$B(t)$ brownian motion, $[Y,Y](t)$ is qudratic variation,prove : 1. $[Y,Y](\infty)<\infty$ 2. for $\omega$ with $\int_0^\infty a(s)^2\, ds=\infty$,

Question

$$ \int_0^t a(s)^2\,ds<\infty, \text{ a.s.} $$

$$ X(t)=\int_0^t a(s)\,dB(s) $$

$$ Y(t)=e^{\int_0^t a(s)\,dB(s)-1/2\cdot\int_0^t a(s)^2\,ds} $$

prove :

1. $[Y,Y](\infty)<\infty$

2. for $\omega$ with $\int_0^\infty a(s)^2 \, ds=\infty$,

$$ Y(t,\omega) \rightarrow 0 $$

i.e $$ \left\{ \omega:\int_0^a a(s,\omega)^2 \, ds=\infty \right\}=\left\{\omega:Y(t,\omega) \rightarrow 0 \right\}, \text{ a.s.} $$

my solution:

Since $Y(t)$ is the stochastic exponential weak solution,

$$X(0)=0,$$

$$ [X,X](t)=a(t)^2\cdot t$$

$$Y(t)=e^{X(t)-X(0)-1/2*[X,X](t)}=e^{X(t)-\frac 1 2 [X,X](t)}$$

$$dY(t)=Y(t)\,dX(t),$$

$$[Y,Y](t)=e^{2X(t)-[X,X](t)}\cdot[X,X](t)$$

$$ =e^{2\int_0^t a(s)\,dB(s)-a(t)^2\cdot t}\cdot a(t)^2\cdot t $$

by $$ \int_0^t a(s)^2\,ds<\infty, \text{ a.s.} $$ can I say $[X,X](t)$ is bounded and $X(t)$ is also bounded such that $[Y,Y](t)$ is bounded?

then what should I do? and the second one i have no clue.

Answer 1

1. Note that $X_t$ is a local martingale by the given condition on $a(t)$ and definition of $X_t$. By the SDE of $Y_t$, $Y_t$ is also a local martingale. WLOG, assume $Y$ is a bounded (by constant C) martingale. Otherwise, localize and the result still holds by taking the localizing sequence to $\infty$. By Fatou's Lemma,

$$E(\liminf\limits_{t\to \infty} [Y,Y]_t) \leq \liminf\limits_{t\to \infty} E([Y,Y]_t) = \liminf\limits_{t\to \infty} E(Y_t^2) \leq C^2 < \infty$$

This shows that $\liminf\limits_{t\to \infty} [Y,Y]_t = \lim\limits_{t\to \infty} [Y,Y]_t =: [Y,Y]_\infty < \infty$ a.s., where the first equality is because $[Y,Y]_t$ is nondecreasing.

2. Since $[X,X]_t = \int_0^t a(s)^2 ds \to \infty$ a.s., by time-change formula, $X_t = B_{[X,X]_t}$, where $B_s$ is some Brownian motion. We use the fact that $$\lim\limits_{t\to \infty} \frac{B_t}{t} = 0$$ Then, $$Y_t = \exp(X_t - \frac{1}{2}[X,X]_t) = \exp([X,X]_t( \frac{B_{[X,X]_t}}{[X,X]_t} - \frac{1}{2})) \to 0$$ since $[X,X]_t \to \infty$ and $\frac{B_{[X,X]_t}}{[X,X]_t} - \frac{1}{2}$ will be negative for sufficiently large t.