What would be the distribution of the following equation:

$$y = a^2 + 2ad + d^2$$

where $a$ and $d$ are independent non-central chi-square random variables with $2 \textbf{M}$ degrees of freedom.

OBS.: The r.v.'s generating both $a$ and $d$ have $\mu = 0$ and $\sigma^2 \neq 1$, let's say $\sigma^2 = c$.

  • 2
    $\begingroup$ 1. How are $a$ and $d$ related? 2. Chi-square random variables already have mean > 0 Why would you need to state it explicitly? (Or are you trying to refer to a non-central chi-square?) $\endgroup$
    – Glen_b
    Commented Dec 27, 2016 at 2:40
  • $\begingroup$ I've just added some more information to the question. They are non-central chi-square r.v.'s as they were generated by non-standard circular symmetric complex Gaussian random variables. $\endgroup$ Commented Dec 27, 2016 at 19:45
  • $\begingroup$ 2M is the degrees of freedom for each of the two? $\endgroup$ Commented Dec 27, 2016 at 21:56
  • 2
    $\begingroup$ Felipe, in your question you state $a$ and $d$ do "have $\mu=0$" but now in your latest comment you state they don't have this property. Which is it?? $\endgroup$
    – whuber
    Commented Dec 27, 2016 at 23:23
  • 2
    $\begingroup$ Thank you for trying to explain, but I still cannot make sense of it. Where you write "$a$ and $d$ are independent non-central chi-square random variables" it sounds like you are summing squares of Normal random variables that have nonzero means, because that's how non-central Chi-squared variables usually arise. But later your write "The r.v.'s generating both $a$ and $d$ have $\mu=0$", which suggests you are working with central Chi-squared variables. I suspect these are the inconsistencies that prompted the initial comment by @Glen_b. Could you show explicitly what $a$ and $d$ are? $\endgroup$
    – whuber
    Commented Dec 28, 2016 at 14:55

2 Answers 2


If $a, d\sim\chi^2_{2M}$ are independent, then $X=a+d$ will have $\chi^2_{4M}$ distribution. Since $X$ is non-negative, CDF of $Y=a^2+2ad+d^2=(a+d)^2=X^2$ can be found by noting $$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq \sqrt{y})=F_X(\sqrt{y}).$$ Therefore, $$f_Y(y)=\frac{1}{2\sqrt{y}}f_X(\sqrt{y})=\frac{1}{2^{2M+1}\Gamma(2M)}y^{M-1}e^{-\sqrt{y}/2}.$$

If $a$ and $d$ are correlated then things are much more intricate. See for example N. H. Gordon & P. F. Ramig's Cumulative distribution function of the sum of correlated chi-squared random variables (1983) for a definition of multivariate chi-squared and distribution of its sum.

If $\mu\neq 2M$ then you are dealing with non-central chi-squared so the above will no longer be valid. This post may provide some insight.

EDIT: Based on the new information it seems $a$ and $d$ are formed by summing up normal r.v. with non-unit variance. Recall if $Z\sim N(0, 1)$ then $\sqrt{c}Z\sim N(0, c)$. Since now $$a=c\sum_{i=1}^{2M}Z_i^2=d,$$ both $a,d$ will have chi-squared distribution scaled by $c$, i.e. $\Gamma(M, 2c)$ distribution. In this case $X=a+d$ will be $\Gamma(2M, 2c)$ distributed. As a result, for $Y=X^2$ we have $$f_Y(y)=\frac{1}{2(2c)^{2M}\Gamma(2M)}y^{M-1}e^{-\sqrt{y}/2c}.$$

  • $\begingroup$ How does mu enter? Is it suppose to be the mean of one of the chi-square variables? I suspect it has nothing to do with the problem. $\endgroup$ Commented Dec 27, 2016 at 0:01
  • $\begingroup$ @MichaelChernick: probably means $a, d$ can be non-central chi-squared? $\endgroup$
    – Francis
    Commented Dec 27, 2016 at 0:06
  • $\begingroup$ I suppose you can make that assumption but the OP does not make any connection. I think you took the right approach, the non-central could not enter into this problem. X is the square of a chi-square here. In the case of independence that you used here what is the this distribution called? $\endgroup$ Commented Dec 27, 2016 at 0:19
  • $\begingroup$ @MichaelChernick I am not sure if there is a special name associated with the distribution. "chi-tesseracted" maybe? $\endgroup$
    – Francis
    Commented Dec 27, 2016 at 0:36
  • $\begingroup$ $a$ and $d$ are non-central chi-squared. $\endgroup$ Commented Dec 27, 2016 at 19:46

Since a non-central chi-square is a sum of independent rv's, then the sum of two independent non-central chi-squares $X = a+b$ is also a non-central chi-square with parameters the sum of the corresponding parameters of the two components, $k_x = k_a+k_b$ (degrees of freedom), $\lambda_x = \lambda_a+\lambda_b$ (non-centrality parameter).

To obtain the distribution function of its square $Y =X^2$ , one can apply the "CDF method" (as in @francis answer),

$$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq \sqrt{y})=F_X(\sqrt{y})$$

and where

$$F_X(x)=1 - Q_{k_x/2} \left( \sqrt{\lambda_x}, \sqrt{x} \right)$$


$$F_Y(y)=1 - Q_{k_x/2} \left( \sqrt{\lambda_x}, y^{1/4} \right)$$

where $Q$ here is Marcum's Q-function.

The above apply to non-central chi-squares formed as sums of independent squared normals each with unitary variance but different mean.


If the base rv's are $N(0,c)$, then the square of each is a $Gamma (1/2,2c)$ see https://stats.stackexchange.com/a/122864/28746 .

So the rv $a \sim Gamma (M, 2c)$ and $b \sim Gamma (M, 2c)$ so also $X = a+b \sim Gamma(2M, 2c)$ (shape-scale parametrization, and see the wikipedia article for the additive properties for Gamma).

Then one can apply again the CDF method to find the CDF of the square $Y = X^2$

  • $\begingroup$ @FelipeAugustodeFigueiredo Sorry, I am not familiar with complex rv's. My answer took as given the fact that we start from non-central chi-squares. $\endgroup$ Commented Dec 27, 2016 at 20:34
  • $\begingroup$ What if the r.v.'s are circular symmetric complex Gaussian random variables with $\mu = 0$ and $\sigma = cI$? $\endgroup$ Commented Dec 27, 2016 at 20:35
  • $\begingroup$ let's forget about complex rv's. What if the r.v.'s generating $a$ and $d$ are Gaussian r.v.'s with $\mu = 0$ and $\sigma \neq 1$? All Gaussian r.v.'s have the same variance, let's call it $c$. $\endgroup$ Commented Dec 27, 2016 at 21:28
  • $\begingroup$ could you please help me with the following question: stats.stackexchange.com/questions/253764/…. Any hint would be very appreciated. Thanks! $\endgroup$ Commented Dec 30, 2016 at 12:41
  • $\begingroup$ @FelipeAugustodeFigueiredo I am afraid I do not have something to offer for that question. $\endgroup$ Commented Dec 30, 2016 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.