Finding the variance of the sum of weighted normal distributions I'm trying to find the variance of the PDF of a "center of mass" calculation, where every (discreet) node has a normal distribution, as:
$\frac{\sum_{i=-\frac{N}{2}}^{\frac{N}{2}} i\mathcal{N}(S_i,\sigma^2)}{\sum_{i=-\frac{N}{2}}^{\frac{N}{2}} S_i}$
I first tried to perform the calculation by treating the i's as constants, changing the variance as: $\sigma^2 \Rightarrow i^2  \sigma^2$ but this means that the location of the node changes its variance, which does not make sense. Furthermore, the variance of the sum should be $\sigma^2=\sum_{i=-\frac{N}{2}}^{\frac{N}{2}}i^2 \sigma^2$, but this gives me results for the variance that are quite different than what my simulations of this scenario yield.
so my question is what am i missing, and how will this new PDF look like?

Let the denom term be $\alpha = \sum_{i=-\frac{N}{2}}^{\frac{N}{2}} S_i$. So 
$$var\Big(\frac{1}{\alpha}\sum_{i=-\frac{N}{2}}^{\frac{N}{2}} i\mathcal{N}(S_i,\sigma^2)\Big) = \frac{1}{\alpha^2}.var\Big(\sum_{i=-\frac{N}{2}}^{\frac{N}{2}} i\mathcal{N}(S_i,\sigma^2)\Big) = \frac{1}{\alpha^2}.var(A)$$
Let $w$ be an $(N+1) \times 1$ vector as : 
$$w = [\mathcal{N}(S_{-\frac{N}{2}},\sigma^2) \ldots \mathcal{N}(S_{+\frac{N}{2}},\sigma^2)]^{T}  $$ and 
 a "counter" vector as 
$$c = [-\frac{N}{2} \ldots \frac{N}{2}]^T$$
So $A = c^Tw$, then 
$$var(A) = var(c^Tw) = E(c^Tww^Tc) = c^TE(ww^T)c = c^T(\sigma^2 \pmb{I})c = \sigma^2c^Tc = \sigma^2\sum\limits_{i=-\frac{N}{2}}^{\frac{N}{2}}i^2 = 2\sigma^2 \sum\limits_{i=1}^{\frac{N}{2}}i^2$$
you can say that $\sum\limits_{i=1}^{\frac{N}{2}}i^2 = \frac{N}{12}(\frac{N}{2} + 1)(N + 2)$
Then the total variance is 
$$\frac{1}{\alpha^2}.var(A) =\frac{\sigma^2}{\alpha^2}.\frac{N}{6}(\frac{N}{2} + 1)(N + 2) $$
