# Covariance between individual IID variables and their sample mean

If we know that $$X_1, X_2, ..., X_n \sim \text{IID N}(\mu, \sigma^2)$$, what is $$\mathbb{Cov}(X_1, \bar{X}_n)$$? Because the set of variables follows an IID distribution, would the covariance just be zero?

The covariance here will not be zero because the sample mean includes the variable $$X_1$$ as part of it; a higher values of the latter variable is going to give you a higher sample mean and a lower value is going to give you a lower sample mean. Consequently, we would expect these two things to be positively correlated.
\begin{align} \mathbb{C}(X_1, \bar{X}_n) &= \mathbb{C} \Bigg( X_1, \frac{1}{n} \sum_{i=1}^n X_i \Bigg) \\[6pt] &= \frac{1}{n} \sum_{i=1}^n \mathbb{C} ( X_1, X_i ) \\[6pt] &= \frac{1}{n} \mathbb{C}(X_1, X_1) \\[6pt] &= \frac{1}{n} \mathbb{V}(X_1) \\[6pt] &= \frac{\sigma^2}{n}. \\[6pt] \end{align}
This confirms that the covariance is positive so long as $$\sigma>0$$. As $$n \rightarrow \infty$$ the sample mean depends less and less on the value $$X_1$$, and the covariance diminishes to zero. (Note also that this result does not depend on the assumption of normality of the underlying variables; it holds for any underlying distribution with finite variance.)
• @COOLSerdash: Since the data values are IID they are uncorrelated, so you have $\mathbb{C}(X_1, X_i) = 0$ for all $i \neq 1$. Consequently, the only non-zero term in the sum is $\mathbb{C}(X_1, X_1)$. – Ben Apr 15 at 1:55