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Is the reciprocal of a Wiener-process is well defined?

More generally, does Stochastic Calculus work with such „reciprocal processess” of the form $\left(\frac{1}{X}\right)_{t}$, where the denominator can be zero with positive probability? If yes, how are they usually managed?

E.g. the following integral process is „correct”? $$\int_{0}^{T}\frac{1}{1-t}dB_{t},\;\;\;T\in\left[0,1\right]$$

(I wrote $T\in\left[0,1\right]$ instead of $T\in\left[0,1\right)$ intentionally.)

Kapes Mate
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    If $X$ is normally distributed then $\frac1 X$ is well defined with probability $1$. However, if $(X_t)$ is the Wiener process then there is no single null set outside which $\frac 1 {X_t}$ is defined for all $t$. – Kavi Rama Murthy Oct 12 '23 at 08:01

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As mentioned here Proof that a standard Brownian motion visits zero infinitely often at the beginning and Zero set of Brownian Motion is uncountable we have

Let $B_t, t\ge0$ be a standard Brownian motion that starts at 0. Prove that, for any $\epsilon > 0$, $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=1.$$

So given a realization $B_{t}(\omega)$ we have that it keeps hitting zero uncountably many times. This is true for almost every realization.

Usually the way to define reciprocal is via the use of the Bessel process i.e. conditioning the Wiener process to avoid the x-axis.

Thomas Kojar
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