Show that $X$ is a submartingale, given some assumptions. Is the following solution correct?

Question

Let $X=(X_n)_{n>0}$ be an increasing sequence of integrable r.v.'s, each $X_n$ being $\mathcal{F}_n$-measurable. Show that $X$ is a submartingale.

MY SOLUTION

What I have to show is that, given that:

$1)$ $X_n(\omega) < X_{n+1}(\omega)$, each $n$ (or, equivalently, $X_m(\omega)\leq X_n(\omega)$, each $m\leq n$);

$2)$ $\mathbb{E}(|X_n|)< \infty$, each $n$;

$3)$ $X_n$ is $\mathcal{F}_n$-measurable, each $n$;

then $X$ is a submartingale, that is:

$1.1)$ $\mathbb{E}(|X_n|)< \infty$, each $n$;

$1.2)$ $X_n$ is $\mathcal{F}_n$-measurable, each $n$;

$1.3)$ $\mathbb{E}(X_n|\mathcal{F}_m) \geq X_m$ a.s., each $m\leq n$.

Clearly, $1.1)$ corresponds to $2)$ and $1.2)$ corresponds to $3)$. Hence, one is left with proving $1.3)$.

To this, one can state that, given assumption $1)$, for each $m\leq n$: \begin{equation} X_n(\omega)\geq X_m(\omega) \end{equation} Then, taking expectation on both sides and conditioning with respect to $\mathcal{F}_m$, taking into account assumption $3)$, one has that: \begin{equation} \mathbb{E}(X_n(\omega)|\mathcal{F}_m) \geq \mathbb{E}(X_m(\omega)|\mathcal{F}_m) = X_m \end{equation} which is exactly point $1.3)$.

Is the above reasoning correct?

Answer 1

Only to answer the question from the comments:

• Integrability and Adaptedness are already part of the assumptions;

• We need to show that $\mathbb E[X_n | \mathcal F_m] \geq X_m$ for each $n \geq m$. By adaptedness and linearity, it is enough to show that $E[X_n-X_m | \mathcal F_m]$ is a non-negative random variable;

• But this is clear : let $Y$ be the event $\{E[X_n-X_m | \mathcal F_m] < 0\}$. Since $E[X_n-X_m| \mathcal F_m]$ is a $\mathcal F_m$ measurable random variable, the event $Y$ belongs in $\mathcal F_m$ i.e. $1_Y$ (the indicator function of $Y$) belongs to $\mathcal F_m$;

• By definition of conditional expectation, $E[(X_n-X_m)1_Y] = E[E[X_n-X_m | \mathcal F_m] 1_Y]$. The LHS of this is non-negative since $X_n \geq X_m$ everywhere, and therefore on $Y$. Therefore, the RHS is non-negative. However, $1_YE[X_n-X_m | \mathcal F_m]$ is a non-positive random variable! So the integral can be non-negative precisely when $1_Y$ is $0$ almost surely i.e. $Y$ has measure zero. This is the same as $E[X_n | \mathcal F_m] \geq X_m$ almost surely.

Finally, all conditions are complete and we have that $X_m$ is an $\mathcal F_m$-submartingale.

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Note that we have proved above a more general statement :

Let $X,Y$ be random variables on a probability space $(\Omega,\mathcal F,P)$ and let $\mathcal G \subset \mathcal F$ be any $\sigma$-algebra. Then, if $X \geq Y$ we have $E[X | \mathcal G] \geq E[Y | \mathcal G]$.

In words, if one random variable dominates another, then even if I provide you with any information, the domination will continue to hold. This is obvious when you think of it.