Consider each $X_i \sim N(0,1)$. Then the random variable $Y=\sum_{i=1}^n X_i^2$ is a $\chi^2$ distribution with $n$ degree of freedom. Is there any probability distribution about a weighted sum of the square of standard normal random variables $Z=\sum_{i=1}^n w_i X_i^2$?
[Math] Generalized $\chi^2$ distribution
probability distributions
Related Solutions
In other words, gives two independent random variables $X$ and $Y$, distributed according to hypoexponential distribution with parameters $\{ w_1, w_2, \ldots, w_n \}$ and $\{v_1,v_2, \ldots, v_m \}$ respectively, you are asking to determine the distribution of $Z=X-Y$.
Let $\Theta_X$, and $\Theta_Y$ denote matrices from the probability density functions of respective hypoexponential distributions, see wiki page: $$ f_X(x) = - \langle\vec{\alpha}_n, \exp(x \Theta_X) \Theta_X, \vec{1}_n \rangle \cdot [ x > 0 ] \qquad \qquad f_Y(y) = - \langle\vec{\alpha}_m, \exp(y \Theta_Y) \Theta_Y, \vec{1}_m \rangle \cdot [ y > 0 ] $$ where $(\alpha_n)_i = \delta_{i,1}$, $ (\vec{1}_n)_i = 1$ and $(\alpha_m)_j = \delta_{j,1}$, $ (\vec{1}_m)_j = 1$, $i=1,\ldots,n$, and $j=1,\ldots,m$.
Then $$ \begin{eqnarray} f_Z(z) &=& \int_{-\infty}^\infty f_X(z+y) f_Y(y) \mathrm{d} y = \int_{\max(-z,0)}^\infty f_X(z+y) f_Y(y) \mathrm{d} y \\ &=& \int_{\max(-z,0)}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(z+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{y \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\ &=& \int_{0}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(\max(z,0)+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{(\max(-z,0) + y) \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\ &=& \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{\max(z,0) \Theta_X} \otimes \mathrm{e}^{(\max(-z,0) ) \Theta_Y} \right) \cdot \left( \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y \right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \end{eqnarray} $$ The formula above tells the density function for $Z$ will be piecewise, much like Laplace distribution, with functional form of $X$ variate for $z>0$ and functional form of $Y$ variate for $z<0$.
Example: Consider an example with $n=2$ and $m=2$, and $\{w_1,w_2\} = \{1,2\}$, and $\{v_1,v_2\} = \{1,1\}$. Corresponding matrices are $$ \Theta_X = \left( \begin{array}{cc} -1 & 1 \\ 0 & -2 \end{array} \right) \qquad \Theta_Y = \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right) $$ Then $$ \exp\left(x \Theta_X \right) = \left( \begin{array}{cc} e^{-x} & e^{-x}-e^{-2 x} \\ 0 & e^{-2 x} \\ \end{array} \right) \qquad \exp\left(y \Theta_Y \right) = \left( \begin{array}{cc} e^{-y} & e^{-y} y \\ 0 & e^{-y} \\ \end{array} \right) $$ $$ \exp\left(x \Theta_X \right) \Theta_X = \left( \begin{array}{cc} -e^{-x} & 2 e^{-2 x}-e^{-x} \\ 0 & -2 e^{-2 x} \\ \end{array} \right) \qquad \exp\left(y \Theta_Y \right) \Theta_Y = \left( \begin{array}{cc} -e^{-y} & e^{-y}-e^{-y} y \\ 0 & -e^{-y} \\ \end{array} \right) $$ Using Kronecker product, $$ \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y = \left( \begin{array}{cccc} \frac{1}{2} & -\frac{1}{4} & -\frac{1}{6} & \frac{7}{36} \\ 0 & \frac{1}{2} & 0 & -\frac{1}{6} \\ 0 & 0 & \frac{2}{3} & -\frac{4}{9} \\ 0 & 0 & 0 & \frac{2}{3} \\ \end{array} \right) $$ Combining things, with little algebra we get: $$ f_Z(z) = \left\{ \begin{array}{cc} \frac{5}{18} & z=0 \\ \frac{1}{18} \mathrm{e}^{-z} \left(9-4 \mathrm{e}^{-z}\right) & z>0 \\ \frac{1}{18} \mathrm{e}^z (5-6 z) & z < 0 \\ \end{array} \right. $$
Since you're probably performing a chi-squared test, then test-statistic $$ X^2=\sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i} $$ follows a $\chi^2(p)$ distribution with $p=n-1$ degrees of freedom. Since large values are critical, the corresponding $p$-value is given by $$ P(\chi^2(p)\geq X^2)=1-F_{\chi^2(p)}(X^2), $$ i.e. the probability of a $\chi^2(p)$-variable being larger than what we have observed. Here $F_{\chi^2(p)}$ is the distribution function of a $\chi^2(p)$ distribution.
This is easily calculated in R with the command 1-pchisq(x,p), where x denotes the test-statistic $X^2$ and p is the number of degrees of freedom.
Best Answer
Distributions of norm squared of multi-normal distribution had been extensively studied in radio-communication theory. See paper of S.O. Rice, "Distribution of Quadratic Forms in Normal Random Variables".
It is quite easy to find the characteristic function of $Z$, since $$ \mathbb{E} \left( \mathrm{e}^{i t X_i^2} \right) = \frac{1}{\sqrt{1-2 i t}} $$ Therefore $$ \mathbb{E} \left( \mathrm{e}^{i t Z} \right) = \left( \prod_{i=1}^n 1-2 i t w_i \right)^{-1/2} $$ Inversion in closed form is possible only in special cases.