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I am currently reading about stochastic processes and Brownian Motion.

When books have notation such as $E[X_t] = 0$ and $Var[X_t] = \sqrt{t}$ this is considered over sample paths.

However, when we consider the distribution of the increment $X_t - X_s \sim \mathcal{N}(0,|t-s|)$ what random variable is this exactly?

Is it:

  1. The increment at a fixed time over possible paths i.e. the set $ [ X(t,\omega_k) - X(s,\omega_k) ]_k $ for fixed $t,s$
  2. The increment over a particular path, but varying the times the increments are taken i.e. the set $ [X(t,\omega) - X(t+\Delta,\omega) ]_{\Delta}$ for fixed $\Delta,\omega$

Does one of these imply the other perhaps? I have tested it empirically and it seems to be true in both cases.

NavStoke
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  • Part of the definition of a process is that for each finite vector of "time" indices $(t_1,t_2,\ldots, t_d),$ the collection $(X_{t_1}, \ldots, X_{t_d})$ is a $d$-variate random variable. Apply this to the case $d=2.$ – whuber Apr 03 '23 at 16:02
  • Can you provide a bit more detail please? – NavStoke Apr 03 '23 at 16:05
  • For the definitions see https://stats.stackexchange.com/a/160733/919. For more, please search our site. – whuber Apr 03 '23 at 16:33
  • Okay, so to quote from the linked post:

    "If you fix one $t \z in T$, and write the particular value $X(\omega)(t)$ on each ticket, you have--obviously--a random variable. Its name is $X_t$."

    Based on this quote, I would say that it is statement 1. that is the correct interpretation of the distribution of increments?

     – NavStoke Apr 03 '23 at 18:13
  • In the original notation, $t$ and $s$ are arbitrary but fixed times, so (2) cannot be correct. (1) is a little strange in its double use of undefined and apparently superfluous subscripts "$k,$" but looks like the right idea. – whuber Apr 03 '23 at 18:59

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