- #1
economicsnerd
- 269
- 24
I'm given a probability measure ##\mathbb P## on ##\Omega = \{f\in C([0,1],\mathbb R): \enspace f(0)=0\}## and told that ##\mathbb P## satisfies i.i.d. increments.
I'm interested in the weakest additional conditions that will ensure that ##\mathbb P## describes a Brownian motion, i.e. that there is some ##\mu\in\mathbb R## and ##\sigma\in\mathbb R_+## such that ##\mathbb P## is the law of ##X## as described by ##dX_t = \mu dt + \sigma dZ_t## for a standard Brownian motion ##Z_t##.
Does it already follow from the assumptions of i.i.d. increments and continuity? It seems like I should be fine (by CLT) as long as increments have well-defined mean and finite variance. Do these properties come for free? If not, can I get away with assuming less? For instance, is it enough to only assume finite variance?
I'm interested in the weakest additional conditions that will ensure that ##\mathbb P## describes a Brownian motion, i.e. that there is some ##\mu\in\mathbb R## and ##\sigma\in\mathbb R_+## such that ##\mathbb P## is the law of ##X## as described by ##dX_t = \mu dt + \sigma dZ_t## for a standard Brownian motion ##Z_t##.
Does it already follow from the assumptions of i.i.d. increments and continuity? It seems like I should be fine (by CLT) as long as increments have well-defined mean and finite variance. Do these properties come for free? If not, can I get away with assuming less? For instance, is it enough to only assume finite variance?