Submartingale Composed with Increasing Convex Function is Submartingale

Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.

Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-submartingale.

Let $f : \R \to \R$ be an increasing convex function such that $\map f {X_t}$ is integrable for each $t \in \hointr 0 \infty$.

Then $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.

Proof
Since $\sequence {X_t}_{t \ge 0}$ is a martingale, we have:


 * $X_t$ is $\FF_t$-measurable

for each $t \in \hointr 0 \infty$.

From Convex Real Function is Measurable and Composition of Measurable Mappings is Measurable:


 * $\map f {X_t}$ is $\FF_t$-measurable

for each $t \in \hointr 0 \infty$.

So $\sequence {\map f {X_t} }_{t \ge 0}$ is $\sequence {\FF_t}_{t \ge 0}$-adapted.

Now let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$.

Let $\expect {\map f {X_t} \mid \FF_s}$ be a version of the conditional expectation of $\map f {X_t}$ given $\FF_s$.

Since $\sequence {X_t}_{t \ge 0}$ is a submartingale, we have:


 * $\expect {X_t \mid \FF_s} \ge X_s$ almost surely.

So, since $f$ is increasing, we have:


 * $\map f {\expect {X_t \mid \FF_s} } \ge \map f {X_s}$ almost surely.

We then have:

So $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.