3 ms·
Let me see if I understood you correctly. Let Y represent a random variable representing stationary time series data. Let g(.) be the forecast function. 1) Log
by wenc 8y ago
Let me see if I understood you correctly. Let Y represent a random variable representing stationary time series data. Let g(.) be the forecast function.
1) Log-transformation: log(Y)
2) Forecast results of transformed data: E[g(log(Y))]
3) Inversion of forecast results: exp(E[g(log(Y))])
4) However, what we really want is the expectation of the transformed-forecasted-inversed results, which is E[exp(g(log(Y)))].
Jensen's inequality states that for a convex function φ,
φ(E[X]) <= E[φ(X)]
And equality is only attained if either X is constant or φ is affine. Since neither is (generally) the case here, therefore
exp(E[g(log(Y))]) < E[exp(g(log(Y)))]
An offset factor ε is needed to correct (3) to (4)
E[exp(g(log(Y)))] = exp(E[g(log(Y))]) * ε
Did I get that right? (Offset changed to multiplicative factor)
- CapmCrackaWaka 8y agoYes, 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.