# 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

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 $$\rho=1$$