Holonomic gradient method for distribution function of a weighted sum of noncentral chi-square random variables

Abstract

We apply the holonomic gradient method to compute the distribution function of a weighted sum of independent noncentral chi-square random variables. It is the distribution function of the squared length of a multivariate normal random vector. We treat this distribution as an integral of the normalizing constant of the Fisher–Bingham distribution on the unit sphere and make use of the partial differential equations for the Fisher–Bingham distribution.

Appendix: A general form of Proposition 3.1

In Proposition 3.1 we assumed $\lambda _1 > \lambda _2$. In this appendix we state the following proposition for the general case $\lambda _1 = \dots = \lambda _m > \lambda _{m+1}$ without a proof. For this case, the integrand for the Fisher–Bingham integral takes its maximum on the $(m-1)$-dimensional sphere $S^{m-1}(1)$, rather than on a finite number of points. However by appropriate choice of coordinates and by multiplication of the volume $\mathrm{Vol}(S^{m-1}(1))$, the derivation of Proposition 5.1 is basically the same as Proposition 3.1.

Proposition 5.1

Assume that

$$\begin{aligned} 0 > \lambda _1 = \dots = \lambda _m > \lambda _{m+1} \ge \dots \ge \lambda _d. \end{aligned}$$

If $0=\tau _1 = \dots = \tau _m$, then as $r\rightarrow \infty$,

$$\begin{aligned} f(\lambda ,\tau ,r)&=r^{m-1} S_{m-1} \exp \left( r^2 \lambda _1 -\sum _{i=m+1}^d \frac{\tau _i^2}{4(\lambda _i - \lambda _1)} \right) \\&\qquad \times \frac{\pi ^{(d-m)/2}}{\prod _{i=m}^d (\lambda _1-\lambda _i)^{1/2}} (1 + o(1)), \\ \partial _{\lambda _j} \, f(\lambda ,\tau ,r)&= \frac{r^2}{m} f(\lambda ,\tau ,r)(1+o(1)), \quad j\le m, \\ \partial _{\tau _j} \, f(\lambda ,\tau ,r)&= 0, \quad j\le m, \\ \partial _{\tau _j} \, f(\lambda ,\tau ,r)&= -\frac{\tau _j}{2(\lambda _j-\lambda _1)} f(\lambda ,\tau ,r)(1+o(1)), \quad j>m, \\ \partial _{\lambda _j} f(\lambda ,\tau ,r)&= \left( \frac{1}{2(\lambda _1-\lambda _j)} + \frac{\tau _j^2}{4(\lambda _j-\lambda _1)^2}\right) f(\lambda ,\tau ,r)(1+o(1)), \quad j>m. \end{aligned}$$

If $(\tau _1,\dots ,\tau _m) \ne (0,\dots ,0)$, define $\gamma = (\tau _1^2 + \dots + \tau _m^2)^{1/2}$. Then, as $r\rightarrow \infty$,

$$\begin{aligned} f(\lambda ,\tau ,r)&= \exp \left( r^2 \lambda _1 + r\gamma -\sum _{i=m+1}^d \frac{\tau _i^2}{4(\lambda _i - \lambda _1)} \right) \left( \frac{2r}{\gamma }\right) ^{(m-1)/2}\\&\quad \times \frac{ \pi ^{(d-1)/2}}{\prod _{i=m}^d (\lambda _1-\lambda _i)^{1/2}}(1+o(1)),\\ \partial _{\tau _j}\, f(\lambda ,\tau ,r)&= r \frac{\tau _j}{\gamma } f(\lambda ,\tau ,r)(1+o(1)), \quad \tau _j \ne 0, \ j\le m, \\ \partial _{\lambda _j}\, f(\lambda ,\tau ,r)&= r^2 \frac{\tau _j^2}{\gamma ^2} f(\lambda ,\tau ,r)(1+o(1)), \quad \tau _j\ne 0,\ j\le m,\\ \partial _{\tau _j}\, f(\lambda ,\tau ,r)&=0, \quad \tau _j=0, \ j\le m,\\ \partial _{\lambda _j}\, f(\lambda ,\tau ,r)&= \frac{r}{\gamma } f(\lambda ,\tau ,r)(1+o(1)), \quad \tau _j=0,\ j\le m, \\ \partial _{\tau _j}\, f(\lambda ,\tau ,r)&= -\frac{\tau _j}{2(\lambda _j-\lambda _1)} f(\lambda ,\tau ,r)(1+o(1)), \quad j>m, \\ \partial _{\lambda _j}\, f(\lambda ,\tau ,r)&= \left( \frac{1}{2(\lambda _1-\lambda _j)} + \frac{\tau _j^2}{4(\lambda _j-\lambda _1)^2}\right) f(\lambda ,\tau ,r)(1+o(1)), \quad j>m. \end{aligned}$$