Variance of summation of Bernoulli variables 
Let $X_1,\ldots,X_n$ be independent Bernoulli variables, with probability of success $p_i$ and let $Y_n =\frac1n\sum\limits^n_{i=1} (X_i - p_i )$
a) find the mean and variance of $Y_n$
b) show that for every $a>0, \lim\limits_{n\to\infty} P(Y_n<a)=1$

Now for the mean, it was quite straightforward: $E[Y_n]=0$, however the variance not so much. they said $Var(Y_N)= \dfrac{\sum_{i=1}^n p_i(1-p_i)}{n^2}$
why did they not take out the summation as I have to do almost every time? is the same as $\dfrac{Var(X_i)}{n^2}$ ?
 A: $$ \begin{align} \mathbb{E}[Y_n] &=\dfrac{\sum^n_{i=1} (\mathbb{E}[X_i] - p_i )}{n}\\
& =0\\\end{align} $$ 
$$ \begin{align} Var[Y_n]&=\mathbb{E}[Y_n^2]-\mathbb{E}^2[Y_n]\\ 
& =\mathbb{E}[Y_n^2]=\mathbb{E}[(\dfrac{\sum^n_{i=1} (X_i - p_i )}n)^2]\\
&=\frac1{n^2}\mathbb{E}[\sum^n_{i=1}\sum^n_{j=1}(X_i-p_i)(X_j-p_j)]\\
&=\frac1{n^2}\sum^n_{i=1}\sum^n_{j=1}\mathbb{E}[(X_i-p_i)(X_j-p_j)]\\
&=\frac1{n^2}(\underbrace{\sum^n_{i=1}\sum^n_{j \ne i,j=1}\mathbb{E}[(X_i-p_i)]\mathbb{E}[(X_j-p_j)]}_{=0 (i \ne j \Rightarrow X_i \bot X_j)}+\sum^n_{i=1}\mathbb{E}[(X_i-p_i)^2]\\
&=\frac1{n^2}\sum^n_{i=1}p_i(1-p_i)
\end{align}
$$
note that $\mathbb{E}[.]$ is a linear operator
A: For a)
Hint 1: The mean of a Bernoulli variable with probability $p$ is $p$ and its variance is $p(1-p)$.
Hint 2: The variance of a sum of independent variables is the sum of the variances, and the variance of a constant times a variable is that constant squared times the variance of the variable.
If each of the $X_i$ had the same $p$, then $\sum\limits_{i=1}^np(1-p)=np(1-p)$ and we could simplify the variance to $np(1-p)/n^2=p(1-p)/n$. However, each $X_i$ has a possibly different $p_i$, so we have to keep the sum.
For b)
Hint: Note that $p(1-p)\le\frac14$ for any $p\in[0,1]$.
