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In the multivariate normal case for $X\sim N_{p}(0,I)$, consider an idempotent matrix $A$ of rank $k<p$. I need to find the distribution of $\frac{X^{t}AX}{X^{t}X}$. I know that $X^{t}AX$ follows chi square with $k$ df. The answer is that the ratio follows beta. But I'm finding it difficulty to see how $X^{t}X$ follows chi square with $(p-k)$ df.

StubbornAtom
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Harry
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    With $X=(X_1,\ldots,X_p)$, you have i.i.d standard normal $X_i$'s. So certainly $X^TX=\sum_{i=1}^pX_i^2\sim\chi^2_p$. – StubbornAtom Jun 14 '19 at 18:20
  • The answer says that the ratio follows $B(k/2,(p-k)/2)$. If $X^{t}X$ follows chi square $p$, then how is (p-k) got? – Harry Jun 15 '19 at 06:28

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You have $$X^T X=X^T AX+X^T(I-A)X$$

By Fisher-Cochran theorem, a necessary sufficient condition for $X^T AX$ and $X^T(I-A)X$ to be independent chi-square variables is $p=\operatorname{rank}(A)+\operatorname{rank}(I-A)$, which is certainly true here because $A$ is idempotent.

So $X^T AX\sim\chi^2_k$ is independent of $X^T(I-A)X\sim \chi^2_{p-k}$.

It follows from a well-known result (also mentioned on wikipedia) that $$\frac{X^TAX}{X^T AX+X^T(I-A)X}\sim\text{Beta}\left(\frac{k}{2},\frac{p-k}{2}\right)$$

StubbornAtom
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  • Related: https://math.stackexchange.com/q/2263496/321264, https://math.stackexchange.com/questions/3508370/if-u-sim-chi-m2-independently-of-v-sim-chi-n2-then-prove-that-fracv?noredirect=1&lq=1. – StubbornAtom Feb 26 '20 at 14:47