I don't understand this one small point about Brownian motion...
Here's the question, my answer and the part I always miss out...
Let be a standard Brownian motion. Show that.
(a) is a standard Brownian motion, where
Question is easy enough. Can show that almost surely and get that for , is distributed with mean and variance and is independent of . (Not ENTIRELY sure why this is though)
Then comes this small statement that I don't know how to show (and hence don't know if it's true).
has continuous paths.
How do I show that? It's only ever stated in answers so I don't need some in depth proof, just a simple, it has continuous paths because...
The reason for the law is because the normal distribution is symmetric. The independence follows rather easily if you use the weak definition (see below), that is for any continuous bounded function f and you have that (by independent increments of BM)
.
Now if you consider any f that is bounded and continuous then is bounded and continuous hence
.
A rather long and boring way of proving what your intuition tells you.