Estimation of the standard deviation of a zero mean distribution with partial information about it

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I'm trying to estimate the standard deviation of a zero-mean distribution of values. Let me start with this brief introduction to the problem:

Let $\{\Omega_{ij}^{\mu \nu}, C_{ij} \in \mathcal{Z}\}$ be uncorrelated random variables with the following probability distribution $$ P(\Omega_{ij}^{\mu \nu}) = \frac{1}{2}[\delta(\Omega_{ij}^{\mu \nu} - 1) + \delta(\Omega_{ij}^{\mu \nu}+1)]$$

$$ P(C_{ij}) = \frac{1}{N}[\delta(C_{ij} - 1)c + \delta(C_{ij})(N-c)]$$

(such that $N,p,c\in \mathcal{N}$, $i,j = 1,2,...,N$ and $\mu,\nu = 1,2,\ldots,p$)

If we take the limit $p\rightarrow +\infty$, $N\rightarrow +\infty$, $c\rightarrow +\infty$ (but $c/N \sim a <1, a \in \mathcal{R}$), a new random variable $R^\nu$ follows a normal distribution with zero mean and $\sigma = \sqrt{p/c}$ standard deviation,

$$R^\nu = \frac{1}{c} \sum_{j=1}^N \sum_{\mu \neq \nu}^p \Omega_{ij}^{\mu \nu} C_{ij} \sim \mathcal{N}(0,\sqrt{\frac{p}{c}})$$

Question

Now, here is the thing. For a given $(p-1)N$ set of values of $\Omega_{ij}^{\mu \nu}$ I can optimize the way in which the $c$ non-zero values of $C_{ij}$ are distributed such that it minimizes the standard deviation of $R^{\nu}$ (keeping zero mean) $$\sigma^2 = \left< \left( \frac{1}{c} \sum_{j=1}^N \sum_{\mu \neq \nu}^p \Omega_{ij}^{\mu \nu}C_{ij}\right)^2\right>$$ I have done this by using a simulated annealing algorithm (Note: the reason why I used an heuristic algorithm to solve this constrained optimization problem is because of the large dimension the problem had). However, it would be interesting to estimate $\sigma$, before doing any optimization, just by knowing that the new $C_{ij}$ distribution will minimize it. For example, this estimation could tell me about the quality of the results obtained in the optimization using simulated annealing (as its results could correspond to local minima). Any information from where to read about to any hint that could help solve this problem (analytically or numerically) will be welcome!