I am trying to solve this question about optimal stopping theory. I don't know how to get started. Any hints would be very helpful

Let $$Z = (Zn)_{n=0,1,...,N}$$ be the Snell envelope of $$X = (Xn)_{n=0,1,...,N}$$ and $$τ ∈ T_{0,N}$$. Let $$Z_n = M_n − A_n$$ be the Doob decomposition of Z, then $$Z_n^τ = M_n^τ − A_n^τ$$ is the Doob decomposition of $$Z_n^τ$$ (do not prove this).

(a) Explain why $$Aτ$$ = $$A_N^τ$$ .

(b) Hence, prove that $$Z_τ$$ is a martingale if and only if $$A_τ$$ = 0.

• Since $\tau \le N$, it is obvious that $A_N^{\tau} = A_{N\wedge \tau} = A_{\tau}$. Given that $Z_{\tau}$ is a single random variable, it does not make sense to say it is a martingale. Do you mean $Z^{\tau} =\{Z_n^{\tau}\}_{n=1}^N$? – Gordon Aug 15 at 19:04

The Snell envelope is the smallest super-martingale that is greater than $$X$$. Since $$\tau \le N$$, it is obvious that $$A_N^{\tau} = A_{N\wedge \tau} = A_{\tau}$$.
For part (b), note that, from the Doob decomposition, $$M$$ is a martingale, $$A$$ is increasing, $$M_0=Z_0$$, and $$A_0=0$$. If $$Z^{\tau}= \{Z_n^{\tau}\}_{n=1}^N$$ is also a martingale, then \begin{align*} E(A_{\tau}) &= E(A_N^{\tau}) \\ &= E(M_N^{\tau} - Z_N^{\tau}) \\ &= M_0 - Z_0=0. \end{align*} Consequently, $$A_{\tau}=0$$, as $$A_{\tau}$$ is non-negative.