# From univariate to multivariate normal distribution and back

A (standard) univariate normal distribution is given by

$$Z \sim \mathcal{N}(0,1),\quad f_Z(z) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}z^2\right)$$

and a (standard) multivariate distribution is given by

$$\mathbf{Z} \sim \mathcal{N}(\mathbf{0},I),\quad f_{\mathbf{Z}}(\mathbf{z}) \frac{1}{(2\pi)^{n/2}}\exp\left(-\frac{1}{2}\mathbf{z}^T\mathbf{z}\right)$$

where $\mathbf{z} = [z_1,\ldots,z_n]^T$ and $z_i \sim \mathcal{N}(0,1)$. How does one get from the univariate to the multivariate formula?

The thing that puzzles me is that if $\phi(z) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{1}{2}z^2\right)$, then the multivariate distribution seems to imply that $\phi(Z) \sim Z$ (when $n=1$, this yields $\mathbf{Z} = [Z]$). Is this really the case?

• If it's standard, $\sigma = 1$. And in the multivariate case, you want $(2 \pi)^{n/2}$, not $\sqrt{2\pi}$, in the denominator. Also, you want a negative sign in the exponential. Dec 28, 2017 at 21:30
• It's wrong to say $\mathcal N(0,1) = \ldots$. Rather, you're writing the density for $Z$ on the right side. Dec 28, 2017 at 21:33
• @RobertIsrael Corrected! Dec 28, 2017 at 21:50

For the multivariate distribution, the components $Z_j$ are independent standard normal random variables. Thus $$f_{\bf Z}({\bf z}) = \prod_{j=1}^n f_Z(z_j)$$
• So it's incorrect to say that $z_i \sim \mathcal{N}(0,1)$; it should be $Z_i \sim \mathcal{N}(0,1)$ ? Dec 29, 2017 at 9:52
• The usual convention is to use upper case for random variables and lower case for numbers (e.g. values of the random variable). It is the random variable $Z_i$ that has the distribution $\mathcal N(0,1)$. Dec 29, 2017 at 18:56