# Removing zero-counts in exponential decay measurement

I'm doing an experiment where I'm measuring the mean-lifetime of muons. I have a set of data points for the number of decays against time, resembling an exponential distribution (of course). But in the experiment, there weren't a lot of counts measured (approximately 500) so for some time-values I got zero counts where it doesn't really fit to the exponential distribution.

I was wondering if it is justifiable to remove these points and what the reasoning behind that is. It's a much better fit if I do remove them compared to the experimental value of the muon lifetime given by e.g. Source- M. Tanabashi et al. (Particle Data Group), Review of Particle Physics, Phys. Rev. D 98, 030001 (2018)

• it's the time between decays that should be exponentially distributed, isn't it? not the number of decays in equal sized time intervals. – innisfree Dec 17 '18 at 20:49
• No it's the number of counts versus the time (intervalls here). The decay for a muon is completely probabilistic with the probability of it decaying being determined by the exponential relation: $P(t) = e^{\frac{-t}{\tau}$ in its own restframe. So we expect the number of counts to be exponentially dependent on the time passed. – DrShellyCooper Dec 17 '18 at 21:48
• Can't you write equations in comments? – DrShellyCooper Dec 17 '18 at 21:51
• @innisfree the muon does not decay twice! – anna v Dec 18 '18 at 5:10
• this set up is relevant physlab.org/wp-content/uploads/2016/04/Muon_cali.pdf – anna v Dec 18 '18 at 5:16

You can't simply drop the zeros.

It is valid, in my opinion, to chop off the tail of the distribution, as if your measuring device could only function at up to some finite delay time.

Among the methods you might want to consider are both least-squares curve fitting, and also the maximum likelihood method. The latter gives you a little more leeway in saying what fits are more likely than others.

Don't forget that ultimately your goal is not 'which result is closest to the accepted value' but 'which result is the one for which I can legitimately claim the smallest experimental uncertainty from my data and my experimental method, including its possible sources of systematic error'.

• all true, but I think there is a mistake in the OP's analysis regarding which quantity is exponentially distributed – innisfree Dec 17 '18 at 20:55
• Well I understand your point. I still feel like the zeroes don't contribute to anything but ruining the fit (at least for points where the neighbouring points are way above zero). Maybe an inprovement would be to sample more data, to possibly reduce the number of zero-points. Thank you anyways. – DrShellyCooper Dec 17 '18 at 21:56
• If you have the opportunity to work with the equipment, you can of course try to discover if it is suffering from some intermittent problem. – Andrew Steane Dec 17 '18 at 22:06
• this is what is the experiment physlab.org/wp-content/uploads/2016/04/Muon_cali.pdf . The zeros are part of the experiment if done correctly with a chi2 fit as you suggest – anna v Dec 18 '18 at 5:36
• All right, thank you all. Yes the experiment was very similar to the one you linked. – DrShellyCooper Dec 18 '18 at 13:57

From your description of the data, it appears you choose some time interval $$T_k$$, and then count the number of decays, $$N_k$$, observed during that interval, and then repeat for many different intervals: $$k \in [1, 2, ..., n]$$.

Then you posit:

$$N_k \propto e^{T_k/\tau}$$

This is not correct (see others' comments).

Given a uniform random decay rate, the time between decays will be exponentially distributed and the number of decays in a fixed time interval with be Poisson distributed:

$$P(N_k) = \frac{\lambda_k^{N_k}e^{-\lambda_k}}{N_k!}$$

where:

$$\lambda_k \equiv \frac{T_k}{\tau}$$

where $$\tau$$ is the lifetime.

At this point, there is nothing to plot (unless all $$T_k$$ are equal, then you should histogram $$N_k$$ and get a Poisson distribution that can be fit). If you have many different $$T_k$$, and:

$$\sum_{k=1}^n{N_k} \approx 500$$

then simultaneously fitting all $$T_k$$ may not be the best approach. Here you want to use maximum-likelihood. Note that in this case, the $$N_k=0$$ data are extremely important, as:

$$P(0) = e^{-T_k/\tau}$$

is a valid, and expected, measurement.

• This is irrelevant to muon decays, muons do not decay twice ! see this physlab.org/wp-content/uploads/2016/04/Muon_cali.pdf – anna v Dec 18 '18 at 5:14
• Indeed, I now agree with @annav. The OP was probably talking about a histogram of observed lifetime, and seeing some empty bins. The wording lead me astray – innisfree Dec 18 '18 at 6:17
• The wording gives the impression that OP observed decays at particular times and then binned them wrt time. And I thought no, it should be counts against time difference. I now don’t think OP did that. And my argument assumes it was a homogenous Poisson process, which is isn’t as a muon can’t decay twice (thanks @annav) – innisfree Dec 18 '18 at 6:27
• Oh I see, my bad for causing confusion @innisfree. – DrShellyCooper Dec 18 '18 at 13:58