I started by modifying the random_gaussian file to allow for a squared gaussian to be computed. I then ran two tests with 100 samples each, with sigmas and the gaussian centers at (2,10) and (2,1). I jackknifed these samples and then ran the average error program on them. Here are the averages and the error bars for each test:
(2,10) :: $102.214186 \pm 3.822489 $
(2,1) :: $4.680213 \pm 0.683018$
The random samples and the jackknifed samples for the first test are HERE and HERE.
The random samples and the jackknifed samples for the second test are HERE and HERE.
Next I calculated the following functions of the data via the jacknife
and found the average and error bars:
$a_1*a_2 = 478.383793 \pm 71.814093$
$a_1+a_2*a_2 = 124.123288 \pm 7.300540$
2. Bootstrap Samples
I modified the gaussian generation code again to generate gaussians raised to the sixth power. I then ran the code to generate 50 samples that were sent through the jackknife and the bootstrap. Here is the average for the jackknife and its error bar:
Jackknife :: $46.745775 \pm 12.451248$
Here is the probability distribution for the bootstrap samples:
As you an see, the peak is around 47, just as the jackknife predicted. However, also note that the error bar on this gaussian would not be symmetric, as the function is not symmetric about the center.
Both of the methods produce similar results, but we should expect the bootstrap to provide a better estimate. This is because the jackknife assumes that we have gaussian distributed numbers and since we are using a sixth degree gaussian and few samples we break this assumption and the central limit theorem.
3. Effective Mass of a Pion
We can use these averages to compute an estimate to the effective mass of a pion. We note that for a large time separation, the pion's correlator function should be $$C(t)=Ae^{-amt}$$ and eolving for $m$ we have $$m(t) = -\mathrm{ln}\left(\frac{C(t+1)}{C(t)}\right)$$.
Data was provided that had the correlator for each time step in a 64 step iteration. 91 values of the correlator were given at each time step.
I wrote a piece of code to extract an arbitrary time step and the next time step and output them to two files: C.dat and Cp.dat respectively. I also wrote a piece of code that computed $$-\mathrm{ln}\left(\frac{C(t+1)}{C(t)}\right)$$ for two input files.
I then wrote a bash script that allowed me to choose a time step, run the extraction program, run the jackknife on each output dat file, combine the two outputs with the above mathematical expression, and then calculate the average and the error bar.
I found that for $t=12$, the bash script gave me an effective mass $m_{eff} = 0.319857 \pm 0.003792$.
