Martingale derivation by direct calculation

Question

I'm reading the proof of a theorem and stumbled across the following derivation which I cannot replicate myself.

Let $W(t)$ be a $Q$-martingale and be given by $W(t) = B(t) + \mu t$ with $B(t)$ a standard brownian motion under the $P$-measure. Let $\Lambda(t) = E_{P}(\Lambda|\mathcal{F}_t)$ with $\Lambda$ defined as:

$\Lambda = \frac{dQ}{dP}(B_{[0,T]}) = e^{-\mu B(T) - \frac{1}{2} \mu^{2} T}$.

The following derivation is the one which I cannot replicate (the second step of it):

$E_{P}[W(t)\Lambda(t)|\mathcal{F}_{s}] = E_{P}\left((B(t)+\mu t) e^{-\mu B(t) - \frac{1}{2} \mu^{2} t } | \mathcal{F}_{s}\right) = W(s)\Lambda (s)$.

The purpose of this is to prove that $W(t) \Lambda(t)$ is a $P$-martingale by direct calculation but I have failed to to so.

Many thanks!

Answer 1

Let $E:=E_P$, $B_t:=B(t)$, $m:=\mu$, $F_s:=\mathcal F_s$. We have to show that $L=R$ if $0\le s\le t$, where $$L:=E((B_t+mt)e^{-mB_t-m^2t/2}|F_s),\quad R:=(B_s+ms)e^{-mB_s-m^2s/2}.$$

By properties of the conditional expectation and the independence of $B_t-B_s$ of $F_s$, we have $$L=E((B_t+mt)e^{-mB_t-m^2t/2}|F_s) \\ =e^{-mB_s-m^2t/2}E((B_t+mt)e^{-m(B_t-B_s)}|F_s) \\ =e^{-mB_s-m^2t/2}[(B_s+mt)Ee^{-m(B_t-B_s)} +E(B_t-B_s)e^{-m(B_t-B_s)}] \\ =e^{-mB_s-m^2t/2}\Big[(B_s+mt)e^{m^2(t-s)/2} -\frac d{dm}\,e^{m^2(t-s)/2}\Big] \\ =e^{-mB_s-m^2t/2}[(B_s+mt)-m(t-s)]e^{m^2(t-s)/2}=R, $$ as desired.