# Difference of two non-central chi squared random variables

I have a problem where I want to find the distribution of the difference of two non-central chi squared random variables (RV), both independent. Given $$X=a+A\\ Y=b+B$$ where $a=a_r+ja_i$, $b=b_r+jb_i$, $a_r,a_i,b_r,b_i\in\mathbb{R}$ are constants, $A=A_r+jA_i$, $B=B_r+jB_i$, $A_r,A_i,B_r,B_i\sim\mathcal{N}(0,1)$, all independent. The problem is calculating $$P(Z>z)=\int_z^{\infty}f_Z(x)dx$$ where $Z=|X|^2-|Y|^2$. I tried the following approaches:

1. $|X|^2=X_r^2+X_i^2$, where $X_r$ and $X_i$ are the real and imaginary parts of $X$ and the same for $Y$. $X_r^2$ and $X_i^2$ are non-central chi squared with non-centrality parameter $\lambda_X=|a|^2$ and $2$ degrees of freedom (DoF). I know that adding two non-central chi squared is also a non-central chi squared, but it is not clear to me what is the difference of two non-central chi squared.
2. Via Moment Generating Function (MGF). The MGF of the difference of two RV is the product of individual MGF, one with negative argument: $$M_{U-V}(t)=M_U(t)M_V(-t)$$ for arbitrary $U$, $V$ RV. Thus, $$M_Z(t)=M_{|X|^2}(t)M_{|Y|^2}(-t)=\frac{e^{\frac{\lambda_X t}{1-2t}-\frac{\lambda_Y t}{1+2t}}}{1-4t^2}$$ The pdf can be found via the inverse of Laplace transform of the MGF, but I cannot see if the $M_Z$ has a closed-form inverse.
3. Convolving. $$f_Z(z)=\int_{-\infty}^{\infty}f_{|X|^2}(x)f_{|Y|^2}(x+z)dx.$$ I tried to solve this integral using:
• The pdf expressed as the modified Bessel function of first kind $I_0$ : $$f_{|X|^2}(x)=\frac{1}{2}e^{-\frac{x+\lambda_X}{2}}I_0\left(\sqrt{\lambda_X x}\right)$$ In this case, the convolution is an integral of an exponential and product of two Bessel functions. I used the book of "Tables of Integrals" of Gradshteyn and Ryzhik but I could not find any solution.
• The expression of Bessel $I_0$ as infinite series. At the end I have an integral of two infinite series. I tried to express this product as a Cauchy product but arguments are different (one is $x$ and the other is $x+z$).
• The pdf expressed as a mixture. But at the end I have a convolution of two infinite series of functions.
4. Developing the squares. $Z$ can be rewritten as $$Z=|a|^2-|b|^2+2\left(a_rA_r+a_iA_i-b_rB_r-b_iB_i\right)+\left(A_r^2+A_i^2-B_r^2-B_i^2\right)=N+L$$ where $N\sim\mathcal{N}\left(|a|^2-|b|^2,|a|^2+|b|^2\right)$ and $L\sim\mathcal{L}(0,2)$ is a Laplace RV. The sum of $N+L$, though, is not clear what distribution is, since $N$ and $L$ are dependent. The MGF of $N+L$ is the previous one, as expected.

I would appreciate any hint you could give me. Thank you.

