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I need to find the distribution of the random variable $Z$

$Z = \frac{(X - \mu_0)^2}{\sigma_0^2} - \frac{(X - \mu_1)^2}{\sigma_1^2}$, where $X \sim \mathcal{N}(\mu_0, \sigma_0)$.

We can find the distribution of each variable. The first variable $\frac{(X - \mu_0)^2}{\sigma_0^2} \sim \chi_1$, and the second variable $\frac{(X - \mu_1)^2}{\sigma_1^2} \sim \frac{\sigma_0^2}{\sigma_1^2}\chi_1(\lambda = \frac{(\mu_0 - \mu_1)^2}{\sigma_0^2})$. Thus, the distribution of $Z$ is weighted sum of non-central $\textbf{dependent}$ chi square distribution. I find a package in R (sadists) for $\textbf{independent}$ weighted sum to do inference. How can I find the distribution of $Z$ or at least a package to do some inference?

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\begin{align} Z&= \frac{(X-\mu_0)^2}{\sigma_0^2}-\frac{(X-\mu_1)^2}{\sigma_1^2}\\ &= \left(\frac{1}{\sigma_0^2} - \frac{1}{\sigma_1^2}\right)X^2 -2\left(\frac{\mu_0}{\sigma_0^2} - \frac{\mu_1}{\sigma_1^2}\right)X +\left(\frac{\mu_0^2}{\sigma_0^2} - \frac{\mu_1^2}{\sigma_1^2}\right)\\ &= \left(\frac{1}{\sigma_0^2} - \frac{1}{\sigma_1^2}\right) \left(X-A\right)^2 + B \end{align} where the last step is obtained via "completing the square" and the values of $A$ and $B$ are left for you to determine (it is a tedious exercise at the level of middle-school algebra). So $Z$ is a displaced and scaled noncentral $\chi_1^2$ random variable. Note that the scale factor is a negative number when $\sigma_0^2 > \sigma_1^2$, and that if $\sigma_0^2 =\sigma_1^2$, then $Z$ is not a $\chi_1^2$ random variable at all but a normal random variable!

You can work with this instead of calling on the sadists (?) who run R for help.

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  • $\begingroup$ In fact, I have : $T = \sum{Z_i}$. Thanks to your response, $Z_i$ is a scaled noncentral $\chi^2_1$ random variable. Thus, $T$ is a weighted sum of independent noncentral chi-squres random variables. Now, I can use "sadists" for T. can I find the distribution of T? or I have to use "sadists"? Thank you very much ! $\endgroup$
    – bassir
    Jun 6 '15 at 8:52
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    $\begingroup$ I am not a masochist and generally avoid dealing with sadists. $\endgroup$ Jun 6 '15 at 13:31
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The distr package for R lets you create distributions based on functions of known distributions. So you could use it to create a difference of 2 Chi-squares and do various computations based on the new distribution.

edit

Here is a basic example with values for the means and standard deviations chosen:

> library(distr)
> 
> x <- Norm(1, 2)
> z <- ((x-1)/2)^2 - ((x-3)/4)^2
> 
> plot(z)
> distr::q(z)(c(0.025,0.5,0.975))
[1] -1.738  0.121  4.612
> p(z)(0)
[1] 0.408
> mean(r(z)(10000))
[1] 0.503
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  • $\begingroup$ Could you please give me an example? which functions that allow me to create weighted sum of many distributions? How can I do inference on the new distribution? Thanks $\endgroup$
    – bassir
    Jun 8 '15 at 14:05
  • $\begingroup$ @bassir, I added an example above. $\endgroup$
    – Greg Snow
    Jun 8 '15 at 19:21

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