Normal distribution X follows a regular normal distribution on V with center $\xi$ and inner product $<\cdot,\cdot>$, and let $\eta \neq 0$ be a vector in V. Show that the reel stochastic variable $Y=<X-\xi,\eta>$ follows a normal distribution with mean 0 and variance $\Vert \eta \Vert ^2$.
Can anyone help me get started with this problem? 
 A: Hint: Linear combinations of independent (or, in general, jointly) Gaussian random variables are also Gaussian. You can easily prove this with characteristic functions in the independent case, and in the jointly Gaussian case it can be the definition.
Note that $Y = \sum_i (X_i - \xi_i) \eta_i = \sum_i \eta_i X_i - \sum_i \xi_i \eta_i$. The first term is a linear combination of independent Gaussians (so $\sum_i \eta_i X_i$ is Gaussian), the second term is a constant (so it just shifts the mean, showing $Y$ is Gaussian). 
As for calculating the mean and variance, you can calculate the mean by using linearity of expectation on $Y$ to get $E[Y] = \sum_i \eta_i E[X_i] - \sum_i \xi_i \eta_i = \sum_i \eta_i \xi_i - \sum_i \xi_i \eta_i =0$. 
Alternatively, in Matrix notation, you can do $E[Y] = E[(X - \xi)^T \eta] = E[(X-\xi)^T] \eta = (E[X-\xi])^T \eta = (E[X] - \xi)^T \eta = (\xi - \xi)^T \eta = 0^T \eta = 0$. 
As for the variance, note that $Var[Y] = E[Y^2] - E[Y]^2 = E[Y^2] = E[((X - \xi)^T \eta)^2] = E[ (\eta^T (X-\xi)) ((X-\xi)^T \eta)] = \eta^T E[(X-\xi) (X- \xi)^T] \eta = \eta^T I \eta = ||\eta||^2$. 
This follows from $(X-\xi)^T \eta = <X-\xi,\eta> =<\eta, X-\xi>= \eta^T (X-\xi)$ and $X-\xi \sim N(0,I)$ since $X \sim N(\xi,I)$. 
