Proof: for martingale that $E(X_{n\geq m}|F_m)=X_m$ (by induction).

Question

I just wanted to know if my proof by induction of this statement $E(X_{n\geq m}|F_m)=X_m$ for a martingal is correct?

initialisation: $E(X_{n+1}|F_n)=X_n$ by def of a martingale
assumption: $\forall k \in \mathbb{N},E(X_{n+k}|F_n)=X_n$
step: As $F_n\subseteq F_{n+k}$ so $E(X_{n+k+1}|F_n)=E(E(X_{n+k+1}|F_{n+k})|F_n)=E(X_{n+k}|F_n)=X_n$
More precisally using this justification:
-$E(X_{n+k+1}|F_{n+k})=X_{n+k}$ by definition of a martingale
-$E(X_{n+k}|F_n)=X_n$ by assumption

Q.E.D
Is it correct?

Answer 1

The proof is correct.

it is correct. but better to write E(Xn|Fm)=Xm∀n≥m. – Kurt G. Sep 28 at 10:56