= Solution
Differentiate the conditional identity from part a. To justify doing so, fix a compact parameter interval $|\lambda|\leq L$. Every $n$th derivative of $M_\lambda(t)$ is a polynomial in $B_t,t,$ and $\lambda$ times $M_\lambda(t)$, and its absolute value is bounded by
$$
C_{n,L,t}(1+|B_t|^n)e^{(L+1)|B_t|}.
$$
This bound is integrable because a Gaussian random variable has every polynomially weighted exponential moment. Dominated differentiation of conditional expectation therefore gives
$$
\mathbb E\left[\frac{\partial^nM_\lambda(t)}{\partial\lambda^n}\,\middle|\,\mathcal F_s\right]
=\frac{\partial^nM_\lambda(s)}{\partial\lambda^n}.
$$
Thus every <parameter derivative of the exponential Brownian martingale> is itself a martingale.
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