# Find C so that $C^n\lambda^{S_n}$ is a martingale

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?

• 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 $C^n$. Nov 29, 2022 at 14:46
• 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. Nov 29, 2022 at 14:56
• i am just a student but i ve found the same result than you. Nov 30, 2022 at 13:45