Finding not Gaussian vector I know from the answer to another question that if $X,Y$ are independent standard Gaussians and $Z=\text{sign}(X)\cdot |Y|$ then $Z+X$ is not Gaussian. So, $(X, Z)$ is not Gaussian vector.
How to prove this fact? I tried to directly find the distribution function for the sum. Then take the derivative to make sure that the random variable is not Gaussian. But this path is difficult. Are there simpler solutions?  Can you help me?
 A: In this case it helps immensely to look at the joint distribution of $(X,Z).$  Here at the left is a plot of a sample generated from this distribution

Obviously this joint distribution is not Normal, because data do not appear in two of the quadrants.
Consider the distribution of $Z+X.$ The symmetry of the plot indicates this distribution will be symmetric about $0,$ which suggests we study instead the distribution of $|Z+X|.$ By definition its distribution function is
$$\begin{aligned}F_{|Z+X|}(\varepsilon) &= \Pr(|Z+X|\le\varepsilon)=\Pr(|\operatorname{sgn}(X)|Y|+X|\le\varepsilon) = \Pr(|X|+|Y|\le \varepsilon)\\
&= F_{|X|+|Y|}(\varepsilon).
\end{aligned}$$
To confirm this, here are their histograms:

At left, for reference, I have drawn the histogram of the absolute value of a Normal variable.  You can see that $|Z+X|$ cannot be normal: its mode is not at zero. Let's prove this fact.  It amounts to showing the density of $|Z+X|$ near zero is too small for $0$ to be a mode of $|Z+X| = |X|+|Y|.$
To this end, refer to the upper right scatterplot in the first figure.  The red triangle shows the event $|X|+|Y|\le\varepsilon:$ its probability is $F_{|X|+|Y|}(\varepsilon).$ Because it is a subset of the square (which also includes the blue triangle) and because $X$ and $Y$ are independent we immediately have
$$\begin{aligned}
F_{|X|+|Y|}(\varepsilon) &\le \Pr(|X|\le\varepsilon,\,|Y|\le\varepsilon)\\&=\Pr(|X|\le\varepsilon)\Pr(|Y|\le\varepsilon)\\
& = (\Phi(\varepsilon) - \Phi(-\varepsilon))^2.
\end{aligned}$$
($\Phi$ is the common (standard Normal) distribution function of $X$ and $Y$.)
When we divide the left side by $\varepsilon$ we have (by definition of the derivative) an approximation of the density of $|X|+|Y|$ near $0.$  Evaluate this approximation using the right hand side and bound the right hand side by
$$\Phi(\varepsilon) - \Phi(-\varepsilon) \le 2\varepsilon \sup_{t\in[-\varepsilon,\varepsilon]}\Phi^\prime(t) = \frac{2\varepsilon} {\sqrt{2\pi}} \le 2\varepsilon.$$
Therefore (at least when $\varepsilon$ is sufficiently small)
$$\begin{aligned}
F^\prime_{|X|+|Y|}(\varepsilon) &\le \frac{1}{\varepsilon}(\Phi(\varepsilon) - \Phi(-\varepsilon))^2 
 \le \frac{1}{\varepsilon}(2\varepsilon)^2
= 4\varepsilon.
\end{aligned}$$
As $\varepsilon$ shrinks down to $0,$ so does $4\varepsilon,$ forcing the density of $|X|+|Y|$ down to zero, QED.
