11. An astronomer makes three one-second measurements of a star\'s brightness, c

Question

11. An astronomer makes three one-second measurements of a star's brightness, count- ing 4 photons in the first, 81 in the second, and 9 in the third. What is the best estimate of the average photon arrival rate and its uncertainty? Compute this (a) by using Equation (2.18), then (b) by noting that a total of 94 photons have arrived in 3 seconds. Explain why these two methods give such different results. Which method is the correct one? Is there anything peculiar about the data that would lead you to question the validity of either method?

Explanation / Answer

Data observations: x1= 4, x2=81, x3 = 9

Simple average = sum(xi)/n = 4+81+9/3 = 94/3 = 31.33

In absence of any other information about uncertainty of an individual observation, let's use observation - sample mean as a proxy for dispersion or variance. Thus,

sigma12= (31.33-4)2 =750.76 , sigma22= (31.33-81)2 =2460.16 , sigma32= (31.33-9)2 = 501.76

Now using formula 2.18 given in the problem, yc = 1/(1/750.76+1/2460.16+1/501.76)[(4/750.76)+(81/2460.16)+(9/501.76)] =15.05

Uncertainty in above = (1/750.76+1/2460.16+1/501.76)-0.5 = 16.37

Now using second approach, avg rate = total photon/total time =94/3 =31.4

Both methods give different answers because first method weights every observation with its variance, however second method gives equal weight to every observation. Data is very disperse which may make results by second method less suitable. Moreover second data point which has the highest variance seem to have little influence on the weighted mean.

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