New answers tagged stochastic-calculus
2
For Itô Processes $dX(t) = \mu(t) \mathrm{d}t + \sigma(t) \mathrm{d}W(t)$ you have the result that (under appropriate assumptions which ensure that the local martingale is a martingale, e.g. $E( (\int \sigma(t)^2 \mathrm{d}t )^{1/2} ) < \infty$, etc.): $X$ is a martingale $\Leftrightarrow$ $\mu(t) = 0$.
So in order to check if a process $X$ is a ...
0
Rather simply and generally when you take the stochastic differential of a process and get no drift term but simply an ito integral, then this process is a martingale.
From memory that's how you retrieve some pde equations whose solutions lead to martingale (take the differential, look at the dt partial differentials term, then look for solution that would ...
1
i picked this off from Shreve.
Start with the definition of sampled quadratic variation: (1) $\frac{1}{2}Q_\pi = \frac{1}{2}\sum\nolimits_{j=0}^{n-1} (W_{j+1}) - W_j)) ^2$ where $\pi$ = {0,1,2...,n} is a partition of $[0,T]$ (Note we took $\frac{1}{2}$ of both sides for reasons that will be clear in the next line.) Now we know (1) is equal to ...
2
"Like" Ito:
$$d (B^2) = B dB + B dB + dB dB$$
That is
$$B dB = \frac{1}{2} d (B^2) - \frac{1}{2} dB dB$$
Integrate. Last term is 1/2 the quadratic variation.
I understand the questions as follows: In iii) one has to define what $dB dB$ stands for and one has to "proof" the first line in my answer. In ii) one may use Ito to "know" that $dB dB = dt$.
Top 50 recent answers are included
