というNBER論文をFrancis X. Dieboldが上げている(原題は「Assessing Point Forecast Accuracy by Stochastic Error Distance」、ungated版)。共著者はMinchul Shin(イリノイ大)。

We propose point forecast accuracy measures based directly on distance of the forecast-error c.d.f. from the unit step function at 0 ("stochastic error distance," or SED). We provide a precise characterization of the relationship between SED and standard predictive loss functions, and we show that all such loss functions can be written as weighted SED's. The leading case is absolute-error loss. Among other things, this suggests shifting attention away from conditional-mean forecasts and toward conditional-median forecasts.



Minchul and I went in trying to escape the expected loss minimization paradigm. We came out realizing that we hadn't escaped, but simultaneously, that not all loss functions are created equal. In particular, there's a direct and natural connection between our stochastic error divergence (SED) and absolute-error loss, elevating the status of absolute-error loss in our minds and perhaps now making it our default benchmark of choice. Put differently, "quadratic loss is for squares." (Thanks to Roger Koenker for the cute mantra.)
Minchulと私は、期待損失を最小化するというパラダイムを逃れようとして研究に取り掛かった。結局逃れるには至らなかったことが分かったが、同時に、すべての損失関数が平等ではないことも分かった。特に、我々の確率誤差乖離*1SED)と絶対誤差損失の間には、直接的かつ自然なつながりがあることが分かった。我々の心の中で絶対誤差損失の地位は上昇し、今ではおそらく選択肢のデフォルトとなっている。換言すれば、「平方損失は四角四面*2なのである」(このキュートなマントラはRoger Koenkerのもの)。


We've all ranked forecast accuracy by mean squared error (MSE) and mean absolute error (MAE), the two great workhorses of relative accuracy comparison. MSE-rankings and MAE-rankings often agree, but they certainly don't have to -- they're simply different loss functions -- which is why we typically calculate and examine both.
Here's a trivially simple question: Under what conditions will MSE-rankings and MAE-rankings agree? It turns out that the answer it is not at all trivial -- indeed it's unknown. Things get very difficult, very quickly.
With N(μ,σ2) forecast errors we have that
E(|e|) = \sigma \sqrt{2/\pi} \exp\left( -\frac{\mu^{2}}{2 \sigma^{2}}\right) + \mu \left[1-2 \Phi\left(-\frac{\mu}{\sigma}\right)\right],
where Φ(⋅) is the standard normal cdf. This relates MAE to the two components of MSE, bias (μ) and variance (σ2), but the relationship is complex. In the unbiased Gaussian case (μ=0), the result collapses to MAE∝σ, so that MSE-rankings and MAE-rankings must agree. But the unbiased Gaussian case is very, very special, and little else is known.
ここでありきたりなほど簡単な質問がある:MSEランキングとMAEランキングが一致するのはどのような条件下だろうか? その答えはありきたりではまったくないことが分かっている。実際のところ、その答えは未知なのだ。この話は急速に非常に難しくなる。
E(|e|) = \sigma \sqrt{2/\pi} \exp\left( -\frac{\mu^{2}}{2 \sigma^{2}}\right) + \mu \left[1-2 \Phi\left(-\frac{\mu}{\sigma}\right)\right],