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Let $X_t$ be defined as follow: $$X_t:=\int_{h=0}^{h=t}W_h^{-1}dW_h$$ Is $X_t$ well defined? Can the integral be computed?

We know that for a general function $y=\frac{1}{x}$, we can integrate to $ln(x)$, so I thought about defining a function $F(W_t):=ln(W_t)$ and applying Ito's lemma to $F$, but the problem is that $W_t$ can go negative, so that makes me think that we cannot do that (i.e. log of negative number is undefined).

As another example, if I were to integrate $X_t:=\int_{h=0}^{h=t}W_h^{-2}dW_h$, I'd proceed as follows:

We have:

$$W_t=W_0+\int_0^ta(W_t,t)_{=0}dh+\int_0^tb(W_t,t)_{=1}dW_h$$

So that $W_t$ is an Ito process. Let $F(W_t):=W_t^{-1}$, then, using Ito's Lemma:

$$F(W_t)=\int_{0}^t \left( \frac{\partial F}{\partial t}_{=0}+\frac{\partial F}{\partial W_t}_{=-W_h^{-2}}*a(W_t,t)_{=0} + 0.5\frac{\partial^2 F}{\partial W_t^2}_{=2W_h^{-3}}*b(W_t,t)^2_{=1} \right)+\int_0^t\left(\frac{\partial F}{\partial W_t}_{=-W_h^{-2}}b(W_t,t)_{=1}\right)dW_h=\\=\int_0^tW_h^{-3}dh-\int_0^tW_h^{-2}$$

So that:

$$\int_0^tW_h^{-2}dW_h=\int_0^tW_h^{-3}dt-W_t^{-1}$$

The problem above is again that $W_t^{-1}$ goes to infinity, whenever $W(t)$ approaches zero, so the integrals might not be well defined in any case and I am not sure if they converge: i.e. if I can even use Ito's Lemma as above?

Are integrals of negative powers of Brownian motion well defined?

Jan Stuller
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    No, they are not, they are just well defined up to a stopping time. – Paresseux Nguyen Dec 05 '20 at 12:02
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    IMHO, concerning Ito's lemma, you can use Ito's lemma the way you do in the above post, but the result of your calculation should only serve a mean for your reasoning, rather than a demonstration step of good mathematical rigour – Paresseux Nguyen Dec 05 '20 at 12:07
  • @ParesseuxNguyen: thank you for the remarks. I'd happily award the bounty if you can show how the integral is well defined up to a stopping time, with a concrete example? – Jan Stuller Dec 05 '20 at 18:13

1 Answers1

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No, this integral is not well-defined. Recall that $\int_0^t H_s dW_s$ is only well-defined if there exist a sequence of stopping times $(\tau_n) \nearrow t$ such that $\mathbb{E}[\int_0^{\tau_n} H_s^2 ds] < \infty$, but we can show that $\int_0^t (W_s^{-1})^2ds = \infty$ almost surely for every $t>0$:

Fix $\omega \in \hat \Omega$, by the law of the iterated logarithm there exists $T = T(\omega) > 0$ such that $\frac{|W_s(\omega)|}{\sqrt{2s \log (\log (1/s))}} \le 2$ for all $s \le T$, where $\hat \Omega \subset \Omega$ is a set with probability 1 on which the law of iterated logarithm holds. Rewriting, this implies $\frac{1}{|W_s(\omega)|^2} \ge \frac 1{8 s \log(\log(1/s))}$ for all $s \le T$ and hence

\begin{align*} \int_0^t |W_s(\omega)^{-1}|^2ds \ge \frac 18 \int_0^t \frac 1{s \log(\log(1/s))} ds = \infty \end{align*}

for any $t \le T$. Since this is the integral of a non-negative function, it is non-decreasing in $t$, so this shows that we actually do have $\int_0^t |W_s(\omega)^{-1}|^2ds = \infty$ for all $t>0$ and $\omega \in \hat \Omega$. In particular, if $\tau$ is any stopping time that is non-zero with positive probability we have $\mathbb{E}[\int_0^{\tau} (W_s^{-1})^2 ds] = \infty$ and therefore $\int_0^t (W_s)^{-1}dW_s$ is not well-defined.

user6247850
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    Perfect answer, +1 – Sarvesh Ravichandran Iyer Dec 10 '20 at 03:41
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    Just wanted to double check: I suppose your result still holds for $\frac{a}{b+W_t}$ - right? – Jan Stuller Jan 07 '21 at 08:33
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    I think that one should be well-defined up to a stopping time. The issue with $b=0$ is that we are then starting at $0$ in the denominator, and the law of the iterated logarithm controls how fast we can back down to finite values. I would have to think about it a little more to give a definitive answer, though. – user6247850 Jan 08 '21 at 21:38