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.


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?

  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$ May 15, 2020 at 13:17
  • $\begingroup$ 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?) $\endgroup$ May 17, 2020 at 12:21
  • $\begingroup$ 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? $\endgroup$ May 17, 2020 at 13:50
  • $\begingroup$ 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 $\endgroup$ May 17, 2020 at 14:44
  • 1
    $\begingroup$ 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". $\endgroup$ May 18, 2020 at 3:35

1 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.

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.


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