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Let $\Bbb Y = (Y_1, \cdots, Y_n)' \sim N_n(\Bbb \theta J_n, \sigma^2I_n)$ be a random vector and $S^2=\frac{1}{n-1}\sum\limits_{i=1}^n(Y_i-\bar Y)^2$. Prove, that $\frac{n-1}{\sigma^2}S^2 \sim \chi^2_{n-1}$.

So I could start with writing that $({n-1})S^2=\sum\limits_{i=1}^n(Y_i-\bar Y)^2=\sum\limits_{i=1}^nY_i^2-n\bar Y^2=Y'Y-\frac{1}{n}Y'J_nJ_n'Y=Y'HY$ where $H$ is a symetric idempotent matrix, so it gives me $(HY)'HY$ and I would be much happier if $HY \sim N(0,I_n)$. If $H$ was orthogonal I could then write that $HY \sim N(H\theta J_n, \sigma^2I_n)$ and dividing it by $\sigma^2$ would almost give me a desired distribution.

The other idea is to somehow standarize my random variable $Y_i \sim N(\theta_i,\sigma^2)$, but I don't really know how it is distributed if I subtract the sample mean instead of $\mu$.

alladinsane
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1 Answers1

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Let $Z \sim N_n(0 J_n, I_n)$ so that $Y = \theta J_n + \sigma Z$.

Note that $Y'HY = \sigma^2 Z' H Z$ because $H J_n = 0$. It then suffices to show $Z'H Z \sim \chi^2_{n-1}$.

Cryo has given a great hint: because $H$ is symmetric and idempotent with trace $n-1$, it is diagonalizable [with respect to an orthonormal basis] with eigenvalues $1$ (multiplicity $n-1$) and $0$ (multiplicity $1$). Let $H = UDU'$ be that diagonalization, where $U$ is orthogonal.

Then $Z' H Z = Z'UDU'Z = (U'Z)' D (U'Z)$. Use rotational invariance of the $N(0, I_n)$ distribution to note that $\tilde{Z} := U'Z$ also follows the $N(0, I_n)$ distribution as well. Then note that $\tilde{Z}' D \tilde{Z}$ is the sum of squares of $n-1$ components of $\tilde{Z}$, so it follows a $\chi^2_{n-1}$ distribution.

angryavian
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