Product of two Gaussian pdf's with different dimensions Consider a two-dimensional normal density $N((x_1,x_2)\vert \mu,\Sigma)$ and a one-dimensional normal density $N(x_1\vert\mu',\Sigma')$.
Question: Is the product $ p(x_1,x_2) = N((x_1,x_2)\vert \mu,\Sigma)\cdot N(x_1\vert\mu',\Sigma')$ normal (up to a proportionality constant) and what are its mean and covariance?
 A: This question has horrendous notation and it is not at all clear whether
the random variables are being multiplied, or just their densities. Nor
is it clear whether we have two random variables or three, and in the former
case, whether the $\mu^\prime$ and $\Sigma^\prime$ are consistent with
$\mu$ and $\Sigma$.
Case I: We have only two jointly normal random variables $X$ and $Y$
and are being asked whether
$f_{X,Y}(x,y)f_X(x)$ is a valid density function (up to a multiplicative
constant). The answer is YES, and the details are as follows.
For ease of expression, assume that $X$ and $Y$ are zero-mean
unit-variance jointly normal random variables with correlation
coefficient $\rho$ where $|\rho| < 1$. Then,
\begin{align}
f_{X,Y}(x,y)f_X(x) 
&\propto \exp\left(\left.\left.-\frac{1}{2(1-\rho^2)}\right[x^2+y^2-2\rho xy\right]\right)
\exp\left(-\frac 12x^2\right)\\
&= \exp\left(\left.\left.-\frac{1}{2(1-\rho^2)}\right[\left(1+(1-\rho^2)\right)x^2+y^2-2\rho xy\right]\right)\\
&= \exp\left(\left.\left.-\frac{1}{2(1-\rho^2)}\right[\left(2-\rho^2\right)x^2+y^2-2\rho xy\right]\right)\tag{1}
\end{align}
If the quadratic in the exponent be expressed in the form
$$\left.\left.-\frac{1}{2(1-\hat{\rho}^2)}\right[\left(\frac{x}{\sigma_X}\right)^2 
+ \left(\frac{y}{\sigma_Y}\right)^2 
-2\hat{\rho}\left(\frac{x}{\sigma_X}\right)\left(\frac{y}{\sigma_Y}\right)\right] \tag{2}$$
for appropriate choices of $\hat{\rho}, \sigma_X$ and $\sigma_Y$,
then we can say that $f_{X,Y}(x,y)f_X(x)$ is proportional
to a bivariate normal density. Comparing the coefficients of $x$, $y$ 
and $xy$ in $(1)$ and the rights side of $(2)$ gives us
\begin{align}
\frac{1}{1-\hat{\rho}^2}\frac{1}{\sigma_X^2} = \frac{2-\rho^2}{1-\rho^2}
&\implies \frac{1}{\sigma_X^2} = (1-\hat{\rho}^2)\frac{2-\rho^2}{1-\rho^2}\tag 3\\
\frac{1}{1-\hat{\rho}^2}\frac{1}{\sigma_Y^2} = \frac{1}{1-\rho^2}
&\implies \frac{1}{\sigma_Y^2} = \frac{1-\hat{\rho}^2}{1-\rho^2}
\tag 4\\
\frac{\hat{\rho}}{1-\hat{\rho}^2}\frac{1}{\sigma_X\sigma_Y} = \frac{\rho}{1-\rho^2}
&\implies \hat{\rho} = \frac{\rho}{\sqrt{2-\rho^2}}.\tag 5
\end{align}
Note that we substituted the values of $\sigma_X$ and $\sigma_Y$ from
$(3)$ and $(4)$ into $(5)$ to determine $\hat{\rho}$ in terms of $\rho$.
Note also that $\hat{\rho}| \leq |\rho|$ is a valid
value for a purported correlation coefficient. Back-substitution gives us that
$$\sigma_X = \frac{1}{\sqrt 2}, \quad \sigma_Y = \sqrt{1-\frac{\rho^2}{2}}. \tag 6$$
What if $X$ and $Y$ have nonzero means and/or non-unit variances? 
Well, a similar analysis holds with messier calculations leading
to the same conclusion about the variances: the scale factors
relating the new variances to the old are as specified in $(6)$ giving
reductions of $\frac{1}{\sqrt 2}$ and $\sqrt{1-\frac{\rho^2}{2}}$
respectively. 
Note, though, that the means remain the same.
What if $|\rho| = 1$? In this case, $X$ and $Y$ do not have a joint
density at all but a degenerate density in which all the probability
mass lies on a straight line (of zero area). The above results are still
applicable in the sense that the product $f_{X,Y}(x,y)f_X(x)$ 
is also (proportional to) a degenerate density (note
that $\hat{\rho} = 1$}, and the
variances reducing by factors of $\frac{1}{\sqrt 2}$
and
$\sqrt{1-\frac{\rho^2}{2}} = \frac{1}{\sqrt 2}$.
Case II: we have three random variables $X,Y,X$ and
want to know whether the the joint density of $(X,Y)Z = (XZ,YZ)$
is normal. The answer here is NO, even in special case
of $Z = X$ when $XZ = X^2$ has a (scaled and/or noncentral)
$\chi^2$ distribution. 
