This was originally a post on the Kaggle Forums for the Rossman Competition. I thought I'd repost it here as well.

Like many, if not most of the people here, I've been using some variation of \(\ln(sales)\) as the parameter I've been fitting. Yesterday, I was wondering if I could come up a transform that more closely matched the RMSPE evaluation. So, I did a little math. If \(p\) is the predicted value, \(t\) is the target value and our transform is \(f(x)\), then: $$ \begin{align} RMSPE(p,t) &= \sqrt{\frac{1}{n}\sum \left(\frac{p - t}{t}\right) ^2} \\ P &\equiv f(p) = f(t + \delta) \approx f(t) + f'(t)\delta + \ldots \\ T &\equiv f(t) \\ \text{where } \delta & \equiv p - t \\ \end{align} $$ Here, \(P\) and \(T\) are the transformed \(p\) and \(t\). Our goal is to choose \(f\) in order to make \(RMS(P,T)\) approximately equal to \(RMSPE(p,t)\): $$ RMS(P,T) = \sqrt{\frac{1}{n}\sum (P - T)^2} \\ $$ All we need to get these two metrics to agree is choose \(f\) such that: $$ P - T \approx \frac{p - t}{t} \\ $$ If we assume that \(\delta\) is small, and drop all the higher order terms of the Taylor series, then we can make \(RMS(P,T)\) approximately equal to \(RMSPE(p,t)\): $$ \begin{align} \Rightarrow f(t) + f'(t)\delta - f(t) &\approx \frac{p - t}{t} \\ \Rightarrow f'(t)\delta &\approx \frac{\delta}{t} \\ \Rightarrow f'(t) &\approx \frac{1}{t} \\ \Rightarrow f(x) &\approx \ln(x) + C \end{align} $$ Here \(C\) is an arbitrary constant. Since it cancels in computation of \(RMS\), we can choose it to be zero.

So the log transform was right all along! At least under the, possibly dubious, assumption that predicted values are close to the target values. I didn't learn anything new, but at least I can feel better about doing what I was already doing.

This may be a well known result, but I hadn't seen it mentioned anywhere so I figured I'd write it up.

In the comments to the Kaggle forum post, BogareTAU pointed out that there is an alternative approach using the mean value theorem. That's a nice approach, with essentially the same result. Check out the comments on the original post if you are interested.