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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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $t+s$.

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  • $\begingroup$ Sorry but this is most confusing. $\endgroup$ – Did May 11 '16 at 12:11
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    $\begingroup$ 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

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