...,n$$

  • 2
    $\begingroup$ 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. $\endgroup$
    – passerby51
    Jan 3, 2021 at 14:48
  • $\begingroup$ @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. $\endgroup$ Jan 3, 2021 at 15:04
  • 1
    $\begingroup$ $2B(s)$ should be $B(2s)$. $\endgroup$
    – passerby51
    Jan 3, 2021 at 15:08
  • $\begingroup$ @passerby51 Now I tried to test the third assumption using general increment. Ant it looks bad :| . $\endgroup$ Jan 3, 2021 at 15:18
  • $\begingroup$ @passerby51 I also tried to test the third assumption. $\endgroup$ Jan 3, 2021 at 15:31

1 Answer 1


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?

  • $\begingroup$ 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? $\endgroup$ Jan 3, 2021 at 21:06
  • $\begingroup$ 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. $\endgroup$
    – passerby51
    Jan 3, 2021 at 21:27

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