If noise is your signal, what is your noise? Consider the following contrived situation. Imagine a Gaussian white noise process $x[t]$, with bandwidth $Δf$, with PSD equal to some quantity $A$ which you would like to measure.

So the way to measure this seems to measure the variance of the process $x[t]$, which by Parseval's theorem will be $AΔf$.

So you measure points with some frequency $f_m$, probably $2Δf$. And at a lower rate, say $f_v$, you compute the variance of the preceding block of $f_m/f_v$ points and take that to be a measurement. What will the noise/variance be in this measurement of the variance of $x[t]$? How can I approach this question?
There is a derivation over on the math.SE site that might be useful.

Provided your data is Gaussian with variance $\sigma^2$.

If you use the unbiased estimator of the population variance:
$$
S^2_{N-1} = \frac{1}{N-1} \sum_{n=0}^{N-1} (x[n]- \bar{x})^2
$$
where $N$ is the number of points in your batch and $\bar{x}$ is the sample mean, then the variance of this estimate will be:
$$
\frac{2 \sigma^4}{N-1}
$$