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I am trying to prove the following are equivalent

  1. $(X_t,\mathcal{F}_t)$ is a submartingale, i.e. $E[X_t|\mathcal{F}_s]\ge X_s\;\forall t\ge s$
  2. $\int_A X_t dP\ge \int_A X_s dP, \;\;\;\forall A\in\mathcal{F}_s$

Is my attempt below correct? Is there any better/quicker way?

edamondo
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1 Answers1

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Using the same reasoning as in this post, we can see that for random variables $Y$, $Z$ which are $\mathcal{G}$-measurable, $Y\ge Z$ iff $E[Y1_A]\ge E[Z1_A]$ for every $A\in\mathcal{G}$.

$1)\implies 2)$ : Suppose $E[X_t|\mathcal{F}_s]\ge X_s$ . Since $1_A$ is positive we get by the monotonicity of expectation that $E[E[X_t|\mathcal{F}_s]1_A]\ge E[X_s1_A]$. But since $E[E[X_t|\mathcal{F}_s]1_A]=E[E[X_t1_A]|\mathcal{F}_s]=E[X_t1_A]$ the result follows.

$2\implies 1)$ : Suppose $E[X_t1_A]\ge E[X_s1_A]$ for every $A\in\mathcal{F}_s$. Now note $E[X_t1_A]=E[E[X_t1_A]|\mathcal{F}_s]=E[E[X_t|\mathcal{F}_s]1_A]$ so we get $E[E[X_t|\mathcal{F}_s]1_A]\ge E[X_s1_A]$ for every $A\in\mathcal{F}_s$ and so by the remark at the beginning combined with the fact that $E[X_t|\mathcal{F}_s]$ is $\mathcal{F}_s$-measurable, we get $E[X_t|\mathcal{F}_s]\ge X_s$.

edamondo
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