Show that $Z_t = \sqrt{\sqrt{2t}e^{-\sqrt{2t}}} B_{e^{\sqrt{2t}}-1}$ solves $dZ_t = f(t)Z_tdt +dM_t$ and find $f(t)$.

Question

Let $B=(B_t)_{t\geq0}$ be a standard Brownian motion started at $0$ and define $M=(M_t)_{t\geq0}$ by

$$M_t = \int_0^{e^{\sqrt{2t}}-1} \sqrt{\frac{\log{(1+s)}}{1+s}} dB_s$$

so that M is also a standard Brownian motion. Show that the process $Z=(Z_t)_{t\geq0}$ defined by

$$Z_t = \sqrt{\sqrt{2t}e^{-\sqrt{2t}}} B_{e^{\sqrt{2t}}-1}$$

solves the following stochastic differential equation

$$dZ_t = f(t)Z_tdt +dM_t$$

with $Z_0 = 0$ and determine the function $f$ explicitly.

I'm not really sure how to go about this, any help would be appreciated, thanks.

Answer 1

Yes first see the first differential from the integral form you have.

$$ dM_t= \sqrt{\dfrac{\log(1+s)}{1+s}}dB_s$$ where $s=e^{\sqrt{2t}-1}$

Then from definition of $Z_t$

$$dZ_t=d(\sqrt{\sqrt{2t}e^{-\sqrt{2t}})}B_{e^{\sqrt{2t}-1}}+dB_{e^{\sqrt{2t}-1}}\sqrt{\sqrt{2t}e^{-\sqrt{2t}}}$$

And from here I think you can end :)