I am quite new to discrete and continuous stochastic processes. It seems there is something I don`t understand about definition of Brownian motion.

Let $\Omega, \mathcal{F}, \mathbb{P}$ be a probability space and $B_t$ be a standard continuous Brownian motion $B_t: \Omega \rightarrow C(\mathbb{R}^+, \mathbb{R})$.

Then the definition requires that $B_t \sim N(0,t)$ and $B_t - B_s \sim B_{t-s} \sim N(0,t-s)$. I don`t see how the first condition is compatible with the last. We know that $B_t \sim N(0,t)$ and $B_s \sim N(0,s)$ so $B_t - B_s \sim N(0,t+s)$ (seen as a sum of two normally distributed random variables) which seems to be in contradiction with independent increments. I guess I am missing something very basic but since I am new to this topic I cant see it. Thanks

update: based on answers below, yes it was indeed very basic, $B_t$ and $B_s$ are not independent as random variables so that is why the usual rule for sum of independent normally distributed random variables dont apply

  • 1
    $\begingroup$ I think you mean $t-s$ in the variance of $B_{t} - B_{s}$. In general, the variance of a sum is not the sum of the variances unless the variables are uncorrelated. $\endgroup$ – Tom Hallward May 11 '16 at 11:37
  • 1
    $\begingroup$ The catch is that $(B_t,B_s)$ is not independent, only $(B_t-B_s,B_s)$ is, hence your argument "variance of the sum equals sum of the variances" fails. Note that $B_t=B_s+(B_t-B_s)$ and that $t=s+(t-s)$.. $\endgroup$ – Did May 11 '16 at 12:12
  • $\begingroup$ Okay now from here with independence we can get that $B_t \sim B_s + (B_t - B_s) \sim N(0,s + t -s) \sim N(0,t)$ $\endgroup$ – Sina May 11 '16 at 12:17
  • $\begingroup$ Moreover, your first tilde is exactly an equals! $\endgroup$ – jdods May 11 '16 at 12:43

Answer: The values of a Brownian motion path at different times are not independent. The variance of $B_t-B_s$ may at first seem like it should be $t+s$, but we can think of its variance being reduced due to $B_t$'s dependence on $B_s$.

Elaboration: We are only looking at a single Brownian motion process here, $B$. The value of $B$ at time $t$ is exactly its value at time $s$ plus some random fluctuation: $B_t(\omega)=B_s(\omega) + $ "the change in value from $s$ to $t$ for path realization $B(\omega)$". The part in quotes is declared to be independent from $B_s$ but also to be normally distributed with mean zero and variance equal to the length of time it covers: $t-s.$

Of course $B_t=B_s+(B_t-B_s)$ with $B_s$ and $B_t-B_s$ being independent. Given that $\text{Var}(B_t)=t$ and $\text{Var}(B_s)=s$, we can deduce that $\text{Var}(B_t-B_s)=t-s$.

If we were allowed to plug in distinct $\omega$'s to evaluate $B_t$ and $B_s$ independently, then the variance would be $t+s$.

  • $\begingroup$ Sorry but this is most confusing. $\endgroup$ – Did May 11 '16 at 12:11
  • 1
    $\begingroup$ The answer is confusing but correctly pointed the mistake I made about independence so I will vote it as the correct answer. $\endgroup$ – Sina May 11 '16 at 12:14
  • $\begingroup$ Glad to be of service! I will clean up the answer to make it better. $\endgroup$ – jdods May 11 '16 at 12:18
  • $\begingroup$ No problem I kind of get what you mean would be better just to write $B_t = B_s + B_{t-s}$ which tells that $B_t$ can not be independent of $B_s$ since $B_s$ is independent of $B_{t-s}$. $\endgroup$ – Sina May 11 '16 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.