Transformations of random variables in the form of polynomials I am trying to find the probability density function (pdf) of $F$. $F$ is the polynomial of three independent random variables $X, Y, Z$ , where $$F = aX+bY+cZ+dX^2+eY^2+fZ^2+gXY+hYZ+iXZ, $$ and $a, b, c, d, e, f, g, h, i$ are constants. We know that the random variables $X, Y, Z$ are independent and each of them satisfies Gaussian distribution. I do not know how to derive the pdf of $F$. If there are general solutions to this kind of problem? Any kind of help would be appreciated.
 A: If your polynomial was  $p(x,y,z)= a_{11}x^2 + a_{22}y^2 + a_{33}z^2 + (2a_{12})xy + (2a_{13})xz + (2a_{23})yz$, then
$$ p(x,y,z) = (X,Y,Z) A(X,Y,Z)^T $$
where $A=((a_{ij})_{1\leq i,j\leq 3})$. Since $A$ is symmetric, its eigenvalues are real and the eigenvectors are orthogonal, i.e.
$$ A = V\Lambda V^T $$
where $\Lambda$ is diagonal, and $V\in O(3)$. Let $(X,Y,Z)^T = V (\xi,\eta,\zeta)^T $, then
$$ (X,Y,Z)A(X,Y,Z)^T= (\xi,\eta,\zeta) V^T  A V (\xi,\eta,\zeta)^T = (\xi,\eta,\zeta) \Lambda (\xi,\eta,\zeta)^T = \lambda_1 \xi^2 + \lambda_2\eta^2 + \lambda_3 \zeta^2 . $$
Now since $V$ is orthogonal $(\xi,\eta,\zeta)\sim N(0,I)$. Hence the distribution of $p(X,Y,Z)$ is some generalized version of $\chi^2$. 
Now if $p(x,y,z)$ is a general bivariate polynomial of degree $2$, then it is possible to write it in the form
$$ p(x,y,z)= (x-a,y-b,z-c) A (x-a,x-b,z-c)^T +d  $$
where $a,b,c,d$ are appropriate constants and $A$ is an appropriate $3\times 3$ symmetric matrix. Since $(X-a,Y-b,Z-c)\sim N((-a,-b,-c), I) $ the closest one can say about the distribution of $p(X,Y,Z)$ is that it is some generalized version of $\chi^2$ with shift $d$.
Update: The above derivation assumed that $(X,Y,Z)$ is a standard Gaussian vector, i.e. its mean is the zero vector and its covariance matrix is $I$, the identity matrix. In case $(X,Y,Z)\sim N(\mu, diag(\sigma_1^2,\sigma_2^2,\sigma_3^2))$ then $diag(1/\sigma_1,1/\sigma_2,1/\sigma_3)\cdot (X-\mu_1,Y-\mu_2,Z-\mu_3) \sim N(0,I) $.
