= Neutrino event classification model
{title2=$K\sim\operatorname{Bin}(N,q)$}
How can the relative abundance of tracks and showers distinguish hypotheses? Consider an original teaching model, \b[with chosen inputs throughout]. Condition on a sample of $N$ selected events, each assigned exactly one of two labels: track or shower. Multitrack events and unclassified events lie outside this model. A shower label need not identify an <Electron>; background can imitate it. Conversely, a true <Electron> interaction need not receive a shower label.
Let $b$ be the background fraction among selected events. Among the selected signal events, let $f$ be the fraction whose associated <charged lepton> is an <Electron>, with the remainder associated with a <muon>. These are fractions after event selection, rather than fractions of the incident <neutrino> <particle flux>. Define the <conditional probabilities>
$$
a=\Pr(\mathrm{shower}\mid e,\mathrm{signal}),\qquad
r=\Pr(\mathrm{shower}\mid\mu,\mathrm{signal}),\qquad
q_b=\Pr(\mathrm{shower}\mid\mathrm{background}).
$$
Here $a$ and $r$ describe classification within the selected sample. They are not measured historical <detection efficiencies>. The <law of total probability> gives the overall shower <probability>
$$
q(f,b)=(1-b)\bigl[af+r(1-f)\bigr]+bq_b.
$$
This formula is the bridge between an interaction hypothesis and a visible label. It keeps beam contamination, background and mistaken classification conceptually distinct: genuine <electron neutrino> interactions belong in $f$, while non-signal events belong in $b$.
Assume <independent random variables> for the event labels, all with this same $q$. If $K$ counts shower labels, every specified pattern with $k$ showers has <probability> $q^k(1-q)^{N-k}$. Counting those patterns with a <binomial coefficient> yields the <binomial distribution>
$$
\Pr(K=k)=\binom Nk q^k(1-q)^{N-k},\qquad
\mathbb E[K]=Nq,\qquad \operatorname{Var}(K)=Nq(1-q).
$$
The track count is $N-K$, so its <expected value> is $N(1-q)$. At fixed $N$, the two counts cannot fluctuate independently. The <variance> above measures sampling fluctuations under the stipulated model, not uncertainty in $a$, $r$, $b$ or $q_b$.
Choose $N=40$, $a=0.90$, $r=0.05$, $b=0.10$ and $q_b=0.40$. Compare a signal with only <muon>-associated interactions ($f=0$) against one with equal selected <Electron>- and <muon>-associated interactions ($f=1/2$). These give $q=0.085$ and $q=0.4675$, respectively. Their expected shower counts are $3.4$ and $18.7$; their expected track counts are $36.6$ and $21.3$. A background-only model with the specified $q_b$ gives $16$ showers and $24$ tracks on average. These are expectations, so they need not be integers.
\Image[/neutrino-track-shower-classification-model.png]
{title=Chosen <binomial distribution> model: expected track and shower counts, with the shower-count distributions}
{description=Original calculations for forty selected events and the stipulated classification and background probabilities. No historical counts, fitted efficiencies or reconstructed significance are plotted.}
For a hypothetical observation $K=4$, the <likelihood ratio> of the two signal models is
$$
\frac{\mathcal L(f=0)}{\mathcal L(f=1/2)}
=\left(\frac{0.085}{0.4675}\right)^4
\left(\frac{0.915}{0.5325}\right)^{36}
\simeq3.18\times10^5.
$$
The <binomial coefficient> cancels because the same hypothetical sample is used in both <likelihoods>. This ratio is not the <probability> that either hypothesis is true. A separate original check gives $\Pr(K\leq5)\simeq0.880$, $4.57\times10^{-6}$ and $1.44\times10^{-4}$ for the muon-associated, equal-mixture and specified background-only models, respectively.
Why are the controls essential? If the background-only model may choose any $q_b$, taking $q_b=K/N$ can fit the observed shower fraction. Counts alone then fail to distinguish it from a signal model. Independent background measurements and calibrated <particle identification> constrain that freedom. Changing the decay path additionally tests where the candidate-producing <particles> originate. The power of the experiment comes from combining apparatus interventions with event classification; the small <binomial distribution> model isolates only the latter part of that reasoning.
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