I was completely stuck. I know, that $X_n^3=M_n+A_n$, where $M_n$ is a martingale. Please help me :(
2026-03-30 20:44:06.1774903446
Let us determine the Doob's decomposition of $(X_n^3)$, where $X_n$ is the one-dimensional symmetric random walk.
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Apply the binomial theorem to one step of the random walk, $$ X_{n+1}^3=(X_n+\Delta X_n)^3=(M_n+A_n)+3X_n^2ΔX_n+3X_n +ΔX_n, $$ using $ΔX_n^2=1$. So with $M_{n+1}=M_n+(3X_n^2+1)ΔX_n$ and $A_{n+1}=A_n+3X_n$ you get a martingale and a predictable process.