I am preparing for one of my exams. I am looking for some help with the following question:
Let B be Standard Brownian Motion, started at $0$, $X=(X_t)_{t \geqslant 0}$ a non-negative stochastic Process solving:
$$dX_t = 2dt + 2\sqrt{X_t} dB_t \ \ \ \ (X_0 =0)$$ For $F(t,x)= tx^2$, $t \geqslant 0$ and $x \in \mathbb{R}$
1) Apply Ito's Formula to $F(t,X_t)$ for $t \geqslant 0$. Then determine a continuous local martingale $(M_t)$ starting at $0$ and a continuous bounded variation Process $(A_t)$ such that $F(t, X_t) = M_t + A_t$
2) Show $M_t$ is a martingale and compute $[M,M]_t$
3) Compute $\mathbb{E} \tau$ for stopping time: $\tau = \inf \{t \geqslant 0 : X_t= 1-t\}$
My attempt:
1) $F(t, X_t) = F(0, X_0) + \int_{0}^{t}F_s ds+ \int_{0}^{t}F_xdX_s +\frac{1}{2} \int_{0}^{t}F_{xx}d[X,X]_s$
First, $[X,X]= [K,K]$, where $K:= \int_0^t 2\sqrt{X_s}dB_s$ i.e $[K,K] = \int_0^t 4X_sds$
Using the SDE,
$\int_0^tF_xdX_s = 2\int_0^tF_xds + 2\int_0^tF_x\sqrt{X_s}dB_s $
$\int_0^tF_{xx}d[X,X]_s = \int_0^t4F_{xx}X_sds$
Plugging both equations into $F(t,X_t)$:
$F(t,X_t)= F(0,X_0) + \int_0^t F_s + 2F_x + 2F_{xx}X_s ds +2\int_0^tF_x\sqrt{X_s}dB_s$
Define $A = F(0,X_0) + \int_0^t F_s + 2F_x + 2F_{xx}X_s ds $ and $M=2\int_0^tF_x\sqrt{X_s}dB_s$
2) To show M is a martingale, it is sufficient to show that (from a lemma in my notes) that $\mathbb{E}[M,M]_t$ is finite.
$M= 4\int_0^t s X_s \sqrt{X_s} dB_s$, so
$[M,M] = 16\int_0^t s^2 X_s^3ds$
...Not too sure how to proceed. Also, I am not sure how to start question 3.
1) and 2) , you can see the comments I wrote.
3) We define the process$$M_t=X_t-2t$$
We have that $$dM_t=dX_t-2dt=2\sqrt{X_t}dW_t$$. It is a driftless process , therefore it is a local martingale.
Actually, we can prove that $M$ is a martingale, indeed, $X_s$ is a positive process by definition, therefore , using Tonelli's theorem, we have that
$$E\left(4\int_{0}^{t}{X_sds}\right)=4\int_{0}^{t}{E(X_s)ds}=4\int_{0}^{t}{2sds}=4t^2<\infty$$
Moreover, $X_t$ is a positive and continuous process, therefore $$\tau \leq1 \ a.s.$$ and we have ,using the optional stopping theorem,
$$E(M_\tau)=M_0=0$$
Finally , we have that $$M_\tau=X_\tau-2\tau=(1-\tau)-2\tau=1-3\tau \ a.s.$$
Thus, $$E(M_\tau)=1-3E(\tau)=0$$
or
$$E(\tau)=\frac{1}{3}$$