# Variance of a sum of identically distributed random variables that are not independent

I am "new" to probability/statistics and I was hoping someone could verify that this is correct. Let $Y_1,\ldots,Y_n$ be random variables that follow a common distribution with mean $\mu$ and variance $\sigma^2$. However, I am not assuming that the variables are independent. I want to compute the expected value and variance of $X=Y_1+\cdots+Y_n$. Then, $$\mathbb{E}(X) = \mathbb{E}\left(\sum_{i=1}^{n} Y_i \right) = \sum_{i=1}^{n} \mathbb{E}\left( Y_i \right) = n \cdot \mu.$$ Moreover, \begin{eqnarray*} \operatorname{Var}(X) &=& \operatorname{Var}\left(\sum_{i=1}^{n} Y_i \right) \\ &=& \sum_{i=1}^{n} \operatorname{Var}\left( Y_i \right) + 2\cdot\left(\sum_{1\leq i<j\leq n} \operatorname{Cov}(Y_i,Y_j) \right) \\ &=& \sum_{i=1}^{n} \operatorname{Var}\left( Y_i \right) + 2\cdot\left(\frac{ n(n -1 )}{2}\right) \cdot C \\ &=& n \cdot \sigma^2 + n(n -1)\cdot C\\ &=& n\cdot ( \sigma^2 + (n-1)\cdot C), \end{eqnarray*} where we have used the properties of the variance, and the fact that for any $i\neq j$, the variables $Y_i,Y_j$ follow the same distribution, and so $\operatorname{Cov}(Y_i,Y_j)=C$ for all $i\neq j$, for some constant $C$.

Question 1: Is this right? I have an uneasy feeling about assuming that all the covariances equal a constant C. Is there some additional hypothesis on the original random variables $Y_1,\ldots,Y_n$ that one needs to assume so that the covariance $\operatorname{Cov}(Y_i,Y_j)=C$ independent of the chosen $i\neq j$?

Question 2: Also, is $C=\operatorname{Cov}(Y_i,Y_j)=\operatorname{Cov}(Y_i,Y_i)=\sigma^2$?

Thanks!

• Without independence, it does not follow that $\textrm{Cov}(Y_i, Y_j) = \mathbb{E}[Y_i Y_j] - \mu^2$ will be the same for every pair of $\{i,j\}$. For instance, suppose $Y_1$ is symmetric about $\mu$, so $Y_1$ and $2\mu-Y_1$ are identically distributed (e.g. if $Y_i \sim N(\mu,\sigma^2)$). Then if we take $Y_2 = Y_1$ and $Y_3 = 2\mu-Y_1$, we would have $\textrm{Cov}(Y_1, Y_2) = \sigma^2$, but $\textrm{Cov}(Y_1, Y_3) = -\sigma^2$. – Shagnik Sep 2 '16 at 16:08
• As for Question 2, the $Y_1, Y_3$ pair above shows that the covariance does not have to be $\sigma^2$. For a simpler example, if the variables were independent, the covariance would be $0$. – Shagnik Sep 2 '16 at 16:09
• Thank you, this is very helpful. – unity Sep 2 '16 at 16:24