# Relationship between martingales and positive submartingales

Suppose $$X_n$$ is a submartingale with $$X_n > 0$$ for all \$n.

There must exist $$A_n$$, $$M_n$$ such that $$A_n$$ is predictable and increasing and $$M_n$$ is a martingale so that $$X_n = A_nM_n$$.

How is it possible to show this?

This is actually quite easy (unfortunately), so I'll answer myself. Put $$A_0 = 1, \quad A_n = \prod_{i=1}^n\frac{E(X_i|\mathcal{F}_{i-1})}{X_{i-1}}$$
$$A_n$$ is clearly well defined as each $$X_i > 0$$, is predictable by the $$\mathcal{F}_{n-1}$$ measurability of both numerator and denominator, and it is increasing since $$A_{n+1} = A_n \cdot \underbrace{\frac{E(X_{n+1}|\mathcal{F}_n)}{X_n}}_{\ge 1 \text{ by submartingale property}}$$
Now define $$M_n = \frac{X_n}{A_n} \text{ so that }\\ E(M_{n+1}|\mathcal{F}_{n}) = E(X_{n+1}|\mathcal{F}_n) \prod_{i=1}^{n+1}\frac{X_{i-1}}{E(X_i|\mathcal{F}_{i-1})} = X_n \prod_{i=1}^{n}\frac{X_{i-1}}{E(X_i|\mathcal{F}_{i-1})} = \frac{X_n}{A_n} = M_n$$
in the case of $$n \ge 1$$ and for $$n = 0$$ the RHS above simply reduces to $$X_0$$ which is the same thing.