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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 ...
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/...