Let $X_1, \ldots , X_n$ be independent random variables having a common density with mean $\mu$ and variance $\sigma^2$, where

Question

Let $X_1, \ldots , X_n$ be independent random variables having a common density with mean $\mu$ and variance $\sigma^2$, where $\bar{X}=\frac1n\sum^n_{k=1}X_k$. Calculate $\operatorname{Cov}(\bar{X}, X_k-\bar{X})$

Attempt

$$\operatorname{Cov}(\bar{X}, X_k-\bar{X})=\operatorname{Cov}(\bar{X}, X_k)-\operatorname{Cov}(\bar{X},\bar{X})=\operatorname{Cov}(\bar{X}, X_k)-\operatorname{Var}(\bar{X})$$

However,

$$\operatorname{Cov}(\bar{X}, X_k)=E(\bar{X}X_k)-E(\bar{X})E(X_k)$$

According to me, you have

$$E(X_k)=\mu, E(\bar{X})=\mu, \operatorname{Var}(\bar{X})=\frac{\sigma^2}{n}$$

So, how do I calculate $E(\bar{X}X_k)$?, should I use the fact that they are independent variables?

Edit

$$E[\bar{X}X_k] = E\left[\frac{1}{n} \sum_{i=1}^n X_i X_k\right]
= \frac{1}{n} \sum_{i=1}^n E[X_i X_k] =\frac{\sigma^2}{n}+\mu^2?$$

Answer 1

$\bar{X}$ and $X_k$ are not independent since they both involve $X_k$. Expand $\bar{X}$ and proceed.

$$E[\bar{X}X_k] = E\left[\frac{1}{n} \sum_{i=1}^n X_i X_k\right] = \frac{1}{n} \sum_{i=1}^n E[X_i X_k] = \cdots$$