Let $Z\sim N(a\iota_{\nu},I_{\nu}), a\in\mathbb{R}$ whereat $$ \iota_{\nu}=\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix},~~~I_{\nu}=\text{diag}(1,\ldots,1). $$ Show that $Z^TZ=(Z-\tilde{Z}_{\iota_{\nu}})^T(Z-\tilde{Z}_{\iota_{\nu}})+\nu\tilde{Z}^2, \tilde{Z}:=\sum_{i=1}^{\nu}Z_i/\nu$. Furthermore, show that $Z-\tilde{Z}_{\iota_{\nu}}$ and $\tilde{Z}$ are stochastically independent and determine the distributions of $Z-\tilde{Z}_{\iota_{\nu}}$ and $\tilde{Z}$.
Hi, I think the first and the last task are not that difficult:
It is $Z^TZ=\sum_{i=1}^{\nu}Z_i^2$ and $$ (Z-\tilde{Z}_{\iota_{\nu}})^T (Z-\tilde{Z}_{\iota_{\nu}})=\sum_{i=1}^{\nu}(Z_i-\tilde{Z})^2=\sum_{i=1}^{\nu}Z_i^2-2\tilde{Z}\sum_{i=1}^{\nu}Z_i+\nu\tilde{Z}^2\\=\sum_{i=1}^{\nu}Z_i^2-2\nu\tilde{Z}^2+\nu\tilde{Z}^2=\sum_{i=1}^{\nu}Z_i^2-\nu\tilde{Z}^2, $$ so $$ (Z-\tilde{Z}_{\iota_{\nu}})^T (Z-\tilde{Z}_{\iota_{\nu}})+\nu\tilde{Z}^2=\sum_{i=1}^{\nu}Z_i^2=Z^TZ. $$
Moreover, the $Z_i, i=1,\ldots,\nu$ are independent (because the covariance matrix is diagonal), so it follows that $$ \tilde{Z}\sim N(a,1/\nu). $$ Write $Z-\tilde{Z}_{\iota_{\nu}}=c+I_{\nu}Z, c:=-\tilde{Z}_{\iota_{\nu}}$, then it follows that $$ Z-\tilde{Z}_{\iota_{\nu}}\sim N(a\iota_{\nu}-\tilde{Z}_{\iota_{\nu}}, I_{\nu}). $$
This should be ok. But how can I show that $Z-\tilde{Z}_{\iota_{\nu}}$ and $\tilde{Z}$ are independent?
Well the mean of $Z-\tilde{Z} {\iota_{\nu}}$ is 0, so the value of $a$ might as well be 0.
You can show independence of $Z-\tilde{Z} {\iota_{\nu}}$ and $\tilde{Z}$ by looking at their covariances, and using the property of Multivariate Normal (uncorrelated $\Rightarrow$ independent).
You can also find the covariance matrix of $Z-\tilde{Z} {\iota_{\nu}}$. I wouldn't call the resulting distribution Multivariate Normal though, because the resulting matrix will not be full rank. Try the 2-dimensional case to see more clearly what's going on.