New answers tagged probability
1
A Lévy process is defined as (Lévy process and Stochastic Calculus, David Applebaum):
Suppose that we are given a probability space $(\Omega, \mathcal{F}, P)$. A Lévy process $X = (X (t), t \geq 0)$ taking values in $\mathbb{R}^d$ is essentially a
stochastic process having stationary and independent increments; we
always assume that $X (0) = 0$ with ...
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 ...
5
For part 1 of your question, the short answer is no, calculating conditional density is a looong way of doing it. Possible but not the easiest. Here is the sketch for a shorter version. We note that $(X_{T/2},X_{T})$ is a jointly Gaussian vector with mean $\mu = (X_0 + aT/2,X_0 + aT)$, and the variance-covariance matrix
$$
\begin{pmatrix}
b^2 T/2 & b^2 T/...
Top 50 recent answers are included
