Particles, Particles, Particles — Question 10

Section 44-1 · Physics pages 7310–7311

Question 10 A rare-channel limit can be inferred even when no events are observed A background-free search observes zero events. For Poisson mean mu, P(0|µ) = e−µ. Define a one-sided 95 percent upper limit by P(0|µ95) = .0500. Exposure is 100 pb−1 and efficiency .250. µ P(0) Schematic; labels, not drawing lengths, specify the data. (a) Find the upper bound on expected signal events and cross section. (b) Find how that bound changes if exposure doubles with zero events again, and explain why zero observed events does not imply zero cross section.
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Question 10 - Solution A rare-channel limit can be inferred even when no events are observed µ P(0) (a) Method and calculation e−µ95 = .0500 ⇒µ95 = −ln(.0500) = 2.99573 . σ95 = µ95/(ϵL) = 2.99573/25 = .119829 pb . (b) Interpretation and further calculation σ95′ = .0599146 pb (L′ = 200 pb−1). P(0|µ) > 0 for every finite µ so zero counts are not a proof of absence. Physical check The confidence construction and zero-background assumption are specified. Uncertain efficiency or nonzero background requires a modified statistical treatment.

Question 10

A rare-channel limit can be inferred even when no events are observed

A background-free search observes zero events. For Poisson mean mu, P(0|μ)=e−μP(0|\mu)=e^{-\mu}. Define a one-sided 95 percent upper limit by P(0|μ95)=.0500P(0|\mu_{95})=.0500. Exposure is 100pb−1100\,\mathrm{pb^{-1}} and efficiency .250.

Diagram shown in the original SVG worksheet.

  1. Find the upper bound on expected signal events and cross section.

  2. Find how that bound changes if exposure doubles with zero events again, and explain why zero observed events does not imply zero cross section.

Question 10 - Solution

A rare-channel limit can be inferred even when no events are observed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

e−μ95=.0500⇒μ95=−ln⁡(.0500)=2.99573.e^{-\mu_{95}}=.0500\Rightarrow\boxed{\mu_{95}=-\ln(.0500)=2.99573}.σ95=μ95/(ϵℒ)=2.99573/25=.119829pb.\boxed{\sigma_{95}=\mu_{95}/(\epsilon\mathcal L)=2.99573/25=.119829\,\mathrm{pb}}.

(b) Interpretation and further calculation

σ95′=.0599146pb(ℒ′=200pb−1).\boxed{\sigma_{95}\prime=.0599146\,\mathrm{pb}}\quad(\mathcal L\prime=200\,\mathrm{pb^{-1}}).P(0|μ)>0 for every finite μso zero counts are not a proof of absence.\boxed{P(0|\mu)>0\text{ for every finite }\mu}\quad\text{so zero counts are not a proof of absence}.

Physical check

The confidence construction and zero-background assumption are specified. Uncertain efficiency or nonzero background requires a modified statistical treatment.