Determining if a process is a martingale

Question

Let $Z(t)$ be a standard brownian motion, given $W(t) = t^{2} Z(t) -2\int_0^t sZ(s) ds$ , determine if this is a martingale process!

We know that a process is martingale if $E[W(t+1)|F(t)] = W(t)$, with $F(t)$ as a filtration process. I think we should start by writing down $E[W(t+1)|F(t)]$ first, which is $E[(t+1)^{2} Z(t+1) -2\int_0^{t+1} sZ(s) ds|F(t)]$. This can be split into two terms:

First term: $E[(t+1)^{2} Z(t+1)|F(t)]$
which I believe would be $(t+1)^2Z(t)$, since a standard brownian motion is a martingale.

Second term: $-2E[\int_0^{t+1} sZ(s) ds|F(t)]$
for this part, i think we need to use integration by parts, but I ended up with $\int Z(s) ds$ and I'm not sure how to proceed with that result.

Am I doing the right steps? Thanks in advance

Answer 1

First of all I have not seen the definition of martingale(Continuous time) to be $E(X_{t+1}|\mathcal{F}_{t})=X_{t}$ .

The definition which is the most standard is that for an adapted process $(X_{t})$ it is a martingale wrt $(\mathcal{F}_{t})$ if for all $s<t$, you have $E(X_{t}|\mathcal{F}_{s})=X_{s}$ . Obviously what you want follows from this definition, but I don't know whether the converse is true.

$E(W(t)|\mathcal{F}_{s})=E(t^{2}Z(t)|\mathcal{F}_{s})-2E(\int_{0}^{t}xZ(x)\,dx|\mathcal{F}_{s})$

Now $2\int_{0}^{t}xZ(x)\,dx = 2\int_{0}^{s}xZ(x)\,dx + 2\int_{s}^{t}xZ(x)\,dx$

So you have \begin{align}2E(\int_{s}^{t}xZ(x)\,dx|\mathcal{F}_{s})=2E((\int_{s}^{t}x(Z(x) -Z(s))\,dx|\mathcal{F}_{s})+(t^{2}-s^{2})Z(s)\end{align} Above, we have used that $E(Z_{s}|\mathcal{F}_{s})=Z_{s}$ and just added and substracted $\int_{s}^{t}x Z(s)\,dx$

Also you have $t^{2}Z(t)=t^{2}(Z(t)-Z(s))+t^{2}Z(s)$ and by Markov property $(Z(u)-Z(s))_{u\geq s}$ is a Standard Brownian Motion indepdendent of $\mathcal{F}_{s}$.

So we have:- $$E(W(t)|\mathcal{F}_{s})=t^{2}E(Z(t)-Z(s))+t^{2}Z(s)-2E(\int_{s}^{t}x(Z(x)-Z(s))\,dx)\\-(t^{2}-s^{2})Z(s)-2\int_{0}^{s}xZ(x)\,dx$$ where we have used independence to get rid of conditional expectation.

Simplyfying the above, and noting that $E(Z(t)-Z(s))=0$ you have

$$ s^{2}Z(s)-2\int_{0}^{s}xZ(x)\,dx-2E(\int_{s}^{t}x(Z(x)-Z(s))\,dx)$$

So it only remains to show that $E(\int_{s}^{t}x(Z(x)-Z(s))\,dx)=0$ but this is just due to Fubini's theorem,

$$E(\int_{s}^{t}xZ(x)-Z(s)\,dx)=\int_{\Omega}\int_{s}^{t}x(Z(x)(\omega)-Z(s)(\omega))\,dx\,d\mathbb{P}(\omega)$$

By Fubini,

$$=\int_{s}^{t}\int_{\Omega}x(Z(x)(\omega)-Z(s)(\omega))\,d\mathbb{P}(\omega)\,dx = 0$$

This is because $\int_{\Omega}(Z(x)(\omega)-Z(s)(\omega))\,d\mathbb{P}=E(Z(x)-Z(s))=0$ for each $x\in (s,t]$

And thats it.

Answer 2

I'd reason like this. Define the function $f(t,Z_t) := t^2 Z_t$. By Ito's formula its differential is

$$ df = 2tZ_tdt + t^2dZ_t \Rightarrow 2\int_{0}^{t}sZ_s ds = t^2Z_t-\int_{0}^{t}s^2dZ_s $$

Plugging this into our function $W_t$ we have

$$ W_t = \int_{0}^{t}s^2dZ_s \Rightarrow dW_t = t^2dZ_t $$

This is a stochastic integral which is martingale since $t^2$ is $\mathcal{F}_t -$measurable and by Ito isometry $\mathbb{E}\bigg( \int_{0}^{t}s^2dZ_s\bigg)^2 < \infty$.

Thus the process is a martingale.