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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?

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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.

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