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Let $(Z_{n})$ be a discrete-time Adapted Process on State Space $\mathbb{R}$ wrt Filtration $(\mathcal{B}{n})$ s.t. $Z{n} \in L^{1}$ $\forall n \in \mathbb{N}$. Then there exists a Martingale $(X_{n})$ and a Predictable Sequence $(U_{n})$ so that $$Z_{n} = X_{n} + U_{n}$$

Let $Z_{0} := 0$ and let $\mathcal{B}_{0}$ be the Trivial Sigma Algebra. Consider the Discrete-Time Processes:

1. $u_{n}$ = $\mathbb{E}(Z_{n}|\mathcal{B}{n-1}) - Z{n-1}$
2. $d_{n} = Z_{n} - \mathbb{E}(Z_{n}|\mathcal{B}_{n-1})$

Let

1. $U_{n} = \sum\limits_{i=1}^{n} u_{n}$
2. $X_{n} = \sum\limits_{i=1}^{n} d_{n}$

We claim that $X_{n} + U_{n}$ is the decomposition we are looking for. Observe that by definition of Conditional Expectation and because $Z_{n-1} \in L^{0}(\mathcal{B}{n-1})$, we know $u{n} \in L^{0}(\mathcal{B}{n-1})$ $\forall n \in \mathbb{N}$. Since $\mathcal{B}{n}$ forms a Filtration, we then know then that $\sigma(u_{i}) \subset \mathcal{B}{i-1} \subset \mathcal{B}{n-1}$ $\forall i \leq n$, so $u_{i} \in L^{0}(\mathcal{B_{n}})$. Thus $U_{n} \in L^{0}(\mathcal{B}{n-1})$ and $U{n}$ is a Predictable Sequence.

Now consider $(d_{n})$. Observe that because Conditional Expectation is Linear and Conditioning on known information is Idempotent, we have that $$\mathbb{E}(d_{n} | \mathcal{B}{n-1}) = \mathbb{E}(Z{n} - \mathbb{E}(Z_{n}|\mathcal{B}{n-1}) | \mathcal{B}{n-1}) = \mathbb{E}(Z_{n} | \mathcal{B}{n-1}) - \mathbb{E}(Z{n}|\mathcal{B}{n-1}) = 0$$ and $(d{n})$ is a Martingale Difference Sequence. Recall that Accumulation of Martingale Difference Sequence is a Martingale, so $X_{n}$ is a Martingale wrt Filtration $(\mathcal{B}_{n})$.

Finally, observe that $$\begin{align*} X_{n} + U_{n} &= \sum\limits_{i=1}^{n} d_{i} + \sum\limits_{i=1}^{n} u_{i}\\ &=\sum\limits_{i=1}^{n} Z_{i} - \mathbb{E}(Z_{i}|\mathcal{B}{i-1}) + \mathbb{E}(Z{i}|\mathcal{B}{i-1}) - Z{i-1}\\ &=Z_{n} - Z_{0}\\ &=Z_{n}, \end{align*}$$ completing the proof. $\blacksquare$

• Random Variable