Q:

problem finding the Variance of dependent variables using covariance and correlation

Hello I have been trying to figure out this question for a few hours and I am very stuck and don't know how to progress but I am pretty sure I am wrong and would greatly appreciate some help.
$X$~$N(\mu, \sigma^2)$
$\sigma = 2$, and we estimate the mean of of the distribution $E(X)=\mu$ using the average of n Random variables denoted $\bar{X} = \frac{1}{n}\Sigma_{i=1}^n X_i, X_i$~$N(\mu, \sigma^)$
assume that $X_1,X_2,...,X_n$ are dependant
If $\text{Corr}(Xi,Xj) =ρ\text{ for }i=1,...,n\text{ and }j=1, ...,n \text{ such that } i\neq j$, find Var$(\bar{X})$ (in terms of n and ρ):
firstly lets find the Cov$(X_i, X_j)$
$\begin{align}
&\text{Corr}(X_i, X_j) = \frac{\text{Cov}(X_i, X_j)}{\sqrt{\text{Var}(X_i)\text{Var}(X_j)}}\\
\rightarrow &\text{Corr}(X_i, X_j)= \frac{\text{Cov}(X_i, X_j)}{\text{Var}(X)}\\
\rightarrow &\text{Corr}(X_i, X_j)\text{Var}(X)= \text{Cov}(X_i, X_j)\\
\rightarrow  &\text{Cov}(X_i, X_j)= \text{Corr}(X_i, X_j)\text{Var}(X)\\
\rightarrow  &\text{Cov}(X_i, X_j)= p\text{Var}(X)
\end{align}$
now find 
$\begin{align}
\text{Var}(\bar{X})&=\text{Var}(\frac{1}{n^2}\Sigma_{i=1}^nX_i)\\
&=\frac{1}{n^2}\Sigma_{i=1}^n\text{Var}(X_i)+2\Sigma_{1\leq i < j\leq n}Cov(X_i, X_j)\\
&=\frac{1}{n^2}\Sigma_{i=1}^n\text{Var}(X_i)+2 {n \choose 2} Cov(X_i, X_j)\\
&=\frac{1}{n^2}\Sigma_{i=1}^n\text{Var}(X_i)+2 {n \choose 2} p\text{Var}(X)\\
&=\frac{1}{n^2}n\text{Var}(X_i)+2 \frac{n(n-1)}{2} p\text{Var}(X)\\
&=\frac{\text{Var}(X_i)}{n}+ n(n-1) p\text{Var}(X)\\
\end{align}
$
I am also asked to prove that if $X_1,X_2, \cdots, X_n$ are perfectly positively correlated that 
$\text{Var}(\bar{X}) = \text{Var}(X_i) = \sigma^2$
but I don't see how that makes sense if my answer for $\text{Var}(\bar{X})$ is correct so I must be wrong

A:

Somewhere you stopped applying $\frac{1}{n^2}$ to the right-hand side of your expressions
You should have ended up with $$\text{Var}(\bar{X}) =\frac{1}{n^2}\left(n\text{Var}(X)+2 \frac{n(n-1)}{2} \rho\text{Var}(X)\right) =\frac{1+(n-1)\rho}{n}  \text{Var}({X})$$
which, as expected, is $\frac{1}{n}  \text{Var}({X})$ when $\rho=0$, and is $ \text{Var}({X})$ when $\rho=1$

