Jtdylktuy

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
$$dX_t = 3 , dt + 2sqrt{X_t} , dB_t qquad(X_0 = 0)$$
and let $$F(t, x) = e^{−t}x, ,,,,,,,,t geq 0,,,, x in R_+$$

1. 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 $t ≥ 0$.
   
   


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


3. 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 $$M_t = int_0^te^{-s}dX_s = 3int_0^t+2int_o^tsqrt{X_s}dB_s$$
I am unsure as how to show this is martingale and to compute $langle M, Mrangle$. And any hints for 3. would be appreciated too.

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

$$begin{align*} g(X_t) = sqrt{X_t} &= frac{1}{2} int_0^t frac{1}{sqrt{X_s}} , dX_s - frac{1}{8} int_0^t frac{1}{X_s^{3/2}} , underbrace{dlangle X rangle_s}_{=4 X_s , ds} \ &= int_0^t dB_s + int_0^t frac{1}{sqrt{X_s}} , ds. end{align*}$$

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.