# Compute mean and covariance matrix of $\bar{X}$ from a simple random sample

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?

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.