Let $B = (B_t)_{t \geq 0}$ be a standard brownian motion and $I_t = \inf_{0 \leq s \leq t} B_s$, show $F(t, B_t, I_t)$ is a martingale.

Question

Let $B = (B_t)_{t \geq 0}$ be a standard brownian motion and $I_t = \inf_{0 \leq s \leq t} B_s$, show that

$F(t, B_t, I_t) = (B_t−I_t)^6−15t(B_t−I_t)^4+45t^2(B_t−I_t)^2−15t^3$

is a martingale. I'm really not sure where to begin with this, any pointers would be appreciated?

Answer 1

The application of Ito's formula may look intimidating, but it's not too bad. The $dB_t$ terms will combine into a martingale, so it suffices to check that the $dt$ and $dI_t$ terms cancel each other out. In fact, the combined $dI_t$ terms vanish because $B_t-I_t$ vanishes at points of decrease of $I_t$ and the partial derivative of $F(t,x,y)$ with respect to $y$ vanishes whenever $x=y$. This leaves only the $dt$ terms $$ F_t(t,B_t,I_t)dt + {1\over 2} F_{xx}(t,B_t,I_t) dt, $$ which I will leave to you.