# Integrating the inverse of a squared bessel process - integrability

Let $X_t$ be a 4-dimension Squared Bessel Process (BESQ-4). Let $M_t$ be a continuous true martingale. Question:

Does $\int_0^t \frac{1}{X_s}dH_s$ exist?

If so, is it only a local or a true martingale? If fear, that it is not even integrable, here is why:

$1/X$ is integrable with respect to $M$, iff $\mathbb{E}\left[\int_0^t \frac{1}{X^2_s}d[M]_s\right]<\infty$ for all $t>0$ (see e.g. Thm 2 in http://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/). Taking e.g. $M_t$ as Brownian motion we have $[M]_t = t$. Now it is known that the second inverse moment of a BESQ-4-process is infinite. Using Fubini-Tonelli, I'd say this doesn't work. Wrong?

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The moment generating function of $\int_0^t dt/X_t^2$ where $X_t$ is a Bessel process is known and analytic for a certain set of parameters, see Craddock and Lennox 2009. You might want to check that paper, it has some very useful information. – Mr_3_7 Oct 9 '15 at 9:17