co-variance between a sample from normal distribution and the sample mean?

Question

$X_1$ is a sample from a normal distribution with mean$=\mu$ and variance $= 1$. The joint distribution of $X_1$ and the sample mean is bivariate normal. I need to find the conditional distribution of $X_1$ given the sample mean. To do this I need to calculate the co-variance between $X_1$ and the sample mean. By doing some simulations I know that that co-variance between $X_1$ and the sample mean is $\frac{1}{n}$, but I'm not sure how to prove it. How do you prove that the co-variance between $X_1$ and the sample mean is $\frac{1}{n}$?

Answer 1

Suppose that $X_1,X_2,\dots,X_n$ are independent random variables, and suppose that the $X_i$ have variance $1$. By the bilinearity of covariance, we have $$\text{Cov}\left(X_1, \frac{1}{n}(X_1+\cdots+X_n)\right)=\frac{1}{n}\sum_1^n\text{Cov}(X_1,X_i).$$ Almost all the covariances on the right are $0$, by independence, and $\text{Cov}(X_1,X_1)=\text{Var}(X_1)=1$. The result follows.