Stopped Supermartingale is Supermartingale/Corollary

Corollary to Stopped Supermartingale is Supermartingale

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.

Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-supermartingale.

Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.

Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.

Then we have that $X_n^T$ is integrable for each $n \in \N$ and:

$\expect {X_n^T} \le \expect {X_0}$ for each $n \in \Z_{\ge 0}$.

Proof

From Stopped Supermartingale is Supermartingale:

$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale

and so:

$\sequence {X_n^T}_{n \ge 0}$ is integrable.

From Expected Value of Supermartingale Less Than or Equal To Initial Expected Value, we have:

$\expect {X_n^T} \le \expect {X_0^T}$

We have that $\map T \omega \ge 0$ for each $\omega \in \Omega$, so $0 \wedge T = 0$.

So, we have:

$X_0^T = X_0$

and so:

$\expect {X_0^T} = \expect {X_0}$

giving the result.

$\blacksquare$