Let $ X_1,X_2,... \sim i.i.d.$ $P(X_n=1)=p=1-P(X_n=-1)=1-q, n=1,2,...$, $S_n=\Sigma_{k=1}^{n}X_K$ and $F_n$ is filtration generated by $X_n$.

For fixed $\lambda>0$ we aim to find cosntant C that $Z_n = C^n\lambda^{S_n}$ is a martingale with respect to filtrafion $F_n$. We have to start with $$ E(Z_n |F_{n-1}) = E(C^n\lambda^{S_n} | F_{n-1}) = C^{n-1}E(\lambda^{X_1}\cdot ... \cdot \lambda^{X_{n-1}} | F_{n-1} )CE(\lambda^{X_n}|F_{n-1}) = Z_{n-1}CE(\lambda^{X_n}) $$ $$ = Z_{n-1}C(\lambda^1p+\lambda^{-1}q)$$ This has to equal $Z_{n-1}$ so $$C=(\lambda^1p+\lambda^{-1}q)^{-1}$$ Is this a great answer?

  • $\begingroup$ The approach is OK but somewhere the $C^n$ disappears and comes back as $C$. And reading the title it is not clear if you are looking for $\lambda,C$ or $C^n$. $\endgroup$
    – Kurt G.
    Nov 29, 2022 at 14:46
  • $\begingroup$ I fixed the problems @KurtG. I think it is more clear now. Firstly we need $C^{n-1}$ to have $Z_{n-1}$ so i broke $C^n= C^{n-1}C$ then using then I used measurability. $\endgroup$
    – nodis6
    Nov 29, 2022 at 14:56
  • 2
    $\begingroup$ i am just a student but i ve found the same result than you. $\endgroup$
    – X0-user-0X
    Nov 30, 2022 at 13:45


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