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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:

  1. 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 $0$

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

  3. 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?)

Brenton
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1 Answers1

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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).

  1. 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.

  2. 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.

  3. 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.

zhoraster
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