Is the product of two Gaussian random variables also a Gaussian? Say I have $X \sim \mathcal N(a, b)$ and $Y\sim \mathcal N(c, d)$.  Is $XY$ also normally distributed?
Is the answer any different if we know that $X$ and $Y$ are independent?
 A: As @Yemon Choi showed in the first question, without any hypothesis the answer is negative since $P(X^2<0)=0$ whereas $P(U<0)\neq 0$ if $U$ is Gaussian.
For the second question the answer is also no. Take $X$ and $Y$ two Gaussian random variables with mean $0$ and variance $1$. Since they have the same variance, $X-Y$ and $X+Y$ are independent Gaussian random variables. Put $Z:=\frac{X^2-Y^2}2=\frac{X-Y}{\sqrt 2}\frac{X+Y}{\sqrt 2}$. Then $Z$ is the product of two independent Gaussian, but the characteristic function of $Z$ is $\varphi_Z(t)=\frac 1{\sqrt{1+t^2}}$, which is not the characteristic function of a Gaussian.
A: For a general answer to your question please refer to the Wishart distribution.
A: You can use moments to see that  the product $XY$
 of independent normals 
 cannot be normal except in trivial cases. By trivial, I mean $\mathbb{V}(X)\mathbb{V}(Y)=0.$ 
Suppose that $X,Y$ are independent normals so that $XY$ is normal. 
Case 1: Suppose that  $\mathbb{E}(X)=0$. By independence, $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)=0$, so $XY$ is mean zero normal, and hence
$$\mathbb{E}((XY)^4)=3\mathbb{E}((XY)^2)^2.$$ By independence we get
$$\mathbb{E}(X^4)\mathbb{E}(Y^4)=3\mathbb{E}(X^2)^2\mathbb{E}(Y^2)^2.$$
Either  $\mathbb{V}(X)=0$, or dividing by $\mathbb{E}(X^4)$ gives $ \mathbb{E}(Y^4)= \mathbb{E}(Y^2)^2.$
This shows that $Y^2$, and hence $Y$, has zero variance. 
Case 2: Suppose that $\mathbb{E}(X^2)>0$ and $\mathbb{E}(Y^2)>0$.  Then, without loss of generality, $\mathbb{E}(X^2)=1$ and $\mathbb{E}(Y^2)=1$.
In this case, we also have $\mathbb{E}((XY)^2)=1$, so 
$$\begin{eqnarray}
\mathbb{E}(X^3)&=&\mathbb{E}(X)(3-2\mathbb{E}(X)^2)\\
\mathbb{E}(Y^3)&=&\mathbb{E}(Y)(3-2\mathbb{E}(Y)^2)\\
\mathbb{E}((XY)^3)&=&\mathbb{E}(XY)(3-2\mathbb{E}(XY)^2)
\end{eqnarray}
$$
Subtracting the product of the first two lines from the third line gives 
$$0=6\mathbb{E}(X)\,\mathbb{E}(Y)\,\mathbb{V}(X)\,\mathbb{V}(Y).$$ 
Either we are in a trivial case, or back to Case 1. 
Thus, the product cannot be normal except in trivial cases. 
A: As pointed out by Davide Giraudo, the characteristic function of Z is
$\varphi_Z (t) = \frac{1}{\sqrt{1+t^2}}$. So, the corresponding probability density is given by:
$$\eqalignno{f_Z(z)&={1\over2\pi}\int_{-\infty}^{+\infty}e^{-itz}\varphi_Z(t)\,dt={1\over2\pi}\int_{-\infty}^{+\infty}{\cos zt-i\,{\rm sen}\  zt\over\sqrt{1+t^2}}\,dt\cr
&={1\over2\pi}\int_{-\infty}^{+\infty}{\cos zt\over\sqrt{1+t^2}}\,dt-{i\over2\pi}\int_{-\infty}^{+\infty}{{\rm sen}\  zt\over\sqrt{1+t^2}}\,dt.}$$
The last of these integrals is null because the integrand is an odd function. Therefore,  ${\Im[f_Z(z)]=0}$ and we have
$$f_Z(z)={1\over\pi}\int_0^{+\infty}{\cos zt\over\sqrt{1+t^2}}\,dt.$$
For $z > 0$ this integral converges and represents the zero order modified Bessel function of second species, usually denoted as $K_{0}(z)$. Thus,
$$f_Z(z)={1\over\pi}K_0(|\,z\,|), \qquad |\,z\,|>0$$
This shows that the random variable $Z$ has a K-form Bessel distribution. For more information see, for instance, Johnson and Kotz. Distributions in Statistics --- Continuous Univariate Distributions. Boston, Houghton-Mifflin, 1970. See also https://en.wikipedia.org/wiki/K-distribution. 
A: The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables:
$$ XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$ 
Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom.
If $X$ and $Y$ are both zero-mean, then
$$ XY \sim c_1 Q - c_2 R$$
where $c_1=\frac{Var(X+Y)}{4}$, $c_2 = \frac{Var(X-Y)}{4}$ and $Q, R \sim \chi^2_1$ are central.
The variables $Q$ and $R$ are independent if and only if $Var(X) = Var(Y)$.
In general, $Q$ and $R$ are noncentral and dependent.
A: The answer about using a Chi Squared doesn't seem helpful here since  and  are almost certainly noncentral and dependent. There is a note about using a Normal Product Distribution which seems highly relevant, though I haven't seen it in practice and is less flexible than what I propose. What I have seen performed is the Delta Method or bootstrap to estimate this distribution.
The Delta Method says that an arbitrary function of normals is also approximately normal and in your case the $Var(XY)\approx Var(X)\hat Y^2 + Var(Y)\hat X^2 + 2Cov(X,Y)\hat X \hat Y$ where $E(X) = \hat X$.
I have seen this used to compute the CI for a ratio of normals X/Y after taking a log transform.
Hope that helps others that stumble upon this as I have, but feel free to correct me if I am wrong.
