An example of strict local martingale

[Lemma 5 of JPS07]. A process ${\{\beta_t: 0\le t <1\}}$ is defined by the solution of

$\displaystyle \beta_t = \beta_0 + \int_0^t \frac{\beta_u}{\sqrt{1 - u}} d B_u, \ \beta_0>0. \ \ \ \ \ (1)$

Show that ${\{\beta_t: 0 \le t \le 1\}}$ is a continuous strict local martingale, with the last element ${\beta_1 := \lim_{t\rightarrow 1^{-}} \beta_t = 0}$ .

Proof: Let

$\displaystyle M_t \triangleq \int_0^t \frac{1}{\sqrt{1-u}} d B_u.$

For any ${T \in (0,1)}$ , because ${f(u) = \frac{1}{\sqrt{1-u}} \in L^2([0,T])}$ , ${M_t}$ is a well-defined Martingale on ${[0,T]}$ by Proposition 3.2.10 of [KS98].

Next, one can show that, on ${[0,1)}$

$\displaystyle \beta_t = \beta_0 \exp\{ M_t - \frac 1 2 [M]_t\}$

is a strong solution of (1), and the uniquness follows by Theorem IX.3.5(ii) of [RY99]. Thus, ${\beta_t}$ is local martingale on ${[0,1)}$ .

With function ${\phi: [0, \infty) \rightarrow [0,1)}$ of ${\phi(t) = 1 - e^{-t}}$ , consider time-changed process

$\displaystyle \hat \beta_t = \beta_{\phi(t)}$

and

$\displaystyle \hat M_t = M_{\phi(t)}.$

One can check ${[\hat M]_t = t}$ and ${\hat M}$ is a continuous local martingale. By Levy’s characterization theorem, ${\hat M_t}$ is a Brownian motion w.r.t. ${\mathcal{G}_t = \mathcal{F}_{\phi(t)}}$ . Note that