Variance of random vector I have to find the variance of $$ y= {\bf w^*n}$$   where ${\bf n}$ is a vector $(m\times 1)$ i.i.d Gaussian random variables with Covariance (Cov) matrix $$\text{Cov}({\bf n})=\sigma^2 {\bf I}$$ ${\bf I}$ being the identity matrix and where ${\bf w}$ is $(m\times 1)$ $${\bf w}=\sum_{i=1}^na(\phi_i)$$ 
Is it correct to say that 
 $$\text{Var}({\bf w^* n})=\sigma^2 {\bf w^* I w}$$
Thanks
 A: The formula you have given is correct, though your notation for $w$ seems a bit over complicated: is your notation
$$w = \sum_{i=1}^n a(\phi_i)$$
justified by some context? Why not write $w = (w_1,\ldots, w_m)$?
To derive the variance formula we use the following claim. If $X_1,\ldots, X_m$ are real valued random variables, and $w_1,\ldots, w_m \in \mathbb R$ are constants, then
$$\text{Var}\Big( \sum_{j=1}^m w_j X_j \Big)  = \sum_{j,\,k=1}^m w_j w_k\, \text{Cov}(X_j,X_k)$$
Assuming this formula to be true then in the case you have given since the $X_j$ are all independent, we have $\text{ Cov}(X_j,X_k) = 0$ if $j\neq k$, and so the formula you have given becomes
\begin{align} \text{Var}\Big( \sum_{j=1}^m w_j X_j \Big) & = \sum_{j=1}^m w_j^2 Var(X_j) \\
& = \sum_{j=1}^m \sigma^2 w_j^2 \\
& = \sigma^2 w^* w.
\end{align}
To justify the variance formula we have used is an exercise in linearity of expectations:
\begin{align}
\text{Var}\Big( \sum_j w_j X_j \Big) & = \textbf E \Big[ \Big( \sum_j w_j X_j \Big)^2 \Big] - \textbf E \Big[ \sum_j w_j X_j \Big]^2 \\
& = \textbf E \Big[ \sum_{j,k} w_j w_k X_j X_k \Big] - \Big( \sum_j \textbf E[ w_j X_j] \Big)^2 \\
& = \textbf E \Big[ \sum_{j,k} w_j w_k X_j X_k \Big] - \sum_{j,k} \textbf E[ w_j X_j]\textbf E[w_k X_k] \\
& = \sum_{j,k} w_j w_k \Big( \textbf E[ X_j X_k] - \textbf E[X_j]\textbf E[X_k] \Big)\\
& = \sum_{j,k} w_j w_k\, \text{Cov}(X_j,X_k).
\end{align}
