Stopping a Snell Envelope and Its Doob Decomposition

Summary

The document considers a finite-horizon Snell envelope, its Doob decomposition into a martingale and an increasing process, and stopping at a bounded stopping time. It explains that the stopped increasing process at the horizon equals its value at the stopping time, since the stopping time cannot exceed that horizon.

For the martingale condition, the answer uses the stopped decomposition and equality of initial values to show that if the stopped Snell envelope is a martingale, the expected accumulated increasing component is zero. Because that component is nonnegative, it must itself be zero. Conversely, if the increasing component at the stopping time is zero, monotonicity and its zero initial value make the stopped component vanish throughout, leaving the stopped envelope equal to the stopped martingale. The argument is a concise theoretical proof and assumes the stated Doob decomposition and relevant expectations are well defined; it does not provide an application to a pricing or exercise problem.

Key ideas

• The Snell envelope is a supermartingale that dominates the underlying process.
• Stopping the increasing component at a bounded stopping time gives its value at that time.
• A stopped Snell envelope that is a martingale has zero expected accumulated increasing component.
• Nonnegativity then implies that the increasing component at the stopping time is zero.
• If that component is zero, the stopped envelope reduces to the stopped martingale.

Tags

# Answer by Gordon (score 5, accepted)


# 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.

## Answer by Gordon (score 5, accepted)

https://quant.stackexchange.com/a/47134

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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