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

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

Is the above reasoning correct?

• Yes, it is correct. May 15, 2020 at 13:17
• You have $X_n(\omega) \geq X_m(\omega)$ at each $\omega$. How does that tell you that $E[X_n | \mathcal F_m ](\omega)\geq E[X_m | \mathcal F_m](\omega)$? You need to explain this. (Essentially, why is the conditional expectation of a non-negative random variable non-negative?) May 17, 2020 at 12:21
• Do you mean that I am not allowed, starting from $X_n(\omega) \geq X_m(\omega)$, to take expectation on both sides and to condition with respect to $\mathcal{F}_m$ finally getting $\mathbb{E}(X_n(\omega)|\mathcal{F}_m) \geq \mathbb{E}(X_m(\omega)|\mathcal{F}_m) = X_m$? Or do you mean that my reasoning is correct BUT I have to clarify why the conditional expectation of a non-negative random variable is non-negative as well? May 17, 2020 at 13:50
• When applying expectation and conditioning with respect to $F_m$ on both sides, I have simply applied monotonicity property of conditional expectation, according to which: if $X_n(\omega)\geq X_m(\omega)$ a.s., then $\mathbb{E}(X_n|\mathcal{F}_m) \geq \mathbb{E}(X_m|\mathcal{F}_m)=X_m$. Hence, why do you stress the importance of explaining why the conditional expectation of a non-negative random variable is non-negative? I have just used monotonicity property of conditional expectation and the random variables are not necessarily non-negative (this is not specified) @астонвіллаолофмэллбэрг May 17, 2020 at 14:44
• The LHS is clearly non-negative, because $X_n-X_m$ is non-negative. However, the right hand side will be negative if $Y$ has non-zero measure (for any random variable, $Z$, $E[Z1_{Z < 0}]$ will be non-positive, and negative if $Z<0$ has positive measure). It follows that the RHS and LHS are $0$ i.e. $Y$ must be of measure zero. This implies $E[X_n|\mathcal F_m] \geq E[X_m | \mathcal F_m]$ almost surely. Finally, this proof is as good as the proof of "conditional expectation preserves monotonicity". May 18, 2020 at 3:35

• 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.

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.