If $Y = \sum_{i=1}^nX_i$, how do I express $Var(\frac{1}{n}Y)$ as a function of $Var(X)$?

Question

To give more context, we have $X_i = 1$ if true, $0$ otherwise. The probability that the outcome will be true is 0.6. The sample is independent. This is my work so far, but I feel like it's wrong.$\DeclareMathOperator{\Var}{Var}$

$\Var(\frac{1}{n}Y) = E[(\frac{1}{n}\sum_{i = 1}^nX_i)^2] - E[\frac{1}{n}\sum_{i = 1}^nX_i]^2$ $ = E[\frac{1}{n}\sum_{i = 1}^nX_i\cdot\frac{1}{n}\sum_{i = 1}^nX_i] - (E[X])^2 = E[X^2] - E[X]^2 = \Var(X).$

Where am I going wrong? Thanks.

EDIT: To get these equivilances, I'm using the linearity of the expected value and the fact that $E[AB] = E[A]E[B]$.

Answer 1

$E[(\frac{1}{n}\sum\limits_{i = 1}^nX_i)^2] - E[\frac{1}{n}\sum\limits_{i = 1}^nX_i]^2$

Up to here it is OK. Then you cannot say that $E[(\frac{1}{n}\sum_{i = 1}^nX_i)^2]=E[\frac{1}{n}\sum\limits_{i = 1}^nX_i]E\cdot[\frac{1}{n}\sum\limits_{i = 1}^nX_i]$. It is just $\frac1{n^2} E(\sum\limits_{i = 1}^nX_i)^2)$.

And $E[\frac{1}{n}\sum\limits_{i = 1}^nX_i]^2=\frac1{n^2}\cdot E[\sum\limits_{i = 1}^nX_i]^2$.

So in total we have $\frac1{n^2}\cdot Var(Y)$. We know that $Var(\frac1n\cdot Y)=Var(\overline X)=\frac{Var(X_i)}{n}$ if $X_i$ are identical and independent distributed. Thus $\frac1{n^2}\cdot Var(Y)=\frac{Var(X_i)}{n}=p\cdot (1-p)=0.6\cdot 0.4=0.24$

Answer 2

For independent RV's one has $$ Var(\sum_1^n X_i) = \sum_1^n Var(X_i)$$ hence $$ Var(\frac1n Y) =\frac1{n^2} n Var(X_1)=\frac1n Var(X_1) $$ since $X_i$'s are identically distributed.