The second set is just the Round-ed version of the first set, which means there is a strong correlation between both sets (correlation close to 1).
A naive and rough (and incorrect) approach would be to say that the variances of both sets will be very close as the differences caused by rounding are relatively small compared to the values involved.
The variance of the difference of two correlated random variables with correlation coefficient $\rho$ is given by:
$ Var(X-Y) = Var(X) + Var(Y) - 2 \rho*SD(X)* SD (Y) $
which, with $Var(X) \approx Var(Y)$ and $\rho=1$, equals
$Var(X) + Var(X) - 2*1*Var(X) = 0 $
To be exact, and in Mathematica code:
Variance@pts1[[All, 1]] + Variance@pts2[[All, 1]] -
2 Correlation[pts1[[All, 1]], pts2[[All, 1]]] StandardDeviation@
pts1[[All, 1]]*StandardDeviation@pts2[[All, 1]]
(* 0.00312381 *)
However, setting both variances equal is actually unacceptable in this case and another approach would be to study the individual differences exactly. If it is just a rounding process then X-Round(X) should be distributed as a UniformDistribution[{-1/2, 1/2}] (unless the decimal digits have some kind of bias, which ideally should not be the case), so the standard deviation will be given by:
StandardDeviation[UniformDistribution[{-1/2, 1/2}]]
1/(2 Sqrt[3])
which is about 0.29.
This is what you would predict. In the concrete case of the given data sets it would be:
StandardDeviation /@ Transpose[pts1 - pts2]
{0.0558911, 0.0832781}
for the first and second coordinates values, respectively.
This deviates considerable from our prediction. The reason is that the fractional part of the numbers is not uniformely distributed as assumed:
DistributionFitTest[#, UniformDistribution[{-1/2, 1/2}]] & /@ Transpose[pts1 - pts2]
{0.0000134701, 0.000166425}
Very low p-values, so it is unlikely the decimals came from a uniform distribution. If these are measured values, perhaps you should start worrying about your data acquisition process.
