Hot answers tagged brownian-motion
2
This remark is a good reason to avoid assuming that $B_0(\omega)=0$ for all $\omega$ in $\Omega$. Instead, one usually defines different measures $\mathbb P^x$ on a common probability space $\Omega$, and each $\mathbb P^x$ sees a different part of $\Omega$.
For example, one can choose for $\Omega$ the space of continuous functions $[0,+\infty)\to\mathbb ...
2
Anyone knows what should I do next?
An approach is to condition on $B_{1/2}$. Slightly more generally, fix some $s\gt0$ and let
$$
A=\left[\min\limits_{s\leqslant t\leqslant2s}B_t\gt0\right],
$$
then $P[A\mid B_s=x]=0$ for every $x\leqslant0$ and $P[A\mid B_s=x]=P[T_x\gt s]$ for every $x\gt0$, where, for every $x$, $T_x=\min\{t\geqslant0\mid B_t=x\}$. ...
1
$W$ is a pair of independent Brownian motions, but $B$ is not. The trick is just to assume that $B_1$ is a certain Brownian motion (e.g. $W_1$) and that $B_2$ is a linear combination of $B_1$ and a Brownian motion that is independent from $B_1$, e.g. $W_2$. Coefficients of this linear combination are computed based on the correlation condition $\mathrm dB_1 ...
1
The Brownian motion has independent increments, it's always a good idea to use this fact if you have to prove a process a martingale.
$$\begin{align*} \mathbb{E}(B_s^4 \mid [B_r]_{[r \leq t]}) &= \mathbb{E}((B_s-B_t+B_t)^4 \mid [B_r]_{r \leq t}) \\ &= \mathbb{E}((B_s-B_t)^4 \mid [B_r]_{[r \leq t]}) + 4 B_t \mathbb{E}((B_s-B_t)^3 \mid [B_r]_{[r \leq ...
1
Hint By definition, we have
$$\phi(X_t) = \exp(-2\mu \cdot (x + B_t + \mu \cdot t))$$
Hence
$$\mathbb{E}(\phi(X_t) \mid \mathcal{F}_s) = \exp(-2\mu \cdot (x +\mu \cdot s +B_s)) \cdot \mathbb{E} \left( \exp( -2 \mu \cdot (B_t-B_s)) \mid \mathcal{F}_s \right)$$
for all $t \geq s$. Use the independence of the increments and the knowledge about exponential ...
1
Let $f_{t}(x) = \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}$ be the $N(0,t)$ density, and let $f_{s,t}$ be the joint density of $(X_s, X_t)$. If $s<t$, then since the bijection
$(X_s,X_t - X_s) \mapsto (X_s,X_t)$ has inverse $(u,v) \mapsto (u, v-u)$ with jacobian determinant $1$,
$$f_{s,t}(x,y) = f_{s,t-s}(x,y-x) = f_s(x)f_{t-s}(y-x).$$
Then
$$
f_{t|s}(x|y)
...
1
There is no need to delve into the specifics of gaussian densities here since, when restricted to some gaussian family, conditionings amount to linear algebra computations (one reason of the success of gaussian models in appplications).
In the present case, the triplet $(X_t,X_{t_1},X_{t_2})$ is centered normal hence, conditionally on $(X_{t_1},X_{t_2})$, ...
1
I think what you're missing here is Levy's characterisation of Brownian motion.
If $X_t$ is a continuous local martingale and $X^2_t - t$ is also a local martingale then $X_t$ is a Brownian motion.
I'll explain a bit more.
Let $(\Omega,\mathcal F_t, P)$ be some filtered space with $X_\circ:\Omega\times\mathbb R\to\mathbb R$
and let $\left(C^0,\mathscr ...
1
"$\Rightarrow$" Let $h>0$ arbritrary, then clearly
$$\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta) = \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D \leq h)}_{=:p_1} + \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D > h)}_{=:p_2}$$
We have $$\begin{align*} p_1 &\leq \mathbb{P}^w \left( \sup_{t \leq h} |B_t-z| \geq \delta ...
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