Showing that this is a martingale.(4.13 in Øksendals SDE)

Question

This is an exercise from Øksendals stochastic differential equations, where I get stuck. It is exercise number 4.13.(I simplified the notation a bit.)

I have that X is an Itô-process where:

$dX_T=u(\omega,t)dt+dB_t$, $t \in [0,T]$, and I assume u is bounded.

I define the Itô-process Z, where:

$dZ_t = -u_t dB_t -\frac{1}{2}u_t^2dt$, $Z_0 = 0$

I am supposed to show that $X_te^{Z_t}$ is a $\mathcal{F}_t$-martingale, $\mathcal{F}_t$ is the filtration generated by the Brownian motion.

Using Itô formula for the multidimensional case(I omit all the algebra) I get that:

$d(X_te^{Z_t})=[e^{Z_t}(1-X_tu_t)]dB_t$.

Hence $X_te^{Z_t}=X_0+\int_0^te^{Z_t}(1-X_tu_t)dB_t$.

From the theory of the construction of the Itô process I know that this is a martingale if: $E[\int_0^t(e^{Z_t}\{1-u_tX_t\})^2dt]<\infty$.

But how do I show this?

What I am able to get is:

$|Z_t| \leq K+ |\int_0^tu_TdB_t|$, so since u is bounded we know that $|\int_0^tu_tdB_t|\in L_2(\omega \times [0,T])$. So $Z_t \in L_2(\Omega \times [0,T])$.

And by the boundedness of u it also follows that $(1-u_tX_t)\in L_2(\Omega \times [0,T])$.

But how do I show that $e^{Z_t} \in L_2(\omega \times [0,T])$, and $e^{Z_t}u_tX_t \in L_2(\omega \times [0,T])$? The $L_2$ vector-space isnt closed under multiplication, and we can not just take the exponential and expect to still be there?

Any hints?

Answer 1

It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that $(e^{Z_t})_t$ is a local martingale and the boundedness of $u$ to conclude that $e^{Z_t} \in L^2$.

In my answer, I'll present a solution which does not require any prior knowledge of stochastic integrals except the definition of the Itô integral.

Suppose that $u$ is a bounded simple function, i.e. that there exist (suitable measurable) random variables $f_j$ such that $K:=\sup_j \|f_j\|_{\infty}<\infty$ and

$$u(s,\omega) = \sum_{j=1}^n f_j(\omega) 1_{[t_{j-1},t_j]}(s).$$

for $s \in [0,t]$. Then

$$\int_0^t u(s) \, dB_s = \sum_{j=1}^n f_j (B_{t_j}-B_{t_{j-1}}).$$

Using the independence of the increments and the boundedness of $u$, we get

$$\mathbb{E} \exp \left( \left| \int_0^t u(s) \, dB_s \right| \right) \leq \prod_{j=1}^n \mathbb{E}\exp(K |B_{t_j}-B_{t_{j-1}})|).$$

Since $(B_t)_{t \geq 0}$ has independent increments, we have $B_{t_{j}} -B_{t_{j-1}} \sim B_{t_j-t_{j-1}}$. Moreover, by Jensen's inequality, $(e^{K |B_t|})_{t \geq 0}$ is a submartingale and therefore

$$\sup_{t \in [0,T]} \mathbb{E}e^{K |B_t|} \leq \mathbb{E}e^{K |B_T|}<\infty.$$

Hence,

$$\mathbb{E} \exp \left( \left| \int_0^t u(s) \, dB_s \right| \right)\leq \left[ \mathbb{E}e^{K|B_R|} \right]^n < \infty $$

for $R$ sufficiently large such that $R \geq |t_j-t_{j-1}|$ for all $j=1,\ldots,n$. In fact, the above estimates show that

$$\sup_{t \in [0,T]} \mathbb{E} \exp \left( \left| \int_0^t u(s) \, dB_s \right| \right) \leq C$$

for some constant $C$ depending only on $K$ and the mesh size $|\Pi|:= \max_j |t_j-t_{j-1}|$. Now if $u \in L^2(\Omega \times [0,T])$ is a bounded function, we can choose a sequence of simple functions $(u_n)_n$ such that $\|u_n-u\|_{L^2} \to 0$. Since $u$ is bounded, say by a constant $K$, we may also assume that each $u_n$ is bounded by $K$ (otherwise we consider $(-K) \vee u_n \wedge K$). By the very definition of Itô's integral, we have $$\int_0^t u_n(s) \, dB_s \xrightarrow[L^2]{n \to \infty} \int_0^t u(s) \, dB_s.$$ In particular, we can choose a subsequence which converges almost surely. Now it follows from Fatous lemma and the first part of this proof that there exist a constant $C>0$ such that

$$\mathbb{E} \exp \left( \left| \int_0^t u(s) \, dB_s\right| \right) \leq \liminf_{n \to \infty} \mathbb{E} \exp \left( \left| \int_0^t u_n(s) \, dB_s \right| \right) \leq C < \infty$$

for all $t \in [0,T]$ for some fixed $T>0$. Obviously, this implies that

$$\mathbb{E} \left[ \int_0^t \exp \left( \left| \int_0^s u(r) \, dB_r \right| \right) \, ds \right] < \infty$$

and using that $u$ is bounded, it follows easily that $e^{Z_t} \in L^2(\Omega \times [0,T])$. Since

$$|X_t| \leq \|u\|_{\infty} \cdot t+ |B_t| $$

and $B_t$ has moments of arbitrary order, we find $X_t \in L^p(\Omega \times [0,T])$ for any $p \geq 1$. Combining this with the fact that $e^{Z_t} \in L^2(\Omega \times [0,T])$, an application of the Cauchy Schwarz inequality shows that

$$e^{Z_t} \cdot X_t \in L^2(\Omega \times [0,T]).$$

Answer 2

Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $

By Cauchy-Schwarz inequality:

$\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $

Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < \infty$.

Hence it is left to show that $\mathbb{E}[\int_0^t e^{4 Z_s}ds] < \infty$.

It is enough to show that $\mathbb{E}[e^{4 Z_t}] <\infty$ for all $t \in [0,T]$.

$e^{4 Z_t} = \exp( -\int_0^t 4u_s dB_s - \frac{1}{2}\int_0^t ({4u_s})^{2} ds)\exp(6 \int_0^t u_s^{2} ds)=M_t \exp(6 \int_0^t u_s^{2} ds)$. (*)

$M_t$ is the Doléans-Dade exponential (applied to the process $-\int_0^t 4u_s dB_s$) and you can test the Novikov condition: $\mathbb{E}[\frac{1}{2} \exp(\int_0^T (4u_s)^{2} ds)] < \infty $ which is true because $u$ is bounded.

Hence by the Novikov condition $M_t$ is a martingale and $\mathbb{E}[M_t]< \infty$.

Using the fact that $u$ is bounded, we conclude by (*) that $\mathbb{E}[e^{4 Z_t}] < \infty$ and that your process is a martingale.