6
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ I'm not statistically literate enough to see what could be wrong with your argument, but it must be wrong. In the special case of dimension 1, this would say the distribution of the product of two Normally distributed variables is Normal, and I'm fairly sure that's false. And I don't think restricting to higher dimensions will help, since the dimension 1 case can be approximated using an $I_p$ that has very small variance in $n-1$ out of $n$ dimensions. $\endgroup$
    – 2'5 9'2
    Jun 2, 2011 at 10:09

3 Answers 3

5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$p(z)$ 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 $p(x'z|z)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

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