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Let $(X_1,X_2,..., X_n)$ be a random sample from a normal population with mean $0$ and variance $1$. I would like to show that

$(n-1)S^2|\bar X \sim \chi^2(n-1).$

I know there is many way to do this, I would like to have your opinion on the following way:

$(n-1)S^2$ can be factorize in a function of $(X_2-\bar X, X_3- \bar X,\ldots, X_n - \bar X)=A$ as follow $(n-1)S^2=(\sum_{i=2}^{n}(X_i- \bar X))^2+\sum_{i=2}^{n}(X_i - \bar X)^2 \tag{1}.$ I can find a conditional probability density function $f(x_2,x_3,\ldots,x_n|\bar x)=P(X_2=x_2, X_3=x_3,...X_n=x_n|\bar X = \bar x)$.

I did the derivation and I found that $f$ simplify to $f(x_2,x_3,...,x_n|\bar x)=ce^{-(n-1)S^2 \over 2}$, where c is a constant such that the density integrates to $1$.

At this point is there any way to infer that $(n-1)S^2|\bar X \sim \chi^2(n-1)$?

I think it's not possible since $(1)$ is not a simple expression of $A$, and makes computation of the density of $(n-1)S^2|\bar X$ from $f$ difficult. But probably there is something I don't see here. Thanks.

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    @Nicolas: Yes, it is possible to use that theory and the theory covers a more general case. Suppose that $X$ is a zero-mean multivariate normal random vector with covariance $\sigma^2 I_n$ and that $A$ is a symmetric idempotent matrix. Then, $X^T A X / \sigma^2$ is a $\chi^2$ random variable with $\nu = \mathrm{rank}(A) = \mathrm{tr}(A)$ degrees of freedom. In your case, what is $A$? This generalizes further to the noncentral chi-square case when we allow the mean vector of the normal to be nonzero. See the texts of G. A. F. Seber & A. Lee (2003) or F. Graybill (2000) for more on this.2011-09-30

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