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.


1 Answer 1


You can write

$P(Z>z) = P\left(\frac{|X|^2}{|Y|^2+z}>1\right)$

with the assumption $|Y|^2+z \ne 0$. Then the problem turns to studying the ratio of two non-central chi-squared distributed random variables $|X|^2$ and $|Y|^2+z$, which follows the non-central F distribution described in http://mathworld.wolfram.com/NoncentralF-Distribution.html (doubly non-central) and https://en.wikipedia.org/wiki/Noncentral_F-distribution (singly non-central).

This question was partially answered before: Ratio of two independent noncentral $\chi^2$ random variables.


  • $\begingroup$ the support of the difference is $(-\infty, \infty)$ while the support of the ratio is $(0, \infty)$ SO how can the problem regarding difference be the same as the ratio? I see the math but I do not see how we can get the support from that ratio. Also the links point to F distribution which is strictly positive. Yet we know that the difference can be negative $\endgroup$
    – Onyambu
    Commented Dec 19, 2021 at 9:11

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