= Solution
Use the <independent increments> of <Brownian motion>, rather than merely checking that an <Itô formula> drift vanishes. For $0\leq s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of $\mathcal F_s$ and has <normal distribution> $N(0,h)$. Its first four <moments> are $0,h,0,3h^2$. Therefore
$$
\begin{aligned}
\mathbb E(B_t^2\mid\mathcal F_s)&=B_s^2+h,\\
\mathbb E(B_t^3\mid\mathcal F_s)&=B_s^3+3hB_s,\\
\mathbb E(B_t^4\mid\mathcal F_s)&=B_s^4+6hB_s^2+3h^2.
\end{aligned}
$$
For the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha(t)$, so $\alpha(t)=-3t$ makes it $\alpha(s)$. For the quartic expression, choose $\beta(t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomes
$$
3(t-s)^2-6t(t-s)+\gamma(t),
$$
which equals $3s^2$ when $\gamma(t)=3t^2$. Thus a standard choice is
$$
\boxed{\alpha(t)=-3t,\qquad\beta(t)=-6t,\qquad\gamma(t)=3t^2.}
$$
The resulting <stochastic processes> are <Hermite polynomial martingales> $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable <martingales>, since Gaussian <moments> are finite at every finite time and the displayed conditional identities establish the <martingale> property directly.
The choice is not unique. Constants $c,d,e\in\mathbb R$ give the valid family $\alpha(t)=c-3t$, $\beta(t)=d-6t$, and $\gamma(t)=3t^2-dt+e$: these add $cB_t$ to the cubic <martingale> and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.
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