I would appreciate it if someone could clear up the martingale problem: The SDE
## X_t = X_0 + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)dB_s ##
as I understand it, is NOT a martingale wrt ## \{ \mathcal{F}_t:t\in [0,T] \} ## unless the drift term ## b(t,x) ## is zero. In Theorem 10.2.2...
Sorry, I was wrong in two spots; Klöden-Platen does not use Doobs martingale inequality, but rather the 'Markov inequality':
## P(|X| > a) \leq \frac{1}{a^r}E[|X|^r] ##
for ##a,r > 0 ##. And thus I don't need to show that ##X_t^n## is a martingale. Moreover, Kuo uses 'Doobs submartingale...
I'm currently working my way through the existence theorem of strong solutions for the stochastic differential equation
## X_t = X_0 + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)Bs ##,
Where ## \int_0^t \sigma(s,X_s)Bs ## is the Ito integral. The assumptions are:
1: ## b,\sigma ## are jointly...
A fellow student of mine asked a question to our teacher in functional analysis, and the answer we got was not very satisfactory. In our discussion on Banach spaces the student asked "Why is it interesting/important for a normed space to be complete?". To my surprise the teacher said something...
Hmmm.. When is it true that A \subseteq B \Rightarrow \bar{A} \subseteq \bar{B} ?
I ask because I've been working on a similar problem: Prove that \bar{\ell}0 = \ell2 in \ell2.
With a similar approach of the attempt above I got \bar{\ell}0 \supseteq \ell2.
Since \ell0 \subseteq \ell2...
Hi there!
I'm working on a random walk-problem my professor gave me.
Given a MC with the following properties:
P0,0 = 1 - p
P0,1 = p
Pi,i-1 = 1 - pi
Pi,i+1 = pi
i \in[1,∞)
Xn: state at step number n
The chain is irreducible and ergodic and 0 < p ≤ 1
introducing N:
N = {min n...
Ok, this should be quite simple. I've been looking at this problem for quite some time now, and I'm tired.. Please help me!
The equation to solve is r' = r(1-r^2)
The obvious thing to do is to do partial fraction expansion and integrate(from r_{0} to r):
∫ (\frac{1}{r} +...
I'm trying to prove eA eB = eA + B using the power series expansion eXt = \sum_{n=0}^{\infty}Xntn/n!
and so
eA eB = \sum_{n=0}^{\infty}An/n! \sum_{n=0}^{\infty}Bn/n!
I think the binomial theorem is the way to go: (x + y)n = \displaystyle \binom{n}{k} xn - k yk = \displaystyle...