1
$\begingroup$

A one-dimensional continuous-time stochastic process is formalised as a bivariate function taking the first argument from $[0,\infty)$ (time) and the second from a probability space $\Omega$ (trajectory), or $$X:[0,\infty)\times\Omega\ni (t,\omega)\mapsto X(t,\omega)\in\Bbb R.$$ From this definition, for any $t$, $X(t,\omega)$ should be a r.v. taking $\omega\in\Omega$ as the argument. Thus the initial value $X(0,\omega)$, generally speaking, should also be random, i.e., may not be constant.

However, sometimes we do see stochastic processes with given initial value, e.g., a Brownian motion $\{B_t\}_{t\ge 0}$ with $B_0=0$, and I wonder what they mean. My understanding is that this should be similar to conditional probabilities, so under the restriction $B_0=0$, the original probability space $\Omega$ is restricted to $\Omega_0$ consisting of all trajectories starting from position $0$, and the new probability measure on $\Omega_0$ should be defined via conditional probability. Also, the original filter (if any) $\{F_t\}$ should be restricted to $\{F_t\cap \Omega_0\}$.

Am I correct so far? If yes, then, for example, how to formally define a Brownian motion $\{B_t\}$ with $B_0=0$? In particular, I'm having trouble with the conditional probability part.


Motivation: I got this confusion from reading "translation variance" of Brownian motions on page 302 of Durrett's Probability: Theory and Examples (ed 4.1), which states

If $\{B_t\}$ is a Brownian motion, then $\{B_t-B_0,t\ge 0\}$ has the same distribtion as a Brownian motion with $B_0=0$.

  • 0
    Constants **are** random variables, so why the fuss?2017-02-15
  • 0
    @Did This I understand. But please read on my question: what I'm curious about is what's happening when we *restrict* the initial values, from a (possibly) wider range, to a constant.2017-02-15
  • 0
    It really depends a bit on what you are "given"; for Brownian motion for instance I usually think of being given a set of functions $B(t)$ all of which satisfy $B(0)=0$, and then being given a measure on those. Alternately we could have a measure on the continuous functions which happens to be concentrated on functions which vanish at $0$. Anyway, in other situations it could make more sense to think of conditioning on the initial value, but that's not any unusual type of conditioning (except perhaps for the usual difficulties about conditioning on a set of measure zero).2017-02-15
  • 0
    @Ian actually I'm also wondering if it's really conventional to just force the initial value to be deterministic as part of the *definition* of a stochastic process, since, for a lot of real life models, like stock prices, temperatures, the stochastic processes modelling them are all like "trees" rooting from one single point i.e. the deterministic initial value. Unfortunately I don't see any formal discussion on this. Neither Wikipedia nor my book has given any clue whether the initial value should be deterministic or not.2017-02-15
  • 1
    Quite canonical, yes. Asking that $B_0=0$ almost surely is to choose the Dirac measure at $0$ as initial distribution.2017-02-15

0 Answers 0