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I think it's differentiable (everywhere). Interchanging operations freely, we have \begin{eqnarray*} \newcommand{\D}{\frac{\mathrm d}{{\mathrm d}x}} \newcommand{\E}{\mathbb E} V'(x) &=& \D \E_x \int_0 …

As long as $\sigma\ne 0$, we get $P(X_t=x)=0$. By the Levy-Ito decomposition there is a Brownian motion $Y$ and an independent process $Z$ with $X=Y+Z$. Then for $t>0$, $$ P(X_t=x)=P(Y_t=x-Z_t)=0 $$ …

Yes, because since $ P(\xi=0)=0$, immediately after time 0 the process is just the deterministic identity function $\xi_t=t $.

No, consider Brownian motion $W_t$ and $$M_t=\frac{W_t^2-t}{2},$$ $$N_t = -M_t.$$ Source: slides by David Heath page 5.

Suppose the $\mu$ and $\sigma$ are constants, i.e., there are constants $\mu_X$, $\mu_Y$, $\sigma_X$, $\sigma_Y$ such that for all $t$, $$ \mu_{X_t} = \mu_X,\qquad \sigma_{X_t} = \sigma_X, $$ $$ \mu_{ …

Shreve, Stochastic Calculus for Finance, volumes 1 and 2.

Karatzas and Shreve Brownian Motion and Stochastic Calculus (Graduate Texts in Mathematics) (Volume 113) http://www.amazon.com/gp/aw/d/0387976558

Let $W_t$ be 1-dimensional Brownian motion and let $$V_t=W_t+\sum_{n\in\mathbb N,\, n\le t} W_n$$ Then \begin{eqnarray} \Delta V_n&=&W_n,\quad\text{whereas}\\ V_{n-}&=&\sum_{m\in\mathbb N,\, m\le n} W …

Assuming there are no arbitrage opportunities, the price of a derivative depends on the prices of other derivatives available in the market. If you introduce a derivative without giving it a price, or …

One way to improve intuition is to work out a couple of Discrete versions of Ito's lemma Øksendal (6th edition) Example 3.1.9: almost surely, $$ B_t^2 - t = \int_0^t 2B_s dB_s $$ This has …

They're the same, $\mathcal G_t=\mathcal F_t$. Indeed, suppose $A\in\mathcal G_t$. So in particular $A\in\bigcap_{n=1}^\infty(\mathcal F_{t+1/n}\vee\mathcal N)$. Note that for any $\sigma$-algebra $ …

A counterexample is to let $X_t$ be Brownian motion with drift. Start at any point $x$ and suppose the drift is negative. Let $N_y$ be the event that $y$ is never hit, i.e., $N_y=\{(\forall t)\, X_t < …

The process $X$ you mention is uniformly continuous on the rationals* in the compact interval $[0,n]$, with probability 1. So you define Brownian motion $B$ to be the unique continuous extension of: $ …