Submartingale and stopping time

Question

Let {$X_1, \dots, X_n$} be a submartingale, and let $T$ be a stopping time for {$X_i, 1\leqslant i \leqslant n$}. Show that $ E(\mid X_T \mid) \leqslant 2E(X_n^+)-E(X_1)$. The corresponding result for supermartingales, which may obtained by replacing $X_1$ by $-X_i$, is $E(\mid X_T \mid) \leqslant 2E(X^-_n)+E(X_1)$.

Answer 1

For each $n$, we have $$|X_n| = X_n^+ + X^-_n = X_n^+ + (X_n^+ - X_n) = 2X_n^+ - X_n.$$ Then $X_n^+\leqslant |X_n|$ so $X_n^+$ is integrable, and \begin{align} \mathbb E[X_{n+1}^+\mid\mathcal F_n] &= \mathbb E[X_{n+1}^+-X_{n+1}^-+X_{n+1}^-\mid\mathcal F_n]\\ &=\mathbb E[X_{n+1}\mid\mathcal F_n] + \mathbb E[X_{n+1}^-\mid\mathcal F_n]\\ &\geqslant X_n + \mathbb E[X_{n+1}^-\mid\mathcal F_n]\\ &\geqslant X_n, \end{align} so that $X_n^+$ is a submartingale. It follows that \begin{align} \mathbb E[|X_T|] &= \mathbb E[2X_n^+ - X_n]\\ &= 2\mathbb E[X_n^+] - \mathbb E[X_n]\\ &\leqslant 2\mathbb E[X_n^+] - \mathbb E[X_1]. \end{align}