Let $\{Y_i\}_{i=1}^n$ be a set of $n$ i.i.d. random variables $Y_i$.
I am trying to prove:
$$\mathrm{Var}\bigg(\frac{1}{n} \sum_{i=1}^nY_i\bigg) = \frac{1}{n}\mathrm{Var}(Y_1)$$
My work:
$$ \begin{align*} \mathrm{Var}\bigg[\frac{1}{n} \sum_{i=1}^nY_i\bigg] &= \frac{1}{n}\sum_{i=1}^n\bigg[\frac{1}{n} \sum_{i=1}^n\bigg(Y_i-\bar Y_n\bigg)\bigg]^2\\\\ &=\tfrac{1}{n} \cdot n \cdot \bigg[\frac{1}{n}\sum_{i=1}^n\big(Y_i-\bar Y_n\big)\bigg]^2\\\\ &=\bigg[\frac{1}{n}\sum_{i=1}^n\bigg(Y_i-\bar Y_n\bigg)\bigg]^2\\\\ &=\frac{1}{n^2}\bigg[\sum_{i=1}^n\big(Y_i-\bar Y_n\big)\bigg]^2 \\\\ &= 0 \end{align*} $$
I know my conclusion is false but I am wondering where my mistake is. Thank you.
Let $\bar Y = \frac{1}{n}\sum_{i=1}^n Y_i$ denote the sample mean. Then
$$ \mathrm{Var}\bigg[\frac{1}{n} \sum_{i=1}^n Y_i\bigg] = \frac{1}{n^2}\sum_{i=1}^n\mathrm{Var}\big[Y_i\big] $$
Since each $Y_i$ is i.i.d. it follows that for distinct $i$ and $j$ where $i,j \in \{1, 2, \dotsc, n\}$,
$$\mathrm{Var}\big[Y_i\big] = \mathrm{Var}\big[Y_j\big]$$
So now we have
$$ \begin{align*} \mathrm{Var}\bigg[\frac{1}{n} \sum_{i=1}^n Y_i\bigg] &= \frac{1}{n^2}\sum_{i=1}^n\mathrm{Var}\big[Y_i\big]\\\\ &= \frac{1}{n^2} \cdot n\mathrm{Var}\big[Y_i\big] \\\\ &= \frac{1}{n} \mathrm{Var}\big[Y_i\big] \end{align*} $$
Because $\mathrm{Var}\big[\frac{1}{n}\sum_{i=1}^n Y_i\big] = \frac{1}{n} \mathrm{Var}[Y_i]$ is true for all $i = 1,2, \dotsc, n$, we choose $i=1$ to obtain
$$\mathrm{Var}\bigg[\frac{1}{n}\sum_{i=1}^n Y_i\bigg] = \frac{1}{n} \mathrm{Var}[Y_1]$$