My task is to prove that $X_t = W_t^4 – 6tW_t^2 + 3t^2$ is a martingale using Itô's lemma.
So far I've proven that it is a local martingale by applying the lemma to the function $f(t,x) = x^4 – 6tx^2 + 3t^2$ and finding that $dX_t = (4W_t^3 – 12tW_t)dW_t = \sigma_t dW_t$, showing that there is no drift term. Is this correct?
If I understand correctly, to prove that this local martingale is a martingale, I need to prove that it's quadratic variation is integrable. This is what I have so far:
$$
\langle X\rangle _t = \int_0^t \sigma_s^2 ds = 16 \int_0^t W_s^6ds – 96\int_0^tW_t^4ds + 144 \int_0^t sW_s ds
$$
Looking at the first integral now. We use integration by parts:
$$
d(W_t^6t) = (W_t^6 + 15W_t^4)dt + 6W_t^5dW_t \\
W_t^6dt = d(W_t^6t) – 15W_t^4dt – 6W_t^5dW_t \\
\int_0^t W_s^6ds = W_t^6t – 15\int_0^tW_s^4ds + 6\int_0^t W_s^5 dW_s
$$
For fixed $t$, the first term is constant. I need to prove that the two integrals exist. My feeling here is that if I can prove that the third integral exists, I can apply the same logic recursively to the second integral. To my understanding:
$$
\int_0^tW_s^5dW_s = \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{n-1}W_{t_{i+1}}^5(W_{t_{i+1}} – W_{t_i})
$$
How do I go about proving that this limit exists? Is what I'm doing so far correct at all?
Best Answer
You have $dX_t=\sigma_t\,dW_t$, and you need to prove that $$ \mathbb E[\langle X\rangle_t]=\int_0^t\mathbb E[\sigma_s^2]\,ds<+\infty. $$
Since $\sigma_s^2=16W_s^6-96sW_s^4+144s^2W_s^2$ and $W_s$ has the same distribution as $\sqrt t G$, for $G\sim\mathcal N(0,1)$, we have that $$ \mathbb E[\sigma_s^²]=\left(16\mathbb E[G^6]-96 \mathbb E[G^4]+144\mathbb E[G^2]\right)s^3, $$ whose integral on $[0,t]$ is clearly finite.