I'm trying to find the probability distribution of a function $q(x, y)=x^2+y^2$, where $x$ and $y$ are normally distributed: $x$~$N[0,\sqrt{V_x}]$ and $y$~$N[0,\sqrt{V_y}]$.
I began by finding the probability distribution of $x^2$ and $y^2$, and this is the distribution I got: $$Z_1=g(x)=x^2$$ $$Z_2=h(y)=y^2$$ $$f_{Z_1}(z_1)=\frac{e^{-\frac{z_1}{2V_x}}}{\sqrt{2 \pi V_x}\cdot\sqrt{z_1}}$$ $$f_{Z_2}(z_2)=\frac{e^{-\frac{z_2}{2V_y}}}{\sqrt{2 \pi V_y}\cdot\sqrt{z_2}}$$
I then try to find the distribution of the sum of the two variables with newly found distributions, by using the convolution of the two PDFs: $$Z=q(x,y)=x^2+y^2$$ $$f_Z(z)=\int_{-\infty}^{\infty}f_{Z_1}(x)f_{Z_2}(z-x)dx$$
But I can't seem to be able to solve this integral. Am I missing something? Is there a way to compute this integral? Or is there some other way to find the distribution of the sum of the two?
The density is $$f_Z(z)=\frac{d}{dz}\int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{z-x^2}}^{\sqrt{z-x^2}}f_X(x)f_Y(y)dydx=\frac{e^{a z}I_0\left(b z \right)}{c }$$ where:
$I_0(x)$ is the modified Bessel function of the first kind
$a=-\frac{1}{4}(\frac{1}{V_Y} + \frac{1}{V_X} )$
$b=\frac{1}{4}(\frac{1}{V_Y} - \frac{1}{V_X} )$
$c=2 \sqrt{V_XV_Y}$
You can find the Taylor series for the cdf by starting with the series for $I_0(x)$ and $e^{x}$, make the appropriate substitutions, integrating term by term: $$F_Z(z)=\frac{1}c t + \frac{a}{2c}t^2 + \frac{2a^2 + b^2}{12c}t^3 + ...$$
Here is a graph of the cdf (blue) and the approximation (orange) above when $V_X=V_Y=1$. You need more terms in the approximation to get better accuracy for $z$ far from $0$.
The same approach works using the convolution formula. The density is
$$f_Z(z)=\int_{0}^{z}f_{Z_1}(x)f_{Z_2}(z-x)dx=\int_{0}^{z}\frac{e^{-\frac{x}{2V_X}}}{\sqrt{2\pi V_X}\sqrt{x}} \frac{e^{-\frac{z-x}{2V_Y}}}{\sqrt{2\pi V_Y}\sqrt{z-x}}dx$$
$$=\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{0}^{z}\frac{e^{2 b x}}{\sqrt{x}\sqrt{z-x}}dx$$ Now, make the substitution $t=\frac{2x-z}{z}$ so that $x=\frac{z}{2}(1+t)$ and $dx=\frac{z}{2}dt$. The integral becomes
$$\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{-1}^{1}\frac{e^{2 b \frac{z}{2}(1+t)}}{\sqrt{1+t}\sqrt{1-t}}dx =\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{-1}^{1}(1-t^2)^{-\frac{1}2}e^{2 b \frac{z}{2}(1+t)}dx$$ Now, use this formula for the modified Bessel function to find the integral $$\int_{-1}^{1}(1-t^2)^{0-\frac{1}2}e^{2 b \frac{z}{2}(1+t)}dx=e^{b z}\pi I_0(b z)$$ After some simplification, you find $$f_Z(z)=\frac{e^{a z}I_0\left(b z \right)}{c}$$