I'm starting / dabbling with equalities of random variables and I'm seeking for help. I haven't study deeply measure theory and probability theory so it is hard for me to go deeper in proofs by myself.
My problem is the following. I'm still not able to grasp the whole concept of sample path of a random variable.
I get the idea that there is a difference between :
1. "the distributional identity :" $$ B_t- B_s \overset{d}{=} B_{t-s}$$ which means both sides share a common distribution.
2. "pathwise identity : " $$\forall \omega \in \Omega \colon B_t (\omega) - B_s (\omega ) = B_{t-s} (\omega) $$
but I don't get whether one is stronger than the other. Is one implying the other ?
I think that 1 does not imply 2 : because you can have the same distribution while not beeing equal as random variable ( example of the accepted answer here : What is a sample path of a stochastic process)
And for 2 implies 1 I think it is true. Is my intuition true ?
If this is true, can you give me a non trivial example where (1) implies (2) ? Or a sufficient condition ?
Finally, my last point is about understanding why (1) does not imply (2) in the case of a Brownian motion (cf. Mikosch ). If one can explain this to me I would be so grateful.
Thank you.
Notice that if $X(\omega )=Y(\omega )$ almost surely, then of course, $X\overset{d}{=}Y$. The converse doesn't hold since $X$ and $Y$ doesn't necessarily leave on the same probability space. Moreover to say that $X(\omega )=Y(\omega )$ for all $\omega \in \Omega $ is a very strong assumption. We normally rather have the equality almost surely.
Now, $B_t(\omega )-B_s(\omega )=B_{t-s}(\omega )$ will normally never hold. Simply because $B_{t-s}$ is $\mathcal F_{t-s}$ measurable (where $\mathcal F_t=\sigma (X_t\mid s\leq t)$ is the natural filtration of $(B_t)$), whereas $B_t-B_s$ is not.