4 ms·
Yes, exactly. However in this case, the offset is multiplicative.
by CapmCrackaWaka 8y ago
Yes, exactly. However in this case, the offset is multiplicative.
- wenc 8y agoUnderstood. 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.