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I have a quick question regarding generating samples from a multivariate normal distribution. When drawing samples from a standard normal distribution we can assume that the samples are i.i.d. However when drawing samples from a MVN with a co variance matrix which is not the identity matrix may we assume that they are i.i.d?

I am aware that the co variance matrix may highlight correlations between samples so I'm unsure if this nullifies the i.i.d property.Any input/explanation would be greatly appreciated.

Thanks in advance

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Let us say that $(X_1, X_2) \sim BVN(\mu, \Sigma)$, where $$ \Sigma= \begin{pmatrix} \sigma^2_X & \rho\sigma_X\sigma_Y\\ \rho\sigma_Y\sigma_X & \sigma_Y^2 \end{pmatrix}, $$ for $\rho \neq 0$ clearly $X_1$ and $X_2$ are dependent variables with a known dependence structure, however the series $\{(x_{1i}, x_{2i})\}_{i=1}^n$ of random draws from it can be viewed as i.i.d. In other words, the i.i.d property holds for the random vectors that you draw and not for elements of the same vector.

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    You can accept the answer if it satisfies you. Anyway, you are welcome. – 2017-02-13
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    You could imagine that you have a multivariate normal distribution for all the data (vectorized by stacking each transposed 2-dimensional row), and if it had a diagonal covariance all samples are i.i.d. If you have dependence between pairs as explained above you have a block diagonal covariance, where each block is 2x2 for this bivariate case. – 2017-12-20