Inequality related to Doob's martingales

Question

I have the following question on Doob's martingales.

Let $A$ be an integrable $\mathcal F$-measurable random variable on the stochastic basis $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$. Define the process $X = (X_t), t\in \mathbb T$ via $X_t := E [A \mid \mathcal F_t ] \forall t\in \mathbb T$.

Using the tower property for conditional expectation, show that $X$ is a martingale. Furthermore, show that $E [\lvert X_t \rvert] ≤ E [\lvert A \rvert ]$ holds for all $t \in \mathbb T$.

Showing $X$ is a martingale is pretty straight-forward, as $$E(X_{t+1}\mid\mathcal F_t)=E( E [A \mid \mathcal F_{t+1} ]\mid\mathcal F_t)=E(A\mid \mathcal F_t)=X_t$$

For the inequality, I tried to use Jensen's inequality in vain: $$E(\lvert X_t \rvert)\ge\lvert E(X_t) \rvert=\lvert E(E(A\mid \mathcal F_t)) \rvert=\lvert E(A) \rvert=\ ?$$ $$ E(\lvert A \rvert)\ge \lvert E(A) \rvert=\lvert E(X_t) \rvert=\ ?$$

Is there a way to get this inequality?

Answer 1

Use Jensen's inequality for conditional expectations, i.e. $$ \varphi({\rm E}[X\mid\mathcal{G}])\leq {\rm E}[\varphi(X)\mid\mathcal{G}] $$ whenever $X$ is integrable and $\varphi$ is convex on $\mathbb{R}$. This immediately yields $$ |X_t|\leq {\rm E}[|A|\mid\mathcal{F}_t] $$ and so $$ {\rm E}[|X_t|]\leq {\rm E}[{\rm E}[|A|\mid\mathcal{F}_t]]={\rm E}[|A|] $$ by the tower property.