3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 19:04
5
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.