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E(x) = c*exposures^a You have some historical policies with trended losses and come up with data points of E[X] and "exposures." Find the parameters c and a that fit that data set best. You can use software to do this or a simpler solver type approach in Excel. The intuition seems to be that mu = a * ln(exposure) + b Is just another way of expressing E[x] = c*exposures^a That gets you directly to the mu parameter We have E[X] = e^(mu + sigma^2/2) We want E[X] = c*exposures^a If mu = a * ln(exposure) + b then E[X] = e^(ln(exposure^a) + b + sigma^2/2) = exposure^a * e^(b + sigma^2/2) So it works out. Can call the last term c. Now you have solved for a and c, how do you get mu and sigma? Well you have mu = a * ln(exposure) + b sigma ... you haven't said how you are getting this but maybe you have some standard deviation from an data set you are matching to And b is solvable from your c and sigma All just guesswork on my part given the other responses in the thread. The lesson is to always document your work so that the people after you don't face this same problem. 
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