# Uniform boundedness of this SDE? And possibly a stochastic Grönwall inequality?

I have a question on Lawler – Notes on the Bessel process, on page 4. Let $$X_t$$ be one-dimensional Brownian motion, and we want to use $$N_t$$ as a measure-changing (local) martingale, defined as $$N_t=\left(\frac{X_t}{X_0}\right)^a\exp\left(-\frac{a(a-1)}{2}\int_0^t\frac{ds}{X_s^2}\right),\quad t

up to the first visit to zero. It satisfies the SDE $$dN_t=\frac{a}{X_t}N_tdX_t,\quad N_0=1.$$

Then it is supposed in the note that $$0<\varepsilon, and $$\tau=T_{\varepsilon}\wedge T_R$$, as up to the first visit to $$\varepsilon$$ or $$R$$. Then it claims that $$N_s$$ stopped by $$\tau$$, $$dN_{t\wedge\tau}=\frac{a}{X_{t\wedge\tau}}1(t<\tau)N_{t\wedge\tau}dX_t,$$

is uniformly bounded. But why is this true?

I have a rough thought that $$N_{t\wedge\tau}=N_0+\int_0^t\frac{a}{X_{s\wedge\tau}}N_{s\wedge\tau}dX_{s\wedge\tau}\leq N_0+\frac{a}{\varepsilon}\int_0^tN_{s\wedge\tau}dX_{s\wedge\tau}$$, a.s. as $$t<\tau$$. Then if we have the stochastic version of Grönwall's inequality, i.e. if we can read $$N_{t\wedge\tau}\leq N_0\exp(\frac{a}{\varepsilon}X_{t\wedge\tau})$$, a.s., then it is uniformly bounded for sure. But I am quite pessimistic about whether this holds or not.

On the time interval $$[0,\tau]$$, the exponential factor in the definition of $$N_t$$ is bounded below by $$0$$ and above by $$\exp(Kt)$$ for some constant $$K=K(\epsilon,R)\ge 0$$. Therefore $$0. This is "uniform" provided you ignore the dependence on $$t$$.
This isn't quite the "stochastic Gronwall" you're looking for, but starting with the equation: $$N_{t\wedge{\tau}}=1+a\int_{0}^{t\wedge{\tau}}\frac{N_{s}}{X_{s}}dX_{s}$$ Taking the expectation of the square of both sides (okay, technically you need to truncate first and later take a limit, but I trust you can fill in the gaps) we get: $$\mathbb{E}N_{t\wedge{\tau}}^{2}\leq{1+a^{2}\int_{0}^{t\wedge{\tau}}\mathbb{E}\big(\frac{N_{s\wedge{\tau}}^{2}}{X_{s\wedge{\tau}}^{2}}\big)ds}\leq{1+\frac{a^{2}}{\varepsilon^{2}}\int_{0}^{t}\mathbb{E}N_{s\wedge{\tau}}^{2}ds}$$ Applying the usual Gronwall inequality to the function $$f(s)=\mathbb{E}N_{s\wedge{\tau}}^{2}$$, we get that the local martingale $$(N_{t\wedge{\tau}})_{t\geq{0}}$$ is square integrable. This is far from the almost sure bound you want, but it's something.