product of densities One can frequently read, that the product of the densities of two INDEPENDENT random variables is also a density - the joint density of the two variables.
(see for example: http://en.wikipedia.org/wiki/Joint_probability_distribution#Joint_distribution_for_independent_variables)
One can also read, that IN GENERAL the product of two normal pdf is a Gaussian, but not a normal pdf.  i.e. one would have to multiply the product with a scaling factor (normalization constant) to get a normal pdf.
(see for example:  http://www.tina-vision.net/docs/memos/2003-003.pdf page 3, first paragraph)
My naive interpretation of this would be, that in case of independence, the normalization constant equals 1.  But the formula given for the scaling factor in the second source does not seem to support this...
Where is my misunderstanding?
Now, for the multivariate case, i.e. the product of two joint pdf, each one being the joint pdf of a vector of jointly normal variables, but the two vectors being independent of each other, one finds:
"The vectors x1, x2 are statistically independent if their joint distribution is f(x1, x2) = f(x1)f(x2) or, equivalently, if f(x1|x2) = f(x1) and f(x2|x1) = f(x2)."
(I do not have enough reputation points to post more than two links, so I replace the "tt" in http with "**":
h**p://www.le.ac.uk/users/dsgp1/COURSES/THIRDMET/MYLECTURES/5XMULTISTAT.pdf, the above quote can be found on page 3, number 6)
On the other hand, it says in another source:  "Suppose f(x) = N (x;1;1) and f(y) = N (x;2;2) are two INDEPENDENT d-dimensional Gaussian densities. Sometimes we want to compute the density which is proportional to the product of the two Gaussian densities, i.e. f(z) = cf(x)f(y), in which  c is a proper normalization constant to make f(z) a
valid density function."
h**p://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=28A2A856531E5FF9FABEC67397A87B5D?doi=10.1.1.1.2635&rep=rep1&type=pdf" - above quote is the first paragraph in section 7 on page 9 - here slightly modified notation.
These two last sources seem to contradict each other.  Is there a reconciling fact I'm missing?
Finally, the previous source given above provides an application in section 8.2 on page 11, where the Bayes theorem is applied to estimate the vector of mean values of a multivariate normal, the Bayes theorem is written there as follows:  
f(y|x) = cf(x|y)f(y)
with c being a "normalization constant".
more "traditional" descriptions of the Bayes theorem look more like this:
f(y|x)f(x) = f(x|y)f(y)
(see e.g.:  "h**p://en.wikipedia.org/wiki/Bayes%27_theorem#For_random_variables" (slightly rearranged here)
This seems to imply, that the "normalization constant" is actually (always) 1/f(x).
Is this correct?
Best,
JQ
 A: 
Where is my misunderstanding?

It's the distinction between $f(x)g(y)$ and $f(x)g(x)$. If $f$ and $g$ are densities, then $h(x,y):=f(x)g(y)$ is the joint density of $(X,Y)$ where $X\sim f$ and $Y\sim g$. This is true always, and no additional normalization is required. OTOH, the papers you are looking at are studying $h(x):=f(x)g(x)$, i.e., they are trying to identify the distribution of a single random variable/vector defined by the product of two densities. In general there is no reason to expect that $f(x)g(x)$ integrates to $1$, so you need to normalize. For example, $f(x):=e^{-x}I(x\ge0)$ and $g(x):=2e^{-2x}I(x\ge0)$ are proper univariate densities but $f(x)g(x)$ is not. [But $f(x)g(y)$ is a proper bivariate density.]

On the other hand, it says in another source: "Suppose f(x) = N (x;1;1) and f(y) = N (x;2;2) are two INDEPENDENT d-dimensional Gaussian densities. Sometimes we want to compute the density which is proportional to the product of the two Gaussian densities, i.e. f(z) = cf(x)f(y), in which c is a proper normalization constant to make f(z) a valid density function."

This source is also discussing $f(x)g(x)$ [N.B. the original text reads $p(x)=\alpha p_1(x)p_2(x)$], but seems full of typographical errors. It writes $p(z)$ where it should read $p(x)$. Also, I don't see where the independence is used, unless "independent" has the sense "possibly different".

This seems to imply, that the "normalization constant" is actually (always) 1/f(x).

You are correct! The normalization doesn't matter in the sequel, so it's written $\alpha$ for convenience.
