Simple proof of special case of Clark-Ocone formula?

Question

I've seen the usual proof of the Clark-Ocone formula in the context of Malliavin calculus, but I'm wondering whether anyone knows or can think of a simple proof of the statement

$$ B_t^n = E[B_t^n] + \int_0^t E[nB_t^{n-1}\mid\mathcal{F}_s]\,\mathrm dB_s. $$

Working it out even for $n=2,3$ is I think a nice exercise for students in an introductory sotchastic calculus class, but I wonder if there is a relatively easy way to deduce this formula, one that would also be appropriate for students.

Thanks.

Answer 1

$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\d{\mathrm{d}}\def\F{\mathscr{F}}\def\ev{\biggr|}\def\peq{\mathrel{\phantom{=}}{}}$Here is a proof using Ito's formula. Consider a fixed $T > 0$.

Step 1: For $t \in [0, T)$,\begin{gather*} E(B_T^n \mid \F_t) = \sum_{m = 0}^n \binom{n}{m} E((B_T - B_t)^m) B_t^{n - m} = \sum_{m = 0}^{\left[ \frac{n}{2} \right]} a_m (T - t)^m B_t^{n - 2m},\tag{1.1} \end{gather*} where $a_m = \dbinom{n}{2m} (2m - 1)!!$ satisfies\begin{gather*} a_0 = 1,\ a_{m + 1} = \frac{(n - 2m)(n - 2m - 1)}{2(m + 1)} a_m \left( 0 \leqslant m < \left[ \frac{n}{2} \right] \right).\tag{1.2} \end{gather*}

Proof: Because $σ(B_T - B_t)$ and $\F_t$ are independent and $B_t$ is $\F_t$-measurable, so\begin{align*} &\peq E(B_T^n \mid \F_t) = E((B_T - B_t + B_t)^n \mid \F_t)\\ &= E\left( \sum_{m = 0}^n \binom{n}{m} (B_T - B_t)^m B_t^{n - m} \,\middle|\, \F_t \right) = \sum_{m = 0}^n \binom{n}{m} E((B_T - B_t)^m B_t^{n - m} \mid \F_t)\\ &= \sum_{m = 0}^n \binom{n}{m} B_t^{n - m} E((B_T - B_t)^m \mid \F_t) = \sum_{m = 0}^n \binom{n}{m} B_t^{n - m} E((B_T - B_t)^m). \end{align*} By Stein's lemma, $E((B_T - B_t)^m) = (m - 1)(T - t) E((B_T - B_t)^{m - 2})$, thus induction shows that $E((B_T - B_t)^{2m - 1}) = 0$ and (1.2) follows.

Step 2: For $k \in \N_+$ and $l \in \N$,\begin{gather*} \int_0^T B_t^l \,\d B_t = \frac{B_T^{l + 1}}{l + 1} - \int_0^T \frac{l}{2} B_t^{l - 1} \,\d t,\\ \int_0^T (T - t)^k B_t^l \,\d B_t = \int_0^T \left( \frac{k}{l + 1} (T - t)^{k - 1} B_t^{l + 1} - \frac{l}{2} (T - t)^k B_t^{l - 1} \right) \,\d t.\tag{2.1} \end{gather*}

Proof: By Ito's formula,$$ \d(B_t^{l + 1}) = (l + 1) B_t^l \,\d B_t + \frac{1}{2} l(l + 1) B_t^{l - 1} \,\d t, $$ thus$$ \int_0^T B_t^l \,\d B_t = \frac{B_t^{l + 1}}{l + 1} \ev_0^T - \int_0^T \frac{l}{2} B_t^{l - 1} \,\d t = \frac{B_T^{l + 1}}{l + 1} - \int_0^T \frac{l}{2} B_t^{l - 1} \,\d t. $$ Also by Ito's formula,\begin{align*} \d((T - t)^k B_t^{l + 1}) &= -k(T - t)^{k - 1} B_t^{l + 1} \,\d t + (l + 1) (T - t)^k B_t^l \,\d B_t\\ &\peq + \frac{1}{2} l(l + 1) (T - t)^k B_t^{l - 1} \,\d t, \end{align*} thus\begin{gather*} {\small \int_0^T (T - t)^k B_t^l \,\d B_t = \frac{1}{l + 1} (T - t)^k B_t^{l + 1} \ev_0^T + \int_0^T \frac{k}{l + 1} (T - t)^{k - 1} B_t^{l + 1} \,\d t - \int_0^T \frac{l}{2} (T - t)^k B_t^{l - 1} \,\d t}\\ = \int_0^T \left( \frac{k}{l + 1} (T - t)^{k - 1} B_t^{l + 1} - \frac{l}{2} (T - t)^k B_t^{l - 1} \right) \,\d t. \end{gather*}

