A single-photon detector produces spurious dark counts with no light present

PHYSICS · Multi-part · Fermi University Physics · Scientific Practice & Experimental Methods · Mathematical Methods for Physics · Data analysis and model fitting in the laboratory

Mathematical Methods for Physics, Ch Estimation, fitting and hypothesis testing

Problem

A single-photon detector produces spurious dark counts with no light present. Eight successive waiting times between dark counts are recorded, in seconds: 0.42,1.15,0.08,2.31,0.67,0.19,1.02,0.550.42, 1.15, 0.08, 2.31, 0.67, 0.19, 1.02, 0.55. The recording is then stopped T=1.50 sT = 1.50\,\mathrm s after the last of these counts, before the next dark count has arrived. Each waiting time is drawn independently from an exponential density f(t;τ)=1τe−t/τf(t;\tau) = \frac{1}{\tau}e^{-t/\tau}, where τ\tau is the mean time between dark counts. ((a)) Explain why the unfinished wait carries information about τ\tau, write its factor in the likelihood, and show that maximising ln⁡L\ln\mathcal{L} gives τ^=(∑jtj+T)/N\hat\tau = \left(\sum_j t_j + T\right)/N, where the sum runs over the N=8N = 8 complete waits. Evaluate τ^\hat\tau. ((b)) Compare τ^\hat\tau with two simpler estimates: the mean of the eight complete waits, and the mean of nine waits that counts the unfinished one as a complete wait of 1.50 s1.50\,\mathrm s. Say why each falls short. ((c)) For the eight complete waits alone, minimising ∑j(tj−τ)2\sum_j(t_j-\tau)^2 over τ\tau gives τ=tˉ\tau=\bar t, the same as maximising their likelihood. Explain why this coincidence does not mean the least-squares construction is the correct one for these data.

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