# Folded Normal and truncated Normal

Suppose to have a vector of random variables $\mathbf{y}$, distributed as a multivariate normal with mean vector $\boldsymbol{\mu}$ and covariance matrix $\boldsymbol{\Sigma}$.

The variable $\mathbf{Z} = | \mathbf{Y} |$ (where the modulo operator is applied to each component of $\mathbf{Y}$) is distributed as a folded normal.

If we set $\boldsymbol{\mu}= \mathbf{0}$, then is $\mathbf{Z}$ distributed as a multivariate truncated normal, with left truncation at $\mathbf{0}$?

• For some analyses of data from folded Normal distributions, please see stats.stackexchange.com/questions/117935/…. I think a quick look at some of these data will make the answer to this question perfectly obvious. – whuber Nov 13 '14 at 4:32

In the univariate case, when we truncate a zero-mean normal at zero (which is the same as taking its absolute value), then the folded-normal and the truncated normal family share a common member, the half-normal.

This is not in general the case in a multivariate setting. Treating the bivariate case, let's see under which conditions the identity nevertheless holds.

The density of the bivariate half normal can be compactly written (adjusting the formulas that can be found here ) as

$$f_H(x,y)=4\frac{\sqrt {1-\rho^2}}{s_xs_y}\cdot\phi(x/s_x)\cdot\phi(y/s_y)\cdot\text{cosh}\left(\frac{\rho xy}{s_xs_y}\right)$$

with

$$s_x = \sigma_x\sqrt {1-\rho^2}, \,s_y = \sigma_y\sqrt {1-\rho^2}$$

and where $\phi()$ is the standard normal PDF and $\text{cosh}$ is the hyperbolic cosine function.

The density of the bivariate normal truncated from below at zero is

$$f_T(x,y) = \left[\int_{0}^{\infty}\frac {1}{\sigma_x}\phi(x/\sigma_x)\Phi\left(\frac {\rho}{\sqrt {1-\rho^2}}\frac {x}{\sigma_x}\right)dx\right]^{-1}\cdot\frac{\sqrt {1-\rho^2}}{s_xs_y}\times\\\times\; \phi(x/s_x)\cdot\phi(y/s_y)\cdot \exp\left\{-\frac{\rho xy}{s_xs_y}\right\}$$

Note that multiplying and dividing the integral in the expression by $2$ we obtain

$$\Big[...\Big] = \frac 12-\frac 12 G\left(0;0,\sigma_x,\frac {\rho}{\sqrt {1-\rho^2}}\right)$$

where $G(x;\text{location},\text{scale}, \text{skew})$ is the Skew Normal CDF, for which we have (see for example here eq. $(18)$)

$$G\left(0;0,\sigma_x,\frac {\rho}{\sqrt {1-\rho^2}}\right) = 2B(0;0;-\rho)$$

where $B()$ is the bivariate standard normal integral. This relation verifies that the integral can equivalently be written in terms of the $y$ variable without imposing equality of variances between $x$ and $y$ -the variance term does not affect its value.

Then, if $\rho =0$, one can verify that the two densities become identical: the reciprocal of the integral equals $4$ and the other generally non-equal terms all become unity, giving

$$\rho =0 \Rightarrow f_H(x,y) = f_T(x,y) = \frac{4}{\sigma_x\sigma_y}\cdot\phi(x/\sigma_x)\cdot\phi(y/\sigma_y)$$

So it appears that the condition needed for the identity of the densities to hold is that the two underlying zero-mean normals are independent, but not in addition that they have the same variance.

Opportunity taken, Horrace 2005 collects many useful results related to multivariate truncated normal distributions. Perhaps the most interesting is the following:

If $\mathbf w$ is a multivariate truncated normal, then the linear combination $\mathbf y = \mathbf D\mathbf w+\mathbf b$ is not multivariate truncated normal, excpet if $\mathbf D = \mathbf I$.

In other words, with truncation present, normality is preserved under shifting but not under scaling.

Also,

If $\mathbf w$ is a multivariate truncated normal vector of correlated random variables, then the marginal distributions are not, in general, truncated normal, but the conditional distributions are truncated normal. If the variables are independent, then the marginal distributions also are truncated normals.

Finally, log-concavity and the multivariate generalization of unimodality characterize the multivariate truncated normal.

• I'm sorry, you are wrong. In a bivariate case, the density of $(x,y)$ is $4\phi_2(z|\mathbf{0}, \Sigma)$ for the truncated normal ($\phi_2(x,y|\mathbf{0}, \Sigma)$ is the bivariate normal pdf with mean $\mathbf{0}$ and covariance matrix $\Sigma$) while is $\phi_2(x,y|\mathbf{0}, \Sigma)+\phi_2(-x,y|\mathbf{0}, \Sigma)+\phi_2(x,-y|\mathbf{0}, \Sigma)+\phi_2(-x,-y|\mathbf{0}, \Sigma)$ for the bivariate half normal. The two are identical if the correlation is 0, i.e $\phi_2(x,y|\mathbf{0}, \Sigma)=\phi_2(-x,y|\mathbf{0}, \Sigma)=\phi_2(x,-y|\mathbf{0}, \Sigma)=\phi_2(-x,-y|\mathbf{0}, \Sigma)$ – niandra82 Nov 13 '14 at 1:24
• @niandra82 Thanks for the input. I am checking this out. – Alecos Papadopoulos Nov 13 '14 at 1:45
• Im pretty sure, but please let me know if I'm wrong – niandra82 Nov 13 '14 at 1:53
• @niandra82 An intermediate question: you obtain the density of the truncated bivariate normal $4\phi_2(x,y|\mathbf{0}, \Sigma)$, under an assumption of independence between the two variables? Because I cannot see how else the normalizing factor would equal $4$. – Alecos Papadopoulos Nov 13 '14 at 2:09
• Ok, in this case you are right, I made a mistake. – niandra82 Nov 13 '14 at 2:14