Step 3:$$ \int_0^T E(B_T^n \mid \F_t) \,\d B_t = \frac{1}{n + 1} (B_T^{n + 1} - E(B_T^{n + 1})). $$

Proof: By (1.1) and (2.1),\begin{align*} &\peq \int_0^T E(B_T^n \mid \F_t) \,\d B_t = \int_0^T \sum_{m = 0}^{\left[ \frac{n}{2} \right]} a_m (T - t)^m B_t^{n - 2m} \,\d B_t = \sum_{m = 0}^{\left[ \frac{n}{2} \right]} a_m \int_0^T (T - t)^m B_t^{n - 2m} \,\d B_t\\ &= \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m \int_0^T \left( \frac{m}{n - 2m + 1} (T - t)^{m - 1} B_t^{n - 2m + 1} - \frac{n - 2m}{2} (T - t)^m B_t^{n - 2m - 1} \right) \,\d t\\ &\peq + \left( \frac{B_T^{n + 1}}{n + 1} - \int_0^T \frac{n}{2} B_t^{n - 1} \,\d t \right)\\ &= {\small \int_0^T \Biggl( \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m · \frac{m}{n - 2m + 1} (T - t)^{m - 1} B_t^{n - 2m + 1} - \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m · \frac{n - 2m}{2} (T - t)^m B_t^{n - 2m - 1} - \frac{n}{2} B_t^{n - 1} \Biggr) \,\d t}\\ &\peq + \frac{B_T^{n + 1}}{n + 1}.\tag{3} \end{align*} Note that\begin{align*} &\peq {\small \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m · \frac{m}{n - 2m + 1} (T - t)^{m - 1} B_t^{n - 2m + 1} - \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m · \frac{n - 2m}{2} (T - t)^m B_t^{n - 2m - 1} - \frac{n}{2} B_t^{n - 1} }\\ &= \Biggl( \frac{a_1}{n - 1} B_t^{n - 1} + \sum_{m = 1}^{\left[ \frac{n}{2} \right] - 1} a_{m + 1} · \frac{m + 1}{n - 2m - 1} (T - t)^m B_t^{n - 2m - 1} \Biggr)\\ &\peq - \Biggl( \sum_{m = 1}^{\left[ \frac{n}{2} \right]} a_m · \frac{n - 2m}{2} (T - t)^m B_t^{n - 2m - 1} \Biggr) - \frac{n}{2} a_0 B_t^{n - 1}\\ &= \left( \frac{a_1}{n - 1} - \frac{n}{2} a_0 \right) B_t^{n - 1} + \sum_{m = 1}^{\left[ \frac{n}{2} \right] - 1} \left( a_{m + 1} · \frac{m + 1}{n - 2m - 1} - a_m · \frac{n - 2m}{2} \right) (T - t)^m B_t^{n - 2m - 1}\\ &\peq - a_{\left[ \frac{n}{2} \right]} · \frac{n - 2\left[ \frac{n}{2} \right]}{2} (T - t)^{\left[ \frac{n}{2} \right]} B_t^{n - 2\left[ \frac{n}{2} \right] - 1}\\ &\stackrel{(1.2)}{=} -a_{\left[ \frac{n}{2} \right]} · \frac{n - 2\left[ \frac{n}{2} \right]}{2} (T - t)^{\left[ \frac{n}{2} \right]} B_t^{n - 2\left[ \frac{n}{2} \right] - 1} = \begin{cases} -\dfrac{1}{2} a_{\frac{n - 1}{2}} (T - t)^{\frac{n - 1}{2}}; & n \text{ is odd}\\ 0; & n \text{ is even} \end{cases}. \end{align*} Thus for odd $n$,\begin{gather*} (3) = \frac{B_T^{n + 1}}{n + 1} - \frac{1}{2} a_{\frac{n - 1}{2}} \int_0^T (T - t)^{\frac{n - 1}{2}} \,\d t = \frac{B_T^{n + 1}}{n + 1} - \frac{1}{2} a_{\frac{n - 1}{2}} · \frac{2}{n + 1} T^{\frac{n + 1}{2}}\\ \stackrel{(1.2)}{=} \frac{B_T^{n + 1}}{n + 1} - \frac{n!!}{n + 1} T^{\frac{n + 1}{2}} = \frac{B_T^{n + 1}}{n + 1} - \frac{E(B_T^{n + 1})}{n + 1}. \end{gather*} For even $n$,$$ (3) = \frac{B_T^{n + 1}}{n + 1} = \frac{B_T^{n + 1}}{n + 1} - \frac{E(B_T^{n + 1})}{n + 1}. $$