Recent content by river_rat

Hi bpet The methodology I know for the brownian bridge goes as follows: first prove the Brownian bridge is a gaussian process, then find an equivalent process that is adapted to the original filtration generated by your brownian motion and that is a scaled ito integral. Then using ito's lemma...

Hi chiro The formal definition is the operator \mathcal{L} where acts on \mathcal{C}^{2, 1} test functions so that \mathcal{L} f(x, t) = \lim_{h\rightarrow 0^+} \frac{\mathbb{E}(f(X_{t+h}) | X_t = x) - f(x)}{h} . For general ito processes or levy processes it is easy to find, but for a...

I hope someone can put me on the right track here. I need to derive the infinitesimal generator for a bridged gamma process and have come a bit stuck (its for a curve following stochastic control problem - don't ask). Any tips, papers, books that could guide me out of my hole would be greatly...

Well for starters that is just a monic cubic in disguise :)

Well you could assume the contrary and first prove that S must be closed and similarly that S must be open. Now which are the only sets in a connected space with that property?

Ok firstly a sphere has zero volume (a sphere is a closed subset of R^3 and every closed subset is measurable under the normal legesgue measure). However volume is not a topological invariant (think length of (0, 1) which is homeomorphic to the entire real line) but is rather invariant under...

Ahh! Didnt notice he had changed his cubics between lines - that should sort out the problem :)

Okay, so (x-5) is a factor of your cubic. Now factorise it, and see what the other factors are. Edit: No wait - checked the other factors, are you sure your equations are correct?

Yes there is, the tensor fields as used in differential geometry are constructed by taking sections of the tensor product of copies of the tangent bundle and cotangent bundle. The tensor product is also important in that it is used as a starting point to define the exterior product of covectors.

Well check if \vec{e_i} \cdot \vec{e_j} = 0 \forall i \neq j . If that is true then your basis is orthogonal relative to that innerproduct. For R^2 a non standard inner product amounts to declaring some other angle to mean " at \frac{\pi}{2} " - so all you have done is shift your axis so...

Okay, first hint || \vec{x} + \vec{y}||^2 = ( \vec{x}+ \vec{y}, \vec{x}+ \vec{y} ) Where (\cdot, \cdot) is the inner product on your inner product space. So you should not have any square roots to worry about. Expand the inner product, then use the Cauchy-Swartz inequality.

Try the The Gram-Schmidt Algorithm - it will construct an basis orthogonal wrt your inner product given any other basis. The inner product of two vectors is a scalar (it is in a sense the angle between the two vectors) - what exactly are you trying to imagine?

Start with ||\vec{u} + \vec{v}|| - ||\vec{v}|| \le ||\vec{u}|| and choose \vec{u} + \vec{v} = \vec{a} The rest should be pretty self evident after that.