I don't really come from a stats heavy background and am having trouble understanding what the right approach to this question is:

A question states that 'a company is trying to increase the accuracy their real-time data by using two instruments instead of one, error data for 50,000 tests using one instrument is given.

How would you calculate and plot the error distribution for an average across two instruments using the data given and quote the percentage of tests which had abs(error)>some value?

The distribution of the data is not normal, with kurtosis > 10 it has heavy tails, but no significant skew. '

My only hint here is that convolution is involved, but from what I understand since there is only data available for one instrument it must be reused. In which case, working out the distribution of the average the convolution of the PDF of X and the PDF of Y (since they're uniform), would give a triangular result. But from this how would I calculate the percentage of tests above certain error?

I have seen a lot about normal distributions, and plotting the data it appears follows a cauchy or lorrentz distribution with a lambda of 0.5 - initially I thought the sum of two of the same distribution would result in the same distribution over twice the range; the mean equaling a sum of the x,y means and variance the sum of the x,y variances, but is this only true for normal distributions?

  • $\begingroup$ Perhaps your data have a Laplace distribution instead of a normal distribution. In that case it might be better to look at the median of the two measurements, instead of their mean. $\endgroup$ – BruceET Jun 6 '19 at 21:30
  • $\begingroup$ The question implicitly uses assumptions that are critical but, in practice, could very well be strongly violated: namely, that the two instruments have no systematic error and produce independent readings. In the real world, what causes one instrument to read high very might cause another similar instrument to read high, and vice versa, creating strong positive correlation among them. Thus, if the answer to this question has any importance to you, abandon your approach now. $\endgroup$ – whuber Jun 6 '19 at 22:05

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