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Given $\{X_\alpha , \alpha =1,...N\}$ a simple random sample obtained from any p-dimensional distribution with mean $\mu$ and covariance matrix $\Sigma$, compute the mean and the covariance matrix of $\bar{X}$.

Using the linearity of the mean and knowing that for the variance we have $Var(cX)=c^2 Var(X)$ where $c$ is a constant and $X$ a random variable/vector, I obtained that the mean of $\bar{X}$ is again $\mu$ and the covariance matrix is $\frac{1}{N} \Sigma$. Is it correct?

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Your conclusion is correct but your reasoning for the variance is incomplete. You must also invoke the fact that $\{X_\alpha\}_{\alpha=1}^N$ comprises independent observations from the same distribution; thus we have in particular $$\operatorname{Var}\left[\sum_{\alpha=1}^N X_\alpha \right] \overset{\text{ind}}{=} \sum_{\alpha=1}^N \operatorname{Var}[X_\alpha] \overset{\text{i.d.}}{=} N \boldsymbol \Sigma,$$ where the first equality follows from the independence property as just stated, and the second equality follows from the fact that the observations are identically distributed with variance/covariance $\boldsymbol \Sigma$. Then, as $\bar X = \frac{1}{N}\sum_{\alpha=1}^N X_\alpha$, its variance is $(N\boldsymbol \Sigma)/N^2 = \boldsymbol \Sigma/N$ using the property you stated.

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  • $\begingroup$ Ok that’s true!... the fact that the sumation comes out from the mean is due to linearity while only for the variance is for independence, right? $\endgroup$ – Maggie94 Dec 10 '18 at 7:07

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