I have a vector of the form \begin{align} {\bf a }= \frac{1}{\sqrt{N}}[1, e^{jA}, e^{j2A},\cdots, e^{j(N-1)A}]^T \end{align} where A and N are constants.
I also have a vector N of i.i.d $\mathcal{N}(0,1)$ or equivalently, $N \sim N(0,\Sigma_N= \sigma^2 {\bf I})$
I am trying to find the variance of $${\bf a ^* N}$$ where * denotes the complex conjugate. I think the answer should be of the form $$\text {Var}({\bf a ^* N}) = {\bf a ^* \Sigma_N a}={\bf a ^* I a} $$
Does anyone know how to further simplify the above?
I got a little confused of those multiple $N$'s, but still, guess that since elements of a random sample (let's call them $n_1,\ldots, n_i,\ldots, n_{N_n}$) are normal and iid, then using the representation of the inner product $S=\vec{a}^{\dagger}\vec{n}=\sum_{i=1}^{N_n}a_i^{*}n_i$. Here $\vec{a}^{\dagger}$ denotes conjugate transpose.
The sum of normal independent random variables is a normal random variable. So $S\sim\mathcal{N}(0,\frac{1}{N_n}\sum_{i=1}^{N_n}|a_i|^2)$ and since you had chosen $\vec{a}$ in such a way then $S\sim\mathcal{N}(0,1)$.