binned into 2ms time bins. Therefore, assuming the same background noise level, the S/N of the *’long’* sample is larger by a factor of $ \sqrt{32} $ then the S/N of the *’short’* sample. To obtain a comparable sample we produced another data sample (denoted *’noisy long’* sample). This data set is produced by adding noise to the *’long’* sample. We treat this sample as if the basic time bin is 2ms and add a Poisson noise accordingly. For a given long burst with counts $ C(t) $ we generate a noisy signal, $ C_{noisy}(t) $, using the following simple procedure. The noisy signal at time $t$, $ C_{noisy}(t)
$, is a Poisson variable ($
P_{(C_{noisy}(t)=i)}=\frac{\lambda(t) ^{i}}{i!}e^{-\lambda(t) }
$) with $ \lambda (t)\equiv C(t)/32 $. This noisy signal is smaller by a factor of $ 64ms/2ms=32 $ than the original signal and its standard deviation is correspondingly smaller by a factor of $ \sqrt{32} $ then the original standard deviation. There is a minor caveat in this procedure. The original signal contains its own noise, but since this original noise is smaller by a factor of $ \sqrt{32} $ then the added noise, it is negligible. The S/N ratio of 