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I am trying to figure out the distribution of $$ (n-1) \sum_{i=1}^n Z_i^2 - \left( \sum_{i=1}^n Z_i \right)^2 \qquad (*) $$ where $Z_i \sim \mathcal{N}(0,1)$, i.i.d. I know that, taking each of the terms separately, $$ \sum_{i=1}^n Z_i^2 \sim \chi^2(n) $$ and $$ \frac{1}{n}\left( \sum_{i=1}^n Z_i \right)^2 \sim \chi^2(1). $$ But I am unsure about the distribution of (*)

Zailei Chen
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    Welcome to our site! Is this a question from a course or textbook? If so, please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). Otherwise it would be interesting to know in what context this problem arose. Thanks for showing us what you have done and where you're stuck – Silverfish Aug 06 '16 at 13:16
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    @Silverfish No, this is just something I'm trying to figure out. I've distilled the problem down to this simple question; my original problem is quite a bit more complicated, I'm afraid! So, I've tried to only ask the part I really need help on. But, if this looks like a textbook problem, I would be very interested to learn from which text, hoping it provides the relevant information :) – Zailei Chen Aug 06 '16 at 13:28
  • That seems sensible! Also, thanks for using the Latex typesetting, we always appreciate the effort – Silverfish Aug 06 '16 at 13:29

1 Answers1

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Here is an attempt:

Consider $Z=X-Y$ such that $X \sim \chi^2(\alpha)$ and $Y \sim \chi^2(\beta)$ with $\alpha \geq \beta$

$$ \mathcal{M}_X(t) = \left(1-2 \, t\right)^{-\alpha/2} $$

$$ \mathcal{M}_Y(t) = \left(1-2 \, t\right)^{-\beta/2} $$

$$ \mathcal{M}_Z(t) = M_X(t)M_Y(-t) = \left(1-2 \, t\right)^{-\alpha/2}\left(1+2 \, t\right)^{-\beta/2} = (1-4t^2)^{-\beta/2} $$

$$ \mathcal{M}_Z(t) = (1-2t)^{-n/2}(1+2t)^{-1/2} = (1-4t^2)^{-1/2} (1-2t)^{-(n-1)/2} $$

I am not sure if it can be reduced to a fathomable MGF.

rightskewed
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    Some discussion of a slightly more general situation appears at http://stats.stackexchange.com/questions/72479, with a reference to a paper that offers an approximation. – whuber Aug 08 '16 at 15:18