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Let $X_1, \cdots ,X_n$ be a random sample from a $N (\mu, \sigma^2)$ distribution, and let $\bar{X}$ and $S^2$ denote sample mean and sample variance. We know that $\bar{X}$ and $S^2$ are independent.

In George Casella's Statistical Inference, page 222, he rewrote $$\bar{X}=\sum_{i=1}^n(\frac{1}{n})X_i,$$ $$X_j-\bar{X}=\sum_{i=1}^n(\delta_{ij}-\frac{1}{n})X_i,$$ where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ otherwise. Then my question is, why $$\textrm{Cov}(\bar{X},X_j-\bar{X})=\sum_{i=1}^n(\frac{1}{n})(\delta_{ij}-\frac{1}{n}).$$ Of course the right hand side equals to $0$, but I don't know why the covariance equal to right hand side.

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You should have mentioned that he's applying a lemma – how did you think we'd be able to guess that without the book?

Here's a copy of the book. This calculation is intended “as an illustration of the application of Lemma $5.3.3$”. Upon comparison with item a. in Lemma $5.3.3$ (p. $220$), it seems that it's a typo and he forgot the factor $\sigma^2$.

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    $\begingroup$ I am sorry I did not notice he used a lemma to prove so. No wondering he said "easy" to show. $\endgroup$ – Unk Jun 22 '18 at 16:40

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