Tag Info

New answers tagged


0/ Let's me use more common notations to avoid misunderstanding. We will consider $B_t^x$ and $B_t^y$ - two correlated Brownian motions, e.g. $<dB_t^x,dB_t^y>=\rho dt$. Just to recall, Ito's process: $$X_t = X_0 + \int_0^t \mu(s,\omega) ds + \int_0^t \sigma(s,\omega) dB_s^x\\ dX_t=\mu(t,\omega) dt + \sigma(t,\omega) dB_t^x$$ 1/ Single BMs: ...


For your first question, your derivative is incorrect. It instead is $\frac{\partial C^2}{\partial x \partial y} = 1+\theta(1-2x-2y+4xy)$. Note also that $x+y-2xy \geq x^2 + y^2 -2xy = (x-y)^2 \geq 0$. That is, $1-2x-2y+4xy \leq 1$. On the other hand, $1-2x-2y+4xy = 2(1-x)(1-y)+2xy - 1 \geq -1$. Then, $\frac{\partial C^2}{\partial x \partial y} \geq 0$, for ...

Top 50 recent answers are included