Is a dot product between two independent multivariate Gaussian random variables also Gaussian random variable? Let $x, z \sim N(0,I_p)$ be two independent multivariate Gaussian random variables. The question is whether the dot product $x'z$ is a Gaussian distributed variable. 
My guess is that it is not. However, I cannot find what is wrong with the following argument. Consider the joint distribution of $(x'z, z)$. We can write $p(x'z,z) = p(x'z|z)p(z)$. Since conditionally $x'z|z$ is a Gaussian and $z$ is Gaussian, the product of two Gaussian densities is a density of a multivariate Gaussian variable. Therefore $(x'z, z)$ are jointly Gaussian, which implies that marginally $x'z$ is also a Gaussian variable. 
 A: Elaborating on alex.jordan's answer. Consider the one-dimensional case, where $X$ and $Z$ are independent ${\rm N}(0,1)$ variables. Then,
$$
p_{XZ,Z} (xz,z) = p_{XZ|Z} (xz|z)p_Z (z).
$$
Now, conditionally on $Z=z$, $XZ$ is normally distributed with mean $0$ and variance ${\rm Var}(Xz)=z^2$.
Hence,
$$
p_{XZ|Z} (xz|z) = \frac{1}{{\sqrt {2\pi z^2 } }}\exp \bigg[ - \frac{{(xz)^2 }}{{2z^2 }}\bigg] = \frac{1}{{\sqrt {2\pi z^2 } }}e^{ - x^2 /2} ,
$$
and in turn,
$$
p_{XZ,Z} (xz,z) = \frac{1}{{\sqrt {2\pi z^2 } }}e^{ - x^2 /2} \frac{1}{{\sqrt {2\pi } }}e^{ - z^2 /2} = 
\frac{1}{{2\pi |z|}}e^{ - (x^2  + z^2 )/2} ,
$$
which is not Gaussian because of the $|z|$ in the denominator.
A: $p(z)$ may be Gaussian.  Then for any particular $z$, $p(x'z|z)$ will be Gaussian too.  But the variance in $p(x'z|z)$ will differ with differing values of $z$.  And that's how $p(x'z,z)$ will end up not being multivariate Gaussian, despite being the product of $p(z)$ and $p(x'z|z)$.
A: Let take the scalar case, and write carefully the probability functions.
Let $A, B \sim N(0,I_p)$, with densities $p_A(A)$, $p_B(B)$ and consider the transformation {$A,B$ } $\to$ {$C,B$ }
with $C=A \cdot B $.
Then $p_{B,C}(B,C) = p_{C|B}(C , B) p_B(B)$
The first factor is, for fixed B, a gaussian - but a gaussian scaled by that factor. Hence
$p_{C|B}(C,B) = \frac{1}{B} p_A(\frac{C}{B}) $
and so
$p_{B,C}(B,C) = \frac{1}{B} p_A(\frac{C}{B}) p_B{B}$ 
is not the product of two gaussian densities, and the conclusion does not follow.
