New answers tagged


You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assumptions. While the first may be negative, the second may not. [Interpretation 1] Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a standard ...


A process indeed depends on time $t$. However, in Ito's lemma, only derivatives with respect to independent time variable $t$ is considered. That is, for a process of the form $f(S_t, t)$, $\frac{\partial f}{\partial t}$ is the derivative with respect to the second, that is, the independent, $t$ variable, however, the parameter $t$ in the process $S_t$ is ...


1. This integral is not Ito's Integral. Indeed $Y_t$ is a random time change with time change rate $\frac{W_t}{1+W_t^2}.$ (Oksendal, Sixth edition,page 147) 2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !. Let $$f''(x)=\frac{-2x}{(1+x^2)^2}$$ then $$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$ and $$f(x)=\...


Hint Let $\,H_0(x,t)=1$ , $H_1(x,t)=x$ and for every $n\ge 2$ set $${{H}_{n}}(x,t)=x {{H}_{n -1}}(x,t)-(n-1)\,t\,{{H}_{n-2}}(x,t)$$ then ${{H}_{n }}(W_t ,t)$ is a Martingale. For exapmple $$H_1(W_t,t)=W_t$$ $$\qquad H_2(W_t,t)=W_t^2-t$$ $$\qquad\qquad H_3(W_t,t)=W_t^3-3tW_t$$ $$\vdots $$

Top 50 recent answers are included