Covariance between partitions of a normal distribution A bit of a contrived example, but if I had a sample of $X_1,\dots,X_n \stackrel{iid}{\sim} N(\mu,\sigma^2)$ (in this case $\mu$ is unknown but $\sigma^2$ is known), and then calculated the arithmetic mean of the sample, how do I find the covariance between $$Cov(X_1,\overline{X})$$
I know (at least I think I do) that it should become equal to Var$(\overline{X})$, which I can state as $\frac{\sigma^2}{n}$ because $\sigma^2$ is known.
I rewrite as:
$$Cov(X_1,\overline{X})=E(X_1\overline{X})-E(X_1)E(\overline{X})$$
which becomes $E(X_1\overline{X})-E(\overline{X})^2$ but evaluating that first part is causing me some issues. Am I on the right track?
Thank you.

You can write
$$
X_1 \bar{X}  = \frac{1}{n}\sum_{i=1}^n X_1 X_i
$$
If you take the expectation you get
$$
\frac{1}{n}\sum_{i=1}^n \mathbb E(X_1 X_i) = \frac{1}{n}\left( \mathbb E(X_1^2) + \sum_{i=2}^n\mathbb E( X_1 X_i)\right)
$$
Since (I assume) $X_i \perp \!\!\! \perp X_1$ for $i \neq 1$ then
$$
 \mathbb E( X_1 X_i) = \mathbb E(X_1) \mathbb E(X_i) = \mu^2
$$
Thus,
\begin{align*}
\frac{1}{n}\sum_{i=1}^n \mathbb E(X_1 X_i) &= \frac{1}{n}\left (\mathbb E(X_1^2) + (n-1) \mu^2 \right ) \\
&= \frac{1}{n}\left( \sigma^2 + \mu^2 + (n-1)\mu^2 \right) \\
&= \frac{\sigma^2}{n} + \mu^2 
\end{align*}
And finally,
\begin{align*}
\text{Cov}(X_1,\bar{X}) &= \frac{\sigma^2}{n} + \mu^2  - \mathbb E(X_1)\mathbb E(\bar{X}) \\
&= \frac{\sigma^2}{n}
\end{align*}

Another possibility to compute $\text{Cov}(X_1,\bar{X})$ is to use the bilinearity of the covariance, i.e for random variables $(X_1,\dots,X_k,Y_1,\dots,Y_n)$ and constants $(\alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_n)$ we have
$$
\text{Cov} \left( \sum_{i=1}^k \alpha_1 X_i, \sum_{j=1}^n \beta_j Y_j \right ) = \sum_{i=1}^k \sum_{j=1}^n \alpha_i \beta_j X_i Y_j
$$
In your case it yields,
$$
\text{Cov}(X_1,\bar{X}) = \frac{1}{n} \sum_{i=1}^n \text{Cov}(X_1,X_i)
$$
By independance of the vector $(X_1,\dots,X_n)$, $\text{Cov}(X_1,X_i) = 0$ for $i \neq 1$ thus
\begin{align*} 
\text{Cov}(X_1,\bar{X}) &= \frac{1}{n} \text{Cov}(X_1,X_1) \\
&= \frac{1}{n} \text{Var}(X_1) \\
&= \frac{\sigma^2}{n}
\end{align*}
