Removing zero-counts in exponential decay measurement

Question

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)

Answer 1

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'.

Answer 2

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.