= Solution
Fix a finite horizon $T$ and define $\tau_n=\inf\{s\ge0:|B_s|\ge n\}$. These <stopping times> increase to infinity almost surely because continuous paths are bounded on each finite interval. Applying the <Itô formula> to $B_{t\wedge\tau_n}^k$, for $k\ge2$, gives
$$
B_{t\wedge\tau_n}^k=k\int_0^t\mathbf1_{\{s\le\tau_n\}}B_s^{k-1}\,dB_s+\frac{k(k-1)}2\int_0^t\mathbf1_{\{s\le\tau_n\}}B_s^{k-2}\,ds.
$$
The first integrand is bounded by $kn^{k-1}$, so its integral is square-integrable and has expectation zero. Thus
$$
\mathbb EB_{t\wedge\tau_n}^k=\frac{k(k-1)}2\mathbb E\int_0^t\mathbf1_{\{s\le\tau_n\}}B_s^{k-2}\,ds.
$$
To justify both limits explicitly, let $S_T=\sup_{s\le T}|B_s|$ and choose an integer $q>\max\{k,1\}$. The Gaussian density gives $\mathbb E|B_T|^q<\infty$ by its exponential tail. The <Doob Lp maximal inequality> states
$$
\mathbb ES_T^q\le\left(\frac q{q-1}\right)^q\mathbb E|B_T|^q<\infty.
$$
Therefore $S_T^k$ and $1+S_T^{k-2}$ are integrable. The stopped terminal powers converge almost surely to $B_t^k$ and are dominated in absolute value by $S_T^k$. The integrands in the time integral converge almost surely and are bounded by $1+S_T^{k-2}$, so their integrals are dominated by $T(1+S_T^{k-2})$. <Dominated convergence theorem> on both sides, and then Fubini justified by the same bound, give the <Brownian moment recursion>
$$
\boxed{\beta_k(t)=\frac{k(k-1)}2\int_0^t\beta_{k-2}(s)\,ds\qquad(k\ge2).}
$$
This uses only finiteness of Gaussian moments, not the closed moment formula requested in the next part.
Back to article page
