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?
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$\begingroup$ 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. $\endgroup$– DanicaCommented Jul 28, 2015 at 5:13
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$\begingroup$ My statistics knowledge is so rusty, I had no idea about this distribution thank you! $\endgroup$– jshe857Commented Jul 28, 2015 at 5:55
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$\begingroup$ 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? $\endgroup$– jshe857Commented Jul 28, 2015 at 6:27
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$\begingroup$ Sorry, I totally forgot last night that the noncentrality would make things annoying. Just answered with more details. $\endgroup$– DanicaCommented Jul 28, 2015 at 18:55
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$\DeclareMathOperator\E{\mathbb E} \DeclareMathOperator\Var{\mathrm{Var}}$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.)
