# Tag Info

6

Using the distribution and independence of increments allows to prove $L^2$ (mean-square) continuity. Proving the a.s. continuity is much harder. Paul Lévy's construction of Brownian motion is related in Le Gall; an alternative is to construct the Brownian motion through Haar wavelet functions or Fourier series.

6

The theory behind the actual reasoning is a bit complicated than the coverage in Hull's, but staying within the simple reasoning, the difference comes down to the following: The Brownian increments over the interval $dt$ are normally distributed with mean zero and variance $dt$, so in terms of distribution, you can express the increments in terms of a ...

4

I happen to agree with your solution $$\mathrm{d}(tB_t)=B_t\mathrm{d}t+t\mathrm{d}B_t.$$ You can either apply Ito's Lemma to $f(t,x)=tx$ as you did or apply the product rule,$$\mathrm{d}X_tY_t=X_t\mathrm{d}Y_t+Y_t\mathrm{d}X_t+\mathrm{d}X_t\mathrm{d}Y_t,$$ where, of course, $\mathrm{d}B_t\mathrm{d}t=0$. There is presumably an error in your book/homework ...

3

It's part of the definition I'd just like to re-iterate the comment by Kevin, (which as far as I can tell is the answer). There are three properties which define a standard Brownian motion / Wiener process: Independent increments. Normally distributed with variance equal to the time increment. The path is continuous. Which hopefully any "standard"...

3

I think the question also brings up a common confusion with notation. I think it is incredibly unfortunate to use notation such as $dW(t)$ (unless it's part of a stochastic integral), and I get upset when I see it being used in textbooks. The definition of Brownian Motion is implicit and goes like this: (i) $W(t=0) = 0$ (ii) $W(t)$ is (almost surely) ...

2

Thought to add this as a comment, but it appears too long. Your question does not appear complete, that is, the rationale for using two Brownian motions is not clear. Note that \begin{align*} dX_t &= \mu dt + \sigma_1 dW^1_t + \sigma_2 dW^2_t \\ &=\mu dt + \sqrt{\sigma_1^2 + \sigma_2^2 + 2 \rho \sigma_1\sigma_2}\frac{\sigma_1 dW^1_t + \sigma_2 dW^...

1

I try to clarify. First, let us be under the real world probability measure $\mathbb{P}$. Say that the process $Y_t$ is a $\mathbb{P}$ standard Brownian motion. Then the following is true by definition (assume wlg. that s < t): $(Y_t - Y_s) \sim N(0, t-s)$ Now, using your definition of $W_t$: \$\mathbb{E} \left[(W_t - W_s)\right] = \mathbb{E} \left[(B_t ...

Only top voted, non community-wiki answers of a minimum length are eligible