Orthogonal transformation of multivariate normal Let $X \sim N_n(\boldsymbol{\mu}, I )$.
Let $O$ be an orthogonal matrix, with the first line $\frac{\boldsymbol{\mu}^T}{\|\boldsymbol{\mu}\|}$, and $Y=OX$.
It can be proved that $E(Y_1)=\boldsymbol{\mu}$ and $E(Y_k)=0$, for $k\geq 2$.
I'm asked to prove that to use this information to conclude that the distribution of $\|\mathbf{X}\|^2$ is the same that of $\|\mathbf{Y}\|^2$, and hence depends on $\boldsymbol{\mu}$ only through $\|\boldsymbol{\mu}\|$. 
Well, we can see that  $\|\mathbf{Y}\|^2=\mathbf{Y}^T\mathbf{Y}=\mathbf{X}^T\mathbf{O}^T\mathbf{O}\mathbf{X}=\|\mathbf{X}\|^2$. So both have the same distribution. I'm having difficulty is understanding is the hence part. 
Any help would be appreciated.
 A: Under an orthogonal transformation, $\boldsymbol{Y}= O\boldsymbol{X}$ we have $\boldsymbol{Y} \sim N_n(O\boldsymbol{\mu}, OIO^T ) \sim N_n(O\boldsymbol{\mu},I ).$  In other words, $\boldsymbol{Y}$ also has a a joint normal distribution with uncorrelated components.
We are given that $O$ has a structure where the first row is $\frac{\boldsymbol{\mu}^T}{\|\boldsymbol{\mu}\|}$, resulting in  
$$Y_1 = \frac1{\|\boldsymbol{\mu}\|}\sum_{j=1}^{n}\mu_jX_j,\\E(Y_1) = \|\boldsymbol{\mu}\|,\\E(Y_k) = 0 \,\,\,\,(k \geq 2).$$
This defines the distribution of $\boldsymbol{Y}$ further as $\boldsymbol{Y} \sim N_n(\|\boldsymbol{\mu}\|\boldsymbol{e_1}, I )$ where $\boldsymbol{e_1}=(1,0,\ldots,0)^T.$
Let 
$$\boldsymbol{Z}=\boldsymbol{Y}-\|\boldsymbol{\mu}\|\boldsymbol{e_1}.$$
The components of $\boldsymbol{Z}$ are independent normally distributed random variables with zero mean. The sum of the squared components has a Chi-squared distribution with $n$ degrees of freedom -- which does not depend on $\boldsymbol{\mu}$.
We now have
$$\|\boldsymbol{X}\|^2=\|\boldsymbol{Y}\|^2=\|\boldsymbol{Z}\|^2+2\|\boldsymbol{\mu}\|Y_1-\|\boldsymbol{\mu}\|^2.$$
Since $Y_1$ has a normal distribution with mean $\|\boldsymbol{\mu}\|$ and variance $1$, the distribution of $\|\boldsymbol{X}\|^2$ depends only on $\|\boldsymbol{\mu}\|$.  
