How to get $E[Y_{n+1}|\mathcal{F}_n]-Y_n=(2b-6)n+b+c-5 ? $

Question

Let $S_n$ be a symmetric simple random walk starting at $0$, and $Y_n=S_n^4-6nS_n^2+bn^2+cn$. How to get $$ E[Y_{n+1}|\mathcal{F}_n]-Y_n=(2b-6)n+b+c-5 ? $$

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It seems the computation is so hard since $$ E[Y_{n+1}|\mathcal{F}_n]-Y_n=E[S_{n+1}^4-6(n+1)S_{n+1}^2+b(n+1)^2+c(n+1)|\mathcal{F}_n]-(S_n^4-6nS_n^2+bn^2+cn) $$

Answer 1

Simply because $Z_n=S_n^4-6nS_n^2+6n^2$ is a martingale, which can be verified by noting that $(\partial_t+\mathcal{L})(x^4-6tx^2+6t^2)=0$ where $\mathcal{L}$ is the infinitesimal operator of the SSRW(it is a difference operator, but it works well just by taking $\mathcal{L}=\frac 12\partial_x^2$ as SSRW is similar to Brownian motion). Thus $E[Z_{n+1}\mid\mathcal{F}_n]=Z_n$ and the desired equality follows.

Then it seems like your equality is kind of incorrect...