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Old 08-25-2018, 10:59 AM
lllj's Avatar
lllj lllj is offline
Join Date: May 2011
Studying for Exam 9
Posts: 5,153

Originally Posted by TDH View Post
Thanks, but this is slightly different no? Our mu here is not the trended losses, it's the parameter for the lognormal; and it's mu = a * log(exposure) + b.

We can repeat your procedure by first parameterising mu, then finding the expected cost using the lognormal e^(mu + 0.5 sigma ^2) and finding the squared difference, but my question is where is the intuition that mu itself is in the form mu = a * log(exposure) + b?
It might be more straightforward to fit using the equation
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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