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I was looking at the Wikipedia article talking about the multivariate normal distribution and specifically I was looking at the section talking about the probability density function of that distribution.

In that article it is stated that if the covariance matrix $\Sigma$ of the p.d.f. of a m.n.d.

$$f_{\vec{X}}(x_1, x_2, \ldots,x_k) = \frac{1}{\sqrt{(2\pi)^k |\Sigma |}} \exp\left(-\frac{1}{2}\cdot (\vec{x} - \vec{\mu})' \cdot \Sigma^{-1} \cdot (\vec{x} - \vec{\mu}) \right)$$

is a $1 \times 1$ matrix, then the p.d.f. of the multivariate normal distribution is the same as the p.d.f. of the univariatate normal distribution, but I am not seeing why. Could you please explain it to me?

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The univariate normal p.d.f. is $$ x \mapsto \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left( \frac{x-\mu} \sigma \right)^2 \right). \tag 1 $$ The expression $\left(\dfrac{x-\mu} \sigma\right)^2$ is the same as $(x-\mu) (\sigma^2)^{-1} (x-\mu)$. The $1\times1$ matrix $x-\mu$ is its own transpose, and the inverse of the $1\times1$ matrix whose entry is the number $\sigma^2$ is the $1\times1$ matrix whose entry is $(\sigma^2)^{-1}$, so $(1)$ becomes $(x-\mu)'(\sigma^2)^{-1}(x-\mu)$, which is then $(\vec x - \vec\mu)'\Sigma^{-1}(\vec x-\vec\mu)$.

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  • $\begingroup$ There's only one thing I am still not sure about, which is that you consider that $\Sigma$ (the covariance matrix) is the variance for the multiple dimensions, which apparently is true, but since I am relatively new to these things, I guess I don't get it so well, since the covariance matrix is composed of covariances... $\endgroup$ – nbro Nov 16 '15 at 23:02
  • $\begingroup$ But you (and Wikipedia) also transparently ignore that $x$ becomes $\vec{x}$ and its mean becomes $\vec{u}$...that's also the point where I had doubts...Wait a minute, if $\Sigma$ is a $1 \times 1$ matrix, and since it depends on $\vec{x}$, then $\vec{x}$ is also scalar...Never mind... $\endgroup$ – nbro Nov 16 '15 at 23:04
  • $\begingroup$ @nbro : An $n\times1$ column vector is a scalar of $n=1$ and an $n\times n$ positive-definite matrix is a positive number if $n=1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 17 '15 at 21:23

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