1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ By a Gaussian RBF you mean $K(x) = \exp\left( - \gamma \lVert x - c \rVert^2 \right)$, right? Hint: the norm part is chi-squared distributed, and then you have a monotonic transformation of that. $\endgroup$ – Dougal Jul 28 '15 at 5:13
  • $\begingroup$ My statistics knowledge is so rusty, I had no idea about this distribution thank you! $\endgroup$ – jshe857 Jul 28 '15 at 5:55
  • $\begingroup$ Reading up on the distribution I see that there is the Generalized and Non-central distribution. Is there similarly analysis for when both the mean and variance for the normal distribution can be non standard? $\endgroup$ – jshe857 Jul 28 '15 at 6:27
  • $\begingroup$ Sorry, I totally forgot last night that the noncentrality would make things annoying. Just answered with more details. $\endgroup$ – Dougal Jul 28 '15 at 18:55
1
$\begingroup$

$\DeclareMathOperator\E{\mathbb E} \DeclareMathOperator\Var{\mathrm{Var}}$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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.