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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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I haven't thought through all of this, so I will just leave it as a comment: Is there anything helpful among the answers to Proof of $\frac{(n-1)S^2}{\sigma^2} \backsim \chi^2_{n-1}$? – Mike Spivey Sep 29 '11 at 17:50
@NicolasEssis-Breton In the factorization formula for $(n-1)S^2$, do you mean the negative sign between terms to be positive one, as in $(n-1)S^2 = \sum_{i=1}^n (X_i-\bar{X})^2 = \left( \sum_{i=2}^n (X_i-\bar{X}) \right)^2 + \sum_{i=2}^n (X_i-\bar{X})^2$, which follows from $\sum_{i=1}^n (X_i-\bar{X}) = 0$. – Sasha Sep 29 '11 at 18:30
$\bar{X}$ and $S^2$ are independent random variables, to you don't need to speak of the conditional distribution given $\bar{X}$. – Michael Hardy Sep 29 '11 at 18:51
The independence of $\bar{X}$ and $S^2$ is easily established with only the most basic facts about the multivariate normal. Note that $\mathbb E \bar{X} (X_i - \bar{X}) = 0$ for all $i$, which implies independence in the case of jointly distributed normal random variables. Hence $\bar{X}$ is independent of $(X_1-\bar{X},\ldots,X_n-\bar{X})$. Since $S^2$ is a deterministic function of the latter vector, then $\bar{X}$ and $S^2$ are independent. – cardinal Sep 29 '11 at 18:56
The distribution of quadratic forms in much more generality than the question of the OP is handled in most any textbook on the statistical theory of linear regression. – cardinal Sep 29 '11 at 19:02

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