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I was wondering if there were a way to transform the process $$ X(t) = w^4(t) $$ with $w(t)$ which is a standard Brownian motion, into a martingale, by adding a deterministic function. That is: $$ Y(t) = w^4(t) + g(t) $$ with g(t) that is a deterministic function.

I tried to apply Ito's Lemma on $Y(t)$ in order to derive a condition on the term associated with $dt$ (which is equal to 0 for a martingale) but this leads to this condition:

$$ \frac{dg}{dt} = -12w^2(t) $$ which does not seem to help much.

What I also know is that $E[w^4(t)] = 3t^2$ but removing this quantity from $X(t)$ does not give a martingale either.

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No, there isn't. Your argument by Itô's lemma is sufficient (in fact, you can apply Itô's lemma to $W_t^4$ directly and get the same result). Here's an alternative argument you might be interested in.

Consider the martingale $M_t = W_t^2 - t$ and set $R_t = M_t^2 = W_t^4 - 2t W_t^2 +t^2$. By definition, the quadratic variation of $M_t$, $\langle M \rangle_t$, is the unique process such that $M_t^2 - \langle M \rangle_t$ is again a martingale. Let's compute $\langle M \rangle_t$: first note that $dM_t = 2W_tdW_t$, so that $(dM_t)^2 = 4W_t^2 dt$, giving us that $$\langle M \rangle_t = \int_0^t 4W_s^2 ds$$ Thus, $$M_t^2 - \langle M \rangle_t = W_t^4 - 2t W_t^2 +t^2 - \int_0^t 4W_s^2 ds$$ is a martingale, but the compensator $- 2t W_t^2 +t^2 - \int_0^t 4W_s^2 ds$ is not deterministic.

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