Stochastic Integration with respect to Cauchy Process? I'm interested in a one-dimensional stochastic process:
$$dX_t = f(X_t)dt + g(X_t) dZ_t$$
where $Z_t$ is a Cauchy process and $f,g$ are nice polynomials (I'm looking at an ODE that gets perturbed by noise but where the noise has large tails). $Z_t$ has stationary, independent increments with the Cauchy distribution:
$$P\left(Z_{t+s} - Z_s \in dx \right) = \frac{t}{\pi(x^2+t^2)}dx$$
It is known that $Z_t$ is a Levy Process and hence a semimartingale, so I can potentially use Ito's Lemma to prove nice things about the process $X_t$.
A few questions:


*

*Is it possible to compute $E\left[\int_0^t g(X_s)dZ_s\right]$? It would be nice if this is $0$ but since $Z_t$ follows a Cauchy distribution with undefined expectation, I doubt it will be $0$

*Is it possible to compute the quadratic variation of $[dZ,dZ]_t$?

*Since perhaps the tails may be too large, perhaps it may be better to use a distribution with finite first and second moments. I believe this one may work: $\frac{t\sqrt2}{\pi (x^4+t^4)} dx$. Thus I'd still have "large" tails but with a few finite moments. Is it possible to compute the expectation of a stochastic integral w.r.t. this process and its quadratic variation? (Does this distribution have a name?)
 A: First, you need to read a book on Lévy processes. 
Note that you should write $g(X_{t-})dZ_t$ unless you are considering left-continuous processes (which you shouldn't do normally). 


*

*The expectation will always(?) be undefined unless the integrand is identically zero. The reason, heuristically, is that $$E[\int_0^t g(X_{s-}) dZ_s] = \int_0^t E[g(X_{s-}) dZ_s] = \int_0^t E[g(X_{s-}) E[dZ_s]]$$ 
since $g(X_{s-})$ is $\mathcal F_{s-}$-measurable, while $dZ_s$ is independent of $\mathcal F_{s-}$. But the inner expectation is undefined.

*Yes, it is possible. It will be equal to the sum of squares of its jumps. This does not sound too exciting, but it becomes more interesting if you consider the quadratic variation (on interval $(0,t)$) as a process. Then this is an increasing Lévy process (subordinator), which is $1/2$-stable.

*This won't work, since this density won't define a Lévy process. But there are many other possibilities. For instance, you can take a stable  Lévy motion with stability index $\alpha>1$. Then it will have a finite expectation but infinite variance. You'll find more examples in books.