I prefer to write the pdf as $\frac{1}{2}e^{-\frac{x+|a|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}x\mid1\right)$ where $_{0}F_{1}$ is the hypergeometric function of order $(0,1)$.
For $z\geq0$, \begin{align*} \Pr\left(|X|^{2}-|Y|^{2}>z\right) &=\int_{0}^{\infty}\Pr\left(v-z>|Y|^{2}\right)\frac{1}{2}e^{-\frac{v+|a|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}v\mid1\right)\mathrm{d}v \\&=\int_{z}^{\infty}\Pr\left(v-z>|Y|^{2}\right)\frac{1}{2}e^{-\frac{v+|a|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}v\mid1\right)\mathrm{d}v \\&=\int_{0}^{\infty}\Pr\left(v>|Y|^{2}\right)\frac{1}{2}e^{-\frac{v+z+|a|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}v \\&=\int_{0}^{\infty}\left(\int_{0}^{v}\frac{1}{2}e^{-\frac{w+|b|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}w\mid1\right)\mathrm{d}w\right) \frac{1}{2}e^{-\frac{v+z+|a|^{2}}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}v \\&=\frac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \int_{0}^{\infty}\left(\int_{0}^{v}e^{-\frac{w}{2}}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}w\mid1\right)\mathrm{d}w\right) e^{-\frac{v}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}v \\&=\frac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \int_{0}^{\infty}\left(\int_{0}^{1}e^{-\frac{vr}{2}}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}vr\mid1\right)v\mathrm{d}r\right) e^{-\frac{v}{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}v \\&=\frac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \int_{0}^{\infty}\int_{0}^{1}ve^{-\frac{v}{2}(1+r)}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}vr\mid1\right)\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}r\mathrm{d}v \end{align*} Now, \begin{align*} _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right) &=\sum_{n=0}^{\infty}\left(\frac{|a|}{2}\right)^{2n}\frac{(v+z)^{n}}{(n!)^{2}} \\&=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{|a|}{2}\right)^{2n} \sum_{k=0}^{n}\frac{z^{k}}{k!}\frac{v^{n-k}}{(n-k)!} \\&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{1}{n!}\left(\frac{|a|}{2}\right)^{2n}\frac{z^{k}}{k!}\frac{v^{n-k}}{(n-k)!} \\&=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\frac{1}{(n+k)!}\left(\frac{|a|}{2}\right)^{2n+2k}\frac{z^{k}}{k!}\frac{v^{n}}{n!} \\&=\sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{k!}\sum_{n=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}v\right)^{n}}{n!}\frac{1}{(n+k)!} \\&=\sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}}\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}v\mid k+1\right) \end{align*} Then \begin{align*} &\Pr\left(|X|^{2}-|Y|^{2}>z\right) \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \int_{0}^{1}\int_{0}^{\infty}ve^{-\frac{v}{2}(1+r)}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}vr\mid1\right)\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}(v+z)\mid1\right)\mathrm{d}v\mathrm{d}r \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \int_{0}^{1}\int_{0}^{\infty}ve^{-\frac{v}{2}(1+r)}\ _{0}F_{1}\left(\tfrac{|b|^{2}}{4}vr\mid1\right)\ _{0}F_{1}\left(\tfrac{|a|^{2}}{4}v\mid k+1\right)\mathrm{d}v\mathrm{d}r \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \int_{0}^{1}\int_{0}^{\infty}ve^{-\frac{v}{2}(1+r)}\ \left(\sum_{m=0}^{\infty}\frac{\left(\tfrac{|b|^{2}}{4}vr\right)^{m}}{(m!)^{2}}\right) \left(\sum_{n=0}^{\infty}\frac{\left(\tfrac{|a|^{2}}{4}v\right)^{n}}{n!(n+k)!}\right)\mathrm{d}v\mathrm{d}r \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{\left(\tfrac{|b|^{2}}{4}\right)^{m}}{(m!)^{2}} \frac{\left(\tfrac{|a|^{2}}{4}\right)^{n}}{n!(n+k)!} \int_{0}^{1}r^{m} \int_{0}^{\infty}v^{n+m+1}e^{-\frac{1+r}{2}v}\ \mathrm{d}v\mathrm{d}r \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \int_{0}^{1}\int_{0}^{\infty}ve^{-\frac{v}{2}(1+r)}\ \left(\sum_{m=0}^{\infty}\frac{\left(\tfrac{|b|^{2}}{4}vr\right)^{m}}{(m!)^{2}}\right) \left(\sum_{n=0}^{\infty}\frac{\left(\tfrac{|a|^{2}}{4}v\right)^{n}}{n!(n+k)!}\right)\mathrm{d}v\mathrm{d}r \\&=\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{\left(\tfrac{|b|^{2}}{4}\right)^{m}}{(m!)^{2}} \frac{\left(\tfrac{|a|^{2}}{4}\right)^{n}}{n!(n+k)!}(n+m+1)! \int_{0}^{1}r^{m}\left(\frac{2}{1+r}\right)^{n+m+2}\mathrm{d}r \\&=e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{\left(\tfrac{|b|^{2}}{2}\right)^{m}}{(m!)^{2}} \frac{\left(\tfrac{|a|^{2}}{2}\right)^{n}}{n!(n+k)!}(n+m+1)! \int_{0}^{\frac{1}{2}}x^{m}(1-x)^{n}\mathrm{d}x \\&=e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{3}} \int_{0}^{\frac{1}{2}}\Psi_{2}\left(\tfrac{|a|^{2}}{2}(1-x),\tfrac{|b|^{2}}{2}x\mid2;k+1,1\right)\mathrm{d}x \end{align*} Where $\Psi_{2}$ is one of the Humbert series
If you were to evaluate the convolution from your post, for $z\geq0$, $$\tfrac{1}{4}e^{-\frac{z+|a|^{2}+|b|^{2}}{2}} \sum_{k=0}^{\infty}\frac{\left(\frac{|b|^{2}}{4}z\right)^{k}}{k!} \Psi_{2}\left(\tfrac{|a|^{2}}{2},\tfrac{|b|^{2}}{2}\mid2;1,k+1\right)$$
• Yes, a seemingly crude bound can be obtained from $\sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{\left(\tfrac{|b|^{2}}{2}\right)^{m}}{(m!)^{2}} \frac{\left(\tfrac{|a|^{2}}{2}\right)^{n}}{n!(n+k)!}(n+m+1)! \int_{0}^{\frac{1}{2}}x^{m}(1-x)^{n}\mathrm{d}x$ by integrating over $[0,1]$ instead of $[0,\tfrac{1}{2}]$. It gives $\sum_{k=0}^{\infty}\frac{\left(\frac{|a|^{2}}{4}z\right)^{k}}{(k!)^{2}} e^{\tfrac{|b|^{2}}{2}}\sum_{n=0}^{\infty} \frac{\left(\tfrac{|a|^{2}}{2}\right)^{n}}{(n+k)!}$.The later series can be dealt with – shdp Mar 14 '17 at 10:07