Show a Continuous Local Martingale is a Martingale Let $B = (B_t)_{t≥0}$ be a standard Brownian motion started at zero, let  $X=(X_t)_{t≥0}$ be a nonnegative stochastic process solving
$$dX_t = 3 \, dt + 2\sqrt{X_t} \, dB_t \qquad(X_0 = 0)$$
and let $$F(t, x) = e^{−t}x, \,\,\,\,\,\,\,\,t \geq 0,\,\,\, x \in R_+$$


*

*Need to apply Ito’s formula to $F(t, X_t)$ for $t ≥ 0$ and determine a continuous
local martingale $(M_t)_{t≥0}$ starting at $0$ and a continuous bounded
variation process $(A_t)_{t≥0}$ such that $F(t, X_t) = M_t+A_t$
for $t ≥ 0$.

*Then Show that $(M_t)_{t≥0}$ is a martingale and compute $\langle M, M\rangle$
for $t ≥ 0$.

*Compute $E(\int_0^\tau(1/\sqrt{X_t}) \, dt$ when $\tau = \inf(t ≥ 0 : X_t = 2)$.



So far I have calculated the continuous local martingale as $$M_t = \int_0^te^{-s}dX_s = 3\int_0^t+2\int_o^t\sqrt{X_s}dB_s$$
I am unsure as how to show this is martingale and to compute $\langle M, M\rangle$. And any hints for 3. would be appreciated too.
 A: Part 1: Itô's formula shows
$$F(t,X_t)  = \int_0^t e^{-s} \, dX_s - \int_0^t e^{-s} X_s \, ds = M_t+A_t$$
where
$$M_t = 2 \int_0^t e ^{-s} \sqrt{X_s} \, dB_s \quad \text{and} \quad A_t :=  \int_0^t e^{-s} (3-X_s) \, ds.$$
Part 2: Set $\tau_r := \inf\{t \geq 0; X_t \geq r\}$. By the very definition of $X$, we have
$$X_{t \wedge \tau_r} = 3 (t \wedge \tau_r) + 2 \int_0^{t \wedge \tau_r} \sqrt{X_s} \, dB_s$$
and so
$$\mathbb{E}(X_{t \wedge \tau_r}) = 3 \mathbb{E}(t \wedge \tau_r) \leq 3t.$$
This implies that $f(s,\omega) := 2 e^{-s} \sqrt{X_s(\omega)}$ satisfies $\mathbb{E}(\int_0^t f(s)^2 \, ds) < \infty$ for any $t>0$, and therefore $$M_t = \int_0^t f(s) \, dB_s = 2 \int_0^t e ^{-s} \sqrt{X_s} \, dB_s$$ is a martingale. It is well known that for any stochastic integral of this form, the quadratic variation $\langle M \rangle$ is given by $$\langle M \rangle_t = \int_0^t f(s)^2 \, ds = 4 \int_0^t e^{-2s} X_s \, ds.$$
Part 3: Set $g(x) := \sqrt{x}$. Applying formally Itô's formula (see the remark below) we find
$$\begin{align*} g(X_t) = \sqrt{X_t} &= \frac{1}{2} \int_0^t \frac{1}{\sqrt{X_s}} \, dX_s - \frac{1}{8} \int_0^t \frac{1}{X_s^{3/2}} \, \underbrace{d\langle X \rangle_s}_{=4 X_s \, ds} \\ &= \int_0^t dB_s + \int_0^t \frac{1}{\sqrt{X_s}} \, ds. \end{align*}$$
Hence,
$$\mathbb{E}(\sqrt{X_{t \wedge \tau}}) = \underbrace{\mathbb{E}(B_{t \wedge \tau})}_{=0} + \mathbb{E} \left( \int_0^{t \wedge \tau} \frac{1}{\sqrt{X_s}} \, ds \right). \tag{1}$$
Since $0 \leq X_{t \wedge \tau} \leq 2$ this implies
$$\frac{1}{\sqrt{2}} \mathbb{E}(t \wedge \tau) \leq \mathbb{E}(X_{t \wedge \tau}) \leq \sqrt{2},$$
and by the monotone convergence theorem this gives $\mathbb{E}(\tau)<\infty$; in particular, $\tau<\infty$ almost surely. Applying the dominated convergence theorem and the monotone convergence theorem, we conclude from $(1)$ that
$$\mathbb{E} \left( \int_0^{\tau} \frac{1}{\sqrt{X_s}} \, ds \right) = \mathbb{E}(\sqrt{X_{\tau}}) = \sqrt{2}.$$
Remark: Since $g(x) = \sqrt{x}$ is not twice continuously differentiable, we cannot apply the classical version of Itô's formula. To make the above calculations rigorous, one has to use a truncation technique. To this end choose a smooth function $\chi$ such that $\chi(x)=0$ for $|x|<1/2$ and $\chi(x)=1$ for $|x| \geq 1$, and define $$\chi_n(x) := \chi \left( n x \right), \qquad x \in \mathbb{R}.$$
Since $x \mapsto g_n(x) := g(x) \chi_n(x)$ is twice continuously differentiable, we may apply Itô's formula and then we can use $g_n(x) \to g(x)$ to obtain the above result.
