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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 $$F(t, x) = e^{−t}x, \,\,\,\,\,\,\,\,t \geq 0,\,\,\, x \in R_+$$

  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 $\langle M, M\rangle$ 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 $$M_t = \int_0^te^{-s}dX_s = 3\int_0^t+2\int_o^t\sqrt{X_s}dB_s$$ 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.

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    $\begingroup$ how did you even derive this expression for $M_t$? Where did $e^t$ go? Can you show your exact derivation process? $\endgroup$
    – Makina
    Commented Dec 31, 2018 at 20:56
  • $\begingroup$ Your process $M_t$ is not a martingale. Note that the stochastic integral $\int_0^t f(s) \, dY_s$ is a (local) martingale only if $(Y_t)_t$ is a (local) martingale... $\endgroup$
    – saz
    Commented Jan 1, 2019 at 9:01

1 Answer 1

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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.

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  • $\begingroup$ in part 2 you claim that the expectation of the stochastic integral is zero. Are you able to do so because of stopping time (which makes the integrand bounded i.e it can't explode to infinity) $\endgroup$
    – Makina
    Commented Jan 4, 2019 at 18:14
  • $\begingroup$ @Makina Yes, exactly. $\endgroup$
    – saz
    Commented Jan 4, 2019 at 18:24

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