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)}


\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

\displaystyle \hat \beta_t = \beta_0 \exp\{\hat M_t - \frac 1 2 [\hat M]_t\} = \beta_0 \exp\{\hat M_t - \frac 1 2 t\} \rightarrow 0

as {t\rightarrow \infty} by Law of Iterated Logarithm. This implies {\beta_t} is strict local martingale with

\displaystyle \beta_1 := \lim_{t\rightarrow 1} \beta_t = \lim_{t \rightarrow \infty} \hat \beta_t = 0 < \beta_0.




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