An unbiased model comparison test using cross-validation

Abstract

Social scientists often consider multiple empirical models of the same process. When these models are parametric and non-nested, the null hypothesis that two models fit the data equally well is commonly tested using methods introduced by Vuong (Econometrica 57(2):307–333, 1989) and Clarke (Am J Political Sci 45(3):724–744, 2001; J Confl Resolut 47(1):72–93, 2003; Political Anal 15(3):347–363, 2007). The objective of each is to compare the Kullback–Leibler Divergence (KLD) of the two models from the true model that generated the data. Here we show that both of these tests are based upon a biased estimator of the KLD, the individual log-likelihood contributions, and that the Clarke test is not proven to be consistent for the difference in KLDs. As a solution, we derive a test based upon cross-validated log-likelihood contributions, which represent an unbiased KLD estimate. We demonstrate the CVDM test’s superior performance via simulation, then apply it to two empirical examples from political science. We find that the test’s selection can diverge from those of the Vuong and Clarke tests and that this can ultