I'm trying to understand the last comment under the answer in this question:

Just to present another approach, we can instead choose to define a transformed multivariate normal variable $v:=A′Y$ and it will still follow the same distribution $\mathcal{N}\left(0, \sigma^{2}I\right)$ if we use the affine property. Then the last fraction$$\frac{Y'ADA'Y}{\sigma^{2}} = \frac{v'Dv}{\sigma^{2}} = \frac{v'\begin{bmatrix} I & 0\\0 & 0\end{bmatrix}v}{\sigma^{2}}= \sum_{i=1}^{\operatorname{tr}D} \left(\frac{v_{i}}{\sigma}\right)^{2} $$

It states that if we take a spectral decomposition of the identity matrix minus the projection matrix, then transforming the $Y$ vector by pre-multiplying by the transpose of the matrix of eigenvectors, then it will have a standard normal distribution.

Question: I can't wrap my head around why the expectation of this transformed $Y$ vector would have a mean of zero. Does anyone know why?

Edit: The context is that we are trying to prove that the residual sum of squares divided by $\sigma^2$ has a $\chi^2$ distribution with $n-k$ degrees of freedom.

  • $\begingroup$ please give some context so that the readers don't have to go back to the original post to understand $\endgroup$ – Antoine Mar 24 '17 at 14:34

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