2
$\begingroup$

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.

  • 0
    Quite right Canardini, I should have put my attempt up. I have edited my question accordingly. Thanks. – 2017-01-07
  • 0
    @Canardini, I'd be so grateful if you could say if what I have done so far is correct. – 2017-01-08
  • 1
    For the martingality, you can calculate $E[M,M]$. Apply Ito's lemma on $X_t^n$ with $n>0$, you get $E(X)=2t$, $E(X^2)=8t^2$, and $E(X^3)=48t^3$. You should find $$E[M,M]=128t^6$$ For the question 3, I did not have a look at it in details, though you can note that $\tau \leq 1$ a.s. – 2017-01-09
  • 0
    Thanks Canardini. Could I possibly say: $\mathbb{E}\tau = \mathbb{B_{\tau}^2}$. The SDE is a Bessel Squared Process hence has solution $X_t = B_t^2, B=(B_t)$ SBM so $\sigma = \inf \{ t : B_t^2 = 1-t \}$ $\mathbb{E}\tau = \mathbb{E}B_t^2 = \mathbb{E}[1-t]$ then sub t for $\tau$...giving $\mathbb{E}\tau = 1/2$ – 2017-01-09
  • 0
    Are you supposed to use question a) to answer b ) ? or are they independent questions ? Yes , it is a squared Bessel process, but your reasoning is confusing because $X_\tau=1-\tau$ a.s. – 2017-01-09
  • 0
    A solution to a squard Bessel process (Where we have 1 dimensional Brownian motion) is $X_t = B_t^2 + X_0 = B_t^2 $ hence $X_{\tau}= 1 - \tau$ becomes $B_{\tau}^2 = 1 - \tau$ Taking Expectation and using the result $\mathbb{E}B_{\tau}^2= \mathbb{E}\tau$ gives 1/2 – 2017-01-09
  • 0
    I wrote an answer below – 2017-01-13

1 Answers 1

1

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

$$E(\tau)=\frac{1}{3}$$

  • 0
    Many thanks Canardini. All makes sense - very kind of you to take the time and effort to help me understand this. – 2017-01-14