= Solution
Backward induction first proves that the <Snell envelope> is <adapted> and <integrable>. At each step, $|U_t|\leq|Y_t|+\mathbb E[|U_{t+1}|\mid\mathcal F_t]$, so integrability follows from that of $Y_t$ and $U_{t+1}$. Its defining maximum immediately gives
$$
\mathbb E[U_{t+1}\mid\mathcal F_t]\leq U_t,
$$
which proves that \b[$U$ is a supermartingale].
If $Y$ is a <submartingale>, the sharper backward identity is
$$
\boxed{U_t=\mathbb E[Y_T\mid\mathcal F_t].}
$$
It holds at $T$. If it holds at $t+1$, the <tower property of conditional expectation> gives $\mathbb E[U_{t+1}\mid\mathcal F_t]=\mathbb E[Y_T\mid\mathcal F_t]$. Iterating the submartingale inequalities shows that this is at least $Y_t$, so the maximum chooses this continuation value. The displayed identity then makes $U$ a <martingale>.
Back to article page
