I think this is probably a very easy question but I haven't worked with $\sigma$-algebras in depth for a long time now so am finding myself a little rusty. Would be very grateful if someone could give me a (careful) proof of the following (I'm pretty sure it's true!). I guess I'm missing the right way of characterising the elements of the join of two $\sigma$-algebras appropriately.

Let $M_t$ be a martingale with respect to the filtration $\mathcal{F}_t$ on some probability space $(\Omega, \mathcal{F}, P)$. Assume that $\mathcal{G}_t \subseteq \mathcal{F}$ where for each $t \geq 0$, $\mathcal{G}_t$ is independent of $\mathcal{F}_t$ and let $\mathcal{H}_t := \mathcal{F}_t \vee \mathcal{G}_t$. Then $M_t$ is also a martingale with respect to $\mathcal{H}_t$.



1 Answer 1


Here $\mathcal F_t\vee\mathcal G_t$ denotes the smallest (for the inclusion) $\sigma$-algebra that contains both $\mathcal F_t$ and $\mathcal G_t$.

Integrability condition is already verified, so we just need to check that for all $s\leqslant t$, we have $$\mathbb E\left[M_t\mid \mathcal F_s\vee\mathcal G_s\right]=M_s.$$ Let $\mathcal B:=\left\{F\cap G,F\in \mathcal F_s,G\in \mathcal G_s \right\}$. It can be checked that $\mathcal B$ is $\pi$-system which generates $\mathcal F_s\vee\mathcal G_s$,. The set of elements $B$ of $\mathcal F$ such that $\int_B M_tdP=\int_BM_sdP$ is a $\lambda$-system. So we just have to show that for all $F\in\mathcal F_s$, $G\in\mathcal G_s$, we have $$\int_{F\cap G}M_tdP=\int_{F\cap G}M_sdP.$$ Since $\mathcal F_t$ is independent of $\mathcal G_t$, $\mathcal F_t$ is also independent of $\mathcal G_s$. Conclude using the fact that $M_t\chi_F$ is $\mathcal F_t$ measurable.

  • $\begingroup$ Do you have a litterature reference for this? $\endgroup$
    – noidea
    Jan 27, 2018 at 20:46
  • $\begingroup$ Dear Davide, what does that inverted wedge between sigma algebras mean? $\endgroup$
    – Nobody
    Jun 9, 2022 at 13:24
  • 1
    $\begingroup$ @Nobody I have edited. $\endgroup$ Jun 10, 2022 at 14:35

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