Error in variance I've been exploring techniques in statistical physics, specifically applying them to spin ices. I'm in the canonical ensemble. By using the fluctuation dissipation theorem you can extract useful properties from the variance in energy and magnetization such as the heat capacity (constant V) and magnetic susceptibility respectively.
I've included the formulas I'm using below:
\begin{align}
C_V & = \frac{\partial\langle E \rangle}{\partial T} \\
& = -\frac{\beta}{T} \frac{\partial\langle E \rangle}{\partial\beta} \\
& = \frac{\beta}{T} \frac{\partial^2\ln Z}{\partial\beta^2} \\
& = \frac{\beta}{T} \frac{\partial}{\partial\beta}\left(\frac{1}{Z} \frac{\partial Z}{\partial\beta}\right) \\
& = \frac{\beta}{T} \left[\frac{1}{Z} \frac{\partial^2Z}{\partial\beta^2} - \frac{1}{Z^2} \left(\frac{\partial Z}{\partial\beta}\right)^2\right] \\
& = \frac{\beta}{T} \left[\langle E^2 \rangle - \langle E \rangle^2\right], \\
\chi & = \frac{\partial\langle M \rangle}{\partial H} \\
& = \beta \left[\langle M^2 \rangle - \langle M \rangle^2\right].
\end{align}
Now for my question, is there another computable statistical property which acts as the error for the variance? Such that I can get errors for the heat capacity and susceptibility.
 A: Turns out there are two methods that I have found out... there many be others:
1) Repeat the simulation n times with different initial conditions and use the usual statistic techniques to calculate the mean and error of the variance quantities.
2) Bootstrapping, sub sample the data as follows:
Randomly choose N frames from M frames, where M > n N [n defined below]. 
Calculate quantities in the set N. 
Repeat n times
Calculate statistical variances among the n sets of calculations.
Source: Dr Peter Olmsted @ Leeds University
A: Since the energy $E$ is a random value, you can define another random variable 
$c_V=(E-<E>)^2$ 
such that the heat capacity is the mean value of this quantity:
$C_V=<c_V>$
Now we can identificate the mean variance of $c_V$ with the variance of $C_V$, we have:
$<\Delta c_V>=\Delta C_V= \sqrt(<E^4>-<E>^4)$
we can compute $<E^4>$ with the formula:
$<E^n>=(-1)^n\frac{1}{Z}\frac{\partial^n}{\partial \beta^n}Z=(-1)^n\frac{1}{Z}\frac{\partial^n}{\partial \beta^n}e^{\ln Z}$
$\frac{\partial}{\partial \beta}\ln Z=-<E>$
after expressing all the values in terms of derivative with respect of the temperature of the mean energy. You will obtain an estimation for the error of variance from derivatives of the mean energy with respect to temperature.
