Show that $M_n = \sum_{i=1}^n \alpha_k(X_k -X_{k-1})$ is a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$. $\textbf{question}$
Let $\{F_n, \ n\in \mathbb{N}\}$ be a filtration and $\{X_n, \ n\in \mathbb{N}\}$ a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$, with $X_0 = 0$ , and such additionally there exists $C \in (0,\infty)$ such that for all $n \in\mathbb{N}$:
$E((X_n - X_{n-1})^2) \leq C$
Additionally, $\{\alpha_k, \ k \in \mathbb{N}\}$ denotes a sequence of real numbers.
Show that $M_n = \sum_{i=1}^n \alpha_k(X_k -X_{k-1})$ is a square integrable martingale w.r.t. $\{F_n, \ n\in \mathbb{N}\}$.
$\textbf{problem}$
I don't understand how to show $M_n$ is a square integrable martingale $\textbf{with respect to}$ $\{F_n, \ n\in \mathbb{N}\}$.
Do I have to show that $E(M_n^2|F_{n-1})=M_{n-1}^2$ in which case I'll be fine.
Or do I have to show:
$sup_{n \geq 0}E(M_n^2)<\infty$
if so can I write:
$sup_{n \geq 0}E[(\sum_{i=1}^n \alpha_k(X_k -X_{k-1}))^2] \leq sup_{n \geq 0}E[(\sum_{i=1}^n \alpha_k^2(X_k -X_{k-1})^2]\leq M sup_{n \geq 0}E[(\sum_{i=1}^n(X_k -X_{k-1})^2]\leq ME[m(X_m -X_{m-1})^2]\leq MmC\leq \infty$
 A: For a process $\{Y_n\}_{n\in\mathcal{N}}$ to be a martingale wrt to a filtration $\mathcal{F}=\{\mathcal{F}_n\}_{n\in\mathcal{N}}$ you have to show that :

*

*$Y_n$ is adapted to $\mathcal{F}$

*$\mathbb{E}[|Y_n|]$ exists for all $n$

*$\mathbb{E}[Y_n|\mathcal{F}_{n-1}] = Y_{n-1}$
So you have to show both basically to answer you question. The inequality shows that your process is $L^1$ for all $n$. The equality that it is a martingale.

Please do not ignore the first bullet point which justify the fact that you can write the left term of my third bullet point.
EDIT:
Sorry I misread the question and realize that only after struggling to prove Bullet 3.
What you have to show is that $\{M_n\}$ is a martingale and justify that it belongs to $L^2$.
Appendix to show integrability of $Mn^2$:
The two following propositions are not equivalent :

*

*$\sup_n \mathbb{E}[M_n^2] < \infty$

*$\mathbb{E}[M_n^2] < \infty,$ $\forall n \in \mathbb{N}$
The first is more restrictive and tells you that as $n$ goes larger, the quantity remain finite which is equivalent to saying that $\lim_{n\rightarrow\infty}\mathbb{E}[M_n^2] < \infty$
The second only tells you that the process is $L^1$ for all integers. So $1 \implies 2$ but nothing else.
To prove (2) $\forall n \in \mathbb{N}$:
$$
\mathbb{E}[(\sum_{k=1}^{n} \alpha_k(X_k - X_{k-1}))^2] \leq \sum_{k=1}^n \alpha_k^2 \cdot \mathbb{E}[\sum_{k=1}^n (X_k - X_{k-1})^2] 
$$
From this I let you continue.
