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Yes, exactly. However in this case, the offset is multiplicative.


Understood. To add one more point, I'm noticing the reason the above works the way it does is because most forecast algorithms output an expected value instead of a random variable, hence the results are E[g(log(Y)] instead of just g(log(Y)).

It strikes me that if you package the entire thing as a random variable:

Z = exp(G(log(Y)))

and use a different kind of forecast function G : Y -> Y' where Y, Y' ~ Normal, then we don't need the multiplicative factor -- which can be difficult to calculate for an arbitrary transformation. We can just get the expected value of Z, ie. E[Z] = E[exp(G(log(Y)))]. This is not done in the article, but in theory it could be.




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