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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.

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  • $\begingroup$ 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 $Z^{\tau} =\{Z_n^{\tau}\}_{n=1}^N$? $\endgroup$ – Gordon Aug 15 at 19:04
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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.

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