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

Question

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 $$N_t=\left(\frac{X_t}{X_0}\right)^a\exp\left(-\frac{a(a-1)}{2}\int_0^t\frac{ds}{X_s^2}\right),\quad t<T_0,$$

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 $0<\varepsilon<X_0<R$, 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.

Answer 1

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 $0<N_{t\wedge\tau}\le(R/\epsilon)^a\exp(K\cdot (t\wedge \tau))$. This is "uniform" provided you ignore the dependence on $t$.

Answer 2

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.