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Cochran's Theorem says that

The conditional distribution of $\chi^2=\sum_{i=1}^nX_i^2$ under $m(<n)$ independent homogeneous linear constraints: $$\begin{align}a_{11}X_1+\cdots\dots +a_{1n}X_n&=0\\\vdots \\a_{m1}X_1+\cdots\dots +a_{mn}X_n&=0\end{align}$$ is $\chi ^2_{n-m}$, provided $X_1,\dots,X_n$ are i.i.d $N(0,1)$ variables.

It is clear that if the constraint is linear then the dergees of freedom(d.f) diminise.If we consider non-linear (like quadratic or cubic) constraint what will be the effect on the d.f of $\chi^2$ distribution.

kjetil b halvorsen
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Argha
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  • I'm not sure to understand. Is it the conditional distribution of the squared norm of the $X_i$'s conditional to the constraints ? This does not sound like Cochran's theorem to me (but I may be wrong) – Stéphane Laurent Sep 27 '12 at 08:26
  • @StéphaneLaurent: My teacher called the theorem Cochran's Theorem.I am also not sure about that.But this theorem is true and conditional distribution of $\chi^2$ means distribution of $\chi^2$ with the linear restrictions. – Argha Sep 27 '12 at 12:37
  • I ask a question here to understand this point http://stats.stackexchange.com/questions/38111/standard-normal-distribution-on-a-subspace – Stéphane Laurent Sep 27 '12 at 13:22
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    To see how difficult the situation is in general, consider the effect of the *single* nonlinear constraint $\sum_{i=1}^n X_i^2=1$: it reduces $\chi^2$ to a constant! – whuber Sep 27 '12 at 21:05
  • @whuber: your comment is very helpful and it clear my problem.Thank you. – Argha Sep 28 '12 at 11:08

1 Answers1

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Let $U$ be the $(n-m)$-dimensional subspace of $\mathbb{R}^n$ defined by the linear constraints. Then, with the help of @whuber's answer to this question, your claim stems from the fact that the conditional distribution of $(X_1,\dots,X_n)$ given $(X_1,\dots,X_n) \in U$ is a standard normal distribution on $U$.

However there's no similar result when $U$ is not a vector subspace. In such a case one cannot expect that the squared norm of $(X_1,\dots,X_n)$ has a chi-squared distribution.

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Stéphane Laurent
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