different between MVG and joint MVG?

Question

Distribution for "joint multi-variate gaussian distribution" (joint MVG):

$$f_{X}(x) = \frac{1}{(2\pi)^{n/2}\prod \limits_{i=0}^{n}\sigma_i} ~~\text{exp}\bigg[-\frac{1}{2} \sum \limits_{i=1}^{n}\Big(\frac{x_i\mu_i}{\sigma_i}\Big)^2 \bigg]\tag{2}$$

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Distribution for "multi-variate gaussian distribution" (MVG):

$$f_{\mathbf{x}}(x) = \frac{1}{(2\pi)^{n/2}~|\text{det }C|^{1/2} }~~\text{exp}\bigg[ -\frac{1}{2} (\mathbf{x}-\mathbf{\mu})^T C^{-1} (\mathbf{x}-\mu)\bigg]\tag{1}$$

Where:

$[n\times1] ~~~\mathbf{x} \Rightarrow \text{sample vector}$

$[n\times1]~~~\mathbf{\mu} \Rightarrow \text{mean vector}$

$[n\times n]~~~ C \Rightarrow \text{covariance matrix (symmetric positively defined matix)}$

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What's the difference between the two?

Answer 1

There's no difference in formula:

Try setting the $C=\text{diag}(\sigma_1^2, \sigma_2^2,\ldots,\sigma_n^2)$ in $(1)$ to see how $(1)$ simplifies to $(2)$. So, your formula $(2)$ is just a special case of $(1)$.

In the diagonal case, the elements of the multivariate distribution are jointly Gaussian, which is much stronger than them all just being Gaussian distributed.

Joint Gaussian distribution has a name of its own (unlike other multivariate distributions with identically distributed elements) because "joint gaussian multivariate random varaible" has the very interesting feature that simple MVG doesn't have: that uncorrelatedness implies independence. This is actually something that makes it useful in communications engineering (because rotation and normalization doesn't change joint gaussianness and doesn't introduce interdependencies between components - so, in an IQ receiver, the noise on I and Q are still independent, even after phase correction, for example).