Martingale convergnce theorem application I'm trying to solve the following exercise which requires the application of martingale convergence thm.

Let $(M_n)_n$ be a martingale w.r.t $(F_n)_n$ and let
$$C_n = M_n - M_{n-1}, \quad n \in \mathbb{N}$$
Prove that if $E[M_0^2] < \infty$ and $\sum_{n \in \mathbb{N}} E[C_n^2] < \infty$, then there exists a r.v. $M$ such that $M_n \rightarrow M$ a.s. and in $L^2$.


What I need to do is to show that $\sup_{n \in \mathbb{N}}E[M_n^2] < \infty$, in order to apply the theorem mentioned above.
I've seen on the inernet that if $(M_n)_n$ is a square integrable martingale, then the increments $C_n$ are orthogonal in $L^2$ and I can write $$E[M_n^2] = E[M_0^2] + \sum_{k=1}^n  E[C_n^2] , \quad (\star)$$.
If I know this, of course I'm done since I can take the supremum both sides and use the convergence of the series and the finitess of $E[M_0^2]$ and I get the thesis.

So, my questions are:

*

*Is my martingale square integrable? I can't show it.

*How the independence of the increments imply ($\star$) ?

*Are there other ways to attack the problem?

 A: 
What I need to do is to show that $\sup_{ n\in \mathbb{N}} \mathbb{E}[M_{n}^{2}]<\infty$, in order to apply the theorem mentioned above.

As already mentioned, it is enough to show that

$\mathbb{E}[M_n^2] = \mathbb{E}[M_0^2] + \sum_{k=1}^n  \mathbb{E}[C_k^2] , \quad (\star)$

because we then simply have 
$$\sup_{n \in \mathbb{N}} \mathbb{E}[M_{n}^{2}] \le \mathbb{E}{M_{0}^{2}}+\sum_{k=1}^{\infty}\mathbb{E}[C_{k}^{2}],$$
which is finite by assumption.

I've seen on the inernet that if $(M_{n})_{n}$ is a square integrable martingale, then the increments $C_{n}$ are orthogonal in $L^{2}$.

OK, but it is also enough to know, that increments are square integrable. Let us explain it in more details.
For simplicity put $M_{0}:=C_{0}$.
$$\mathbb{E}[M_{n}^{2}]=\mathbb{E}[(C_{0}+\cdots C_{n})^2]=\mathbb{E}[M_{0}^2]+\sum_{k=1}^{n}\mathbb{E}[C_{k}^2]+\sum_{0\le i \not{=}j \le n}\mathbb{E}{C_{i}C_{j}}.$$
We will show that for every $i\not= j$ we have $\mathbb{E}[C_{i}C_{j}]=0$, which will prove $(\star)$.
Because $\sum_{k=0}^{\infty} \mathbb{E}[C_{k}^{2}] <+\infty$, then $\mathbb{E}[C_{k}^2]$ is obviously finite for all $k$. 
Suppose that $i<j$, then $$\mathbb{E}{C_{i}C_{j}}= \mathbb{E}\Big[\mathbb{E}[C_{i}C_{j}|\mathcal{F}_{i} ] \Big],$$ 
Note that we are using here tower property of expectation and finiteness of second moments 


*

*Cauchy- Schwarz inequality tells us that $C_{i}C_{j}$ is in fact integrable and the whole business with conditional expectations makes sense.


Let us now look closer into inside expectation. Since $C_{i}$ is $\mathcal{F}_{i}$ measurable, we have
$$\mathbb{E}[C_{i}C_{j}|\mathcal{F}_{i}]=C_{i}\cdot \mathbb{E}[C_{j}|F_{i}].$$
We will now use matringale property to see, that last term is equal to zero.
$$\mathbb{E}[C_{j}|F_{i}]=\mathbb{E}\Big[\mathbb{E}[C_{j}|F_{j-1}]\Big|F_{i}\Big]=0.$$
In last equality we have used the following:


*

*$i<j$ implies that $i\le j-1.$

*$\mathbb{E}[C_{j}|F_{j-1}]=0$. 
To see the second dot, simply write:
$\mathbb{E}[C_{j}|F_{j-1}]= \mathbb{E}[M_{j}-M_{j-1}|F_{j-1}]=0 \iff \mathbb{E}[M_{j}|F_{j-1}]=M_{j-1}.$
Now for the questions:

Is my martingale square integrable? I can't show it.

Now we can say that it is. 

How the independence of the increments imply (⋆) ?

I think it is visible in the posted proof, but be aware that independence is something different then $\mathbb{E}[C_{i}C_{j}]=0$, that we have actually used.
