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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 $\sigma^2 = c$.

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    $\begingroup$ 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 Dec 27 '16 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. $\endgroup$ – Felipe Augusto de Figueiredo Dec 27 '16 at 19:45
  • $\begingroup$ 2M is the degrees of freedom for each of the two? $\endgroup$ – Alecos Papadopoulos Dec 27 '16 at 21:56
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    $\begingroup$ 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 Dec 27 '16 at 23:23
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    $\begingroup$ 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 Dec 28 '16 at 14:55
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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 $$f_Y(y)=\frac{1}{2(2c)^{2M}\Gamma(2M)}y^{M-1}e^{-\sqrt{y}/2c}.$$

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  • $\begingroup$ 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. $\endgroup$ – Michael Chernick Dec 27 '16 at 0:01
  • $\begingroup$ @MichaelChernick: probably means $a, d$ can be non-central chi-squared? $\endgroup$ – Francis Dec 27 '16 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? $\endgroup$ – Michael Chernick Dec 27 '16 at 0:19
  • $\begingroup$ @MichaelChernick I am not sure if there is a special name associated with the distribution. "chi-tesseracted" maybe? $\endgroup$ – Francis Dec 27 '16 at 0:36
  • $\begingroup$ $a$ and $d$ are non-central chi-squared. $\endgroup$ – Felipe Augusto de Figueiredo Dec 27 '16 at 19:46
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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

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

so

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

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.

ADDENDUM RESPONDING TO QUESTION'S EDIT

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 $Y = X^2$

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  • $\begingroup$ @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. $\endgroup$ – Alecos Papadopoulos Dec 27 '16 at 20:34
  • $\begingroup$ What if the r.v.'s are circular symmetric complex Gaussian random variables with $\mu = 0$ and $\sigma = cI$? $\endgroup$ – Felipe Augusto de Figueiredo Dec 27 '16 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 $c$. $\endgroup$ – Felipe Augusto de Figueiredo Dec 27 '16 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! $\endgroup$ – Felipe Augusto de Figueiredo Dec 30 '16 at 12:41
  • $\begingroup$ @FelipeAugustodeFigueiredo I am afraid I do not have something to offer for that question. $\endgroup$ – Alecos Papadopoulos Dec 30 '16 at 18:58

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