RBF transformation on a Normally Distributed Random Variable I have a random vector $\mathbf{X} \sim \mathcal{N}(\mathbf{m,\Sigma})$ which is transformed by a Gaussian Radial Basis Function into the random variable $\mathbf{Y} = K(\mathbf X) = \exp(-\lambda ||\mathbf X||^2)$ is there an analytic expression for the PDF or atleast the mean and variance of this new variable?
 A: $\DeclareMathOperator\E{\mathbb E}
\DeclareMathOperator\Var{\mathrm{Var}}$As
you noted, $\lVert X \rVert^2$ will have a generalized chi-squared distribution.
If you want a cdf or pdf computationally, the best way is probably to go through that distribution and do a change of variables using the transformation $g(x) = \exp(- \lambda x)$. Since $g$ is monotonic, we have
$g^{-1}(y) = - \frac{1}{\lambda} \log(y)$
and so the density of $Y$ is
$$
f_Y(y)
= \left\lvert \frac{\mathrm d}{\mathrm{d}y}g^{-1}(y) \right\rvert \, f_{\lVert X \rVert^2}(g^{-1}(y))
= \frac{1}{\lambda y} f_{\lVert X \rVert^2}\left( - \frac{1}{\lambda} \log(y) \right)
$$
where $f_{\lVert X \rVert^2}$ is the pdf of $\lVert X \rVert^2$.
I cataloged some papers related to approximating the distribution here, where @caracal points out the R package CompQuadForm implements some of the approximations.

The mean and variance, though, are available analytically, if in a somewhat inconvenient form:
Note that
$\E Y = \E \exp(- \lambda \lVert X \rVert^2) = M(- \lambda)$, where $M$ denotes the moment-generating function of $\lVert X \rVert^2$. Assuming $\Sigma$ is nonsingular, that mgf is, by Theorem 3.2a.1 (page 40) of Mathai and Provost, Quadratic Forms in Random Variables, CRC Press 1992 (free scan from Mathai available on researchgate):
$$
M(t)
=
\lvert I - 2 t \Sigma \rvert^{-\frac12} \exp\left( - \tfrac12 m^T \left[ I - (I - 2 t \Sigma)^{-1} \right] \Sigma^{-1} m \right)
.$$
If $\Sigma$ is singular with $\Sigma = B B^T$, then Theorem 3.2a.3 (page 45) gives
$$
M(t)
=
\lvert I - 2 t B^T B \rvert^{-\frac12}
\exp\left(
t \lVert m \rVert^2
+ 2 t^2 m^T B (I - 2 t B^T B)^{-1} B^T m
\right)
.$$
Also note that
$$
\E Y^k = \E \exp\left( - \lambda \lVert X \rVert^2 \right)^k = \E \exp\left( - k \lambda \lVert X \rVert^2 \right)
$$
and so
$$
\Var Y
= \E Y^2 - (\E Y)^2
= M(- 2 \lambda) - M(\lambda)^2
.$$
(Hat tip to this answer by @NRH for the pointer to that book.)
