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I have been trying to derive the formula for $\chi^2$ distribution with $n-1$ degrees of freedom, but I am still having trouble. Assume $A$ is an orthogonal matrix with first row inputs $A_{1i}=n ^ {-1/2}$ for $1 \leq i \leq n$. $Z_1, ..., Z_n$ are i.i.d $N(0, 1)$, and $W=AZ$. Also $\sum_{i = 2}^{n} W_i^{2}$ has a $\chi_{n-1} ^ 2$. How can I show the following?

$\sum_{i = 2}^{n} W_i^{2} = \sum_{i = 1}^{n} (Z_i - \bar{Z})^{2}$

kjetil b halvorsen
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user300652
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    https://math.stackexchange.com/q/2952346/321264 – StubbornAtom Oct 22 '20 at 07:21
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    Hint: use the definition of orthogonality to express $$\sum_{i=2}^n W_i^2 = \sum_{i=1}^n W_i^2 - W_1^2 = \sum_{i=1}^n Z_i^2 - \left(\sum_{i=1}^n Z_i/\sqrt{n}\right)^2=\sum_{i=1}^n Z_i^2 - n(\bar Z)^2$$ and apply a little algebra to re-express the right hand side in terms of the squares of the $Z_i-\bar Z.$ – whuber Oct 22 '20 at 13:36
  • @whuber this makes so much sense! Thank you. How do you re-express the right hand side in terms of the square of $Z_i - \bar{Z}$ though? – user300652 Oct 22 '20 at 14:59
  • Once you work it out in the case $n=2$ it should become clear. – whuber Oct 22 '20 at 15:00

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