= Solution
Interpret the ordered <Iterated Itô integral> recursively as $I_0(t)=1$ and
$$
I_n(t)=\int_0^t I_{n-1}(s)\,dB_s.
$$
The diagonal boundary of the ordered region does not affect this recursive definition. For $n=1$, $I_1(t)=B_t$. For $n=2$, part (a)(iii) gives $I_2(t)=\tfrac12(B_t^2-t)$. For $n=3$, use parts (b)(ii) and (b)(iii):
$$
\begin{aligned}
I_3(t)
&=\frac12\int_0^t(B_s^2-s)\,dB_s\\
&=\frac12\left(\frac13B_t^3-\int_0^tB_s\,ds
-tB_t+\int_0^tB_s\,ds\right)
=\frac16(B_t^3-3tB_t).
\end{aligned}
$$
Thus the three <Probabilists' Hermite polynomials> give
$$
\boxed{\begin{aligned}
1!I_1(t)&=B_t=t^{1/2}h_1(B_t/\sqrt t),\\
2!I_2(t)&=B_t^2-t=t\,h_2(B_t/\sqrt t),\\
3!I_3(t)&=B_t^3-3tB_t=t^{3/2}h_3(B_t/\sqrt t).
\end{aligned}}
$$
The right sides are interpreted as their polynomial extensions at $t=0$. These <space-time Hermite polynomials> arise here directly from ordered <Iterated Itô integrals>.
Back to article page
