Martingale - integrability condition

Question

We know that for a process $X=(X_t)_{t \in \mathbb R}$ to be a martingale has to satisfy also the integrability condion, namely that $E[|X_t|]< +\infty$ for all t.

1) I guessed that the absolute value is necessary for avoiding the fact that the expectation could be $-\infty$. Is this the reason?

2) However I also read that we can replace the integrability condition with the assumption that X is positive (generalized positive martingale). Why this is the case? I thought that is connect to the constraction of the expectation, namely $E[X]=E[X^+]-E[X^-]$. And that if it is positive there is not the possibility to have the problem of $+\infty -(+\infty)$ but if this is the reason, then I can still have $E[X]=\infty$

Answer 1

We can define martingales by assuming that $EX_t <\infty$ for all $t$ or $EX_t >-\infty$ for all $t$. We usually assume that $E|X_t| <\infty$ for all $t$ for the sake of simplicity. The following consideration are needed if we do nor assume that $E|X_t| <\infty$ for all $t$.

For $X \geq 0$ it is possible define $E(X|\mathcal G)$ through a limiting process: $E(X|\mathcal G)=\lim_n E(X\wedge n|\mathcal G)$ (which gives an extended real valued measurable function); hence we can define positive martingales without assuming that the expectations are finite. This also allows us to define $E(X|\mathcal G)$ with just the assumption $EX <\infty$ (or $EX >-\infty)$.

Answer 2

(1) Assuming $\mathsf{E}|X_t|<\infty$ guarantees that $\mathsf{E}X_t$ is well-defined and is finite (in this case $\mathsf{E}X_t^{+} \vee \mathsf{E}X_t^{-}<\infty$). This condition is necessary for the existence of conditional expectations, i.e. $\mathsf{E}[X_t\mid \mathcal{F}_s]$, that are typically defined for integrable random variables.

(2) $\mathsf{E}[X_t\mid \mathcal{F}_s]$ can be defined as $$ \mathsf{E}[X_t^{+}\mid \mathcal{F}_s]-\mathsf{E}[X_t^{-}\mid \mathcal{F}_s] $$ provided that at least one (conditional) expectation is a.s. finite. This is satisfied in the case of positive $X_t$ (although, $\mathsf{E}[X_t\mid \mathcal{F}_s]$ can be infinite).