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different between MVG and joint MVG?

Signal Processing Asked on December 14, 2021

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

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


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

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

Where:

$[ntimes1] ~~~mathbf{x} Rightarrow text{sample vector}$

$[ntimes1]~~~mathbf{mu} Rightarrow text{mean vector}$

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


What’s the difference between the two?

One Answer

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).

Answered by Marcus Müller on December 14, 2021

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