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?

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 28, 2017 at 21:30
  • 1
    $\begingroup$ It's wrong to say $\mathcal N(0,1) = \ldots$. Rather, you're writing the density for $Z$ on the right side. $\endgroup$ Dec 28, 2017 at 21:33
  • $\begingroup$ @RobertIsrael Corrected! $\endgroup$
    – Frank Vel
    Dec 28, 2017 at 21:50

1 Answer 1


For the multivariate distribution, the components $Z_j$ are independent standard normal random variables. Thus $$f_{\bf Z}({\bf z}) = \prod_{j=1}^n f_Z(z_j)$$

  • $\begingroup$ So it's incorrect to say that $z_i \sim \mathcal{N}(0,1)$; it should be $Z_i \sim \mathcal{N}(0,1)$ ? $\endgroup$
    – Frank Vel
    Dec 29, 2017 at 9:52
  • $\begingroup$ 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 $\mathcal N(0,1)$. $\endgroup$ Dec 29, 2017 at 18:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .