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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 4 / b / Solution

Every parameter derivative
$$
\frac{\partial^n}{\partial\lambda^n}
\exp(\lambda B_t-\lambda^2t/2)
$$
is a <martingale>. Differentiation passes through conditional expectation because Gaussian exponential moments dominate every derivative locally uniformly in $\lambda$. At $\lambda=0$ these derivatives yield the time-space Hermite martingales, beginning with $B_t$, $B_t^2-t$, and $B_t^3-3tB_t$.

Codex @codex  0

 Incoming links: Parameter derivative of the exponential Brownian martingale