Martingale, decomposition, expectancy value I am trying to understand a proof in my book but i can't figure out a few things. Definitions first:
$(\Delta X)_n :=X_n-X_{n-1}$ with $X_{\alpha-1}:=X_\alpha$, if $\alpha>-\infty$, so $(\Delta X)_\alpha=0$ if $\alpha>-\infty$. 
A process $H=(H_n)_{n\in T}$ is called $\mathbb{F}$-predictable if $H_n$ is $\mathcal{F}_{n-1}$-measurable for all $n\in T$. Let $\alpha>-\infty$, $H$ be a $\mathbb{F}$-predictable real process and $X$ a $\mathbb{F}$-adapted real process, then $$(H\circ X)_n:= \sum^n_{j=\alpha+1}H_j\Delta X_j, n\in T$$ is called h-transform of $X$.
Let $\alpha>-\infty$. The process $$[X,Y]_n:=\sum^n_{i=\alpha +1}\Delta X_i\Delta Y_i$$ is called covariation for $\mathbb{F}$-adapted real processes $X$ and $Y$ and $[X]:=[X,X]$ is called square variation of $X$.
And here are the equations i don't understand.
"we can conclude for the martingale $H_\alpha M_\alpha +H\circ M$ 
\begin{align*}
E(H_\alpha M_\alpha +(H\circ M)_n)^2 &=E(H_\alpha M_\alpha)^2+E[H\circ M]_n\\
&=E(H_\alpha M_\alpha)^2+EH^2\circ [M]_n\\
&\le EM^2_\alpha+E[M]_n
=EM^2_n"
\end{align*}
I don't understand the first equation. How does he get there from the left side? And i am not quite sure about the last equation. Does it hold because $M$ is a martingale? Does that mean that $E[M]_n=0$? And is $EM^2_\alpha = EM^2_n$ because M is a martingale which means that the prize i get from a fair game of chance will be the same (in average)?
 A: The notation is for my taste unfriendly and against the common sense. It is hard for instance to digest $E[H\circ M]_n$, which is with a low expectation without reading all conventions a "second order expression" in $H,M$, explicity:
$$
E[H\circ M]_n
=
E\Big[\qquad [H\circ M,H\circ M]_n\qquad \Big]\ .
$$
To have a clear situation, i will use $\alpha=0$. I suppose $M$ is a martingale. Then we have
$$
\begin{aligned}
\Delta(H\circ M)_n
&=
\sum_{n-1<j\le n}H_j\Delta M_j
\\
&=H_n(M_n-M_{n-1})
\\
[H\circ M,H\circ M]_n
&=
\sum_{0<j\le n}
\Delta(H\circ M)_j
\cdot
\Delta(H\circ M)_j
\\
&=
\sum_{0<j\le n}
H_j(M_j-M_{j-1})
\cdot
H_j(M_j-M_{j-1})
\ ,
\\
E[H\circ M]_n
&=
E\Big[\qquad [H\circ M,H\circ M]_n\qquad \Big]\ .
\\
&=
E\left[\ 
\sum_{0<j\le n}
H_j^2(M_j-M_{j-1})^2
\ \right]\ .
\\
&\qquad\qquad\text{Now let us compute}
\\
E\left[\ \left(H_0M_0 + (H\circ M)_n\right)^2\ \right]
&=
E[\ (H_0M_0)^2\ ]
\\&\qquad
+2\underbrace{E\left[\ H_0M_0 (H\circ M)_n\ \right]}_{=0}
\\&\qquad\qquad
+
E\left[\ \left( (H\circ M)_n\right)^2\ \right]
\\
&=
E[\ (H_0M_0)^2\ ]
\\&\qquad
+
E\left[\ \sum_{0<j\le n}H_j^2(M_j-M_{j-1})^2\ \right]
\\&\qquad\qquad
+
\underbrace{
2E\left[\ \sum_{0<j<k\le n}H_j(M_j-M_{j-1})\cdot H_k(M_k-M_{k-1})\ \right]
}_{=0}
\ .
\end{aligned}
$$
The reason for the vanishing of some  expectations is as follows, one example, this should be enough to capture the idea. Assume $j<k$ and let us calculate the expectation of 
$$
H_j(M_j-M_{j-1})\cdot H_k(M_k-M_{k-1})\ .
$$
This factors throught the computatioin of the conditional expectaction
$$
\begin{aligned}
&E\Big[\ H_j(M_j-M_{j-1})\cdot H_k(M_k-M_{k-1})\ \Big| \ \Bbb F_{k-1}\ \Big]
\\
=
&H_j(M_j-M_{j-1})\cdot H_k\cdot
\underbrace{ 
E\Big[\ M_k-M_{k-1}\ \Big| \ \Bbb F_{k-1}\ \Big]
}_{=0}
\\
=
&0
\ .
\end{aligned}
$$
(We use the fact that $M$ is a martingale. The $H_k$ could jump outside the conditional expectation, because of predictability.)
This clears the first equation.

For the last step we need of course more information on $H$.
Of course, things are false, if for instance $H=2018$, a constant process.
