I have troubles understanding the main idea used to prove the following theorem, which should be a coupling argument.
My question refers to the following notes, at page 125.
http://www.win.tue.nl/~rhofstad/NotesRGCN2011.pdf
Theorem For all $k\geq 1$ $$\mathbb{P}_{np}[T^*\geq k]=\mathbb{P}_{np}[\mid C(1)\mid\geq k]\leq\mathbb{P}_{n,p}[T^{\geq}\geq k]$$ where $\mathbb{P}_{n,p}$ is a $\text{Binom}(n,p)$-Branching process, $T^{\geq}$ is the total progeny of a binomial branching process with parameters $n$ and $p$, $\mathbb{P}_\lambda$ denotes the law of $ER_n(\lambda /n) \ (\lambda=np)$, also note:
$$C(v)= \{x\in [n]:v\leftrightarrow x\}.$$ We denote the size of $C(v)$ by $\mid C(v)\mid$, i.e., $\mid C(v)\mid$ denotes the number of vertices connected to $v$. The largest connected component $C_\text{max}$ is equal to any cluster $C(v)$ for which $\mid C (v)\mid$ is maximal, so that $$\mid C_\text{max}\mid = \max_{v\in[n]} \{\mid C (v)\mid\},$$ $[n]=$Set of vertices of $ER_n(p)$.
Hier is the proof. I don't understand basically the argument of bringing the new r.v. $Y_i$ and taking the sum $X^\geq$.
You want to know something about $\sum_iX_i$. However, $X_i$ has a different distribution for different values of $i$, so this is difficult. Estimating $\sum_i X_i+Y_i$ is easier, since $X_i+Y_i$ has a $Bin(n,p)$ distribution for all $i$. So $Y_i$ is just an auxiliary variable to show that $X_i$ is dominated by a $Bin(n,p)$ variable.
I would say that there is a mistake in the proof though, I think that it should say $X_i\leq X_i^{\geq}$.
Then $P(\sum_i X_i-(i-1)\geq 0)\leq P(\sum_i X_i^{\geq}-(i-1)\geq 0)$, since $X_i\leq X_i^{\geq}$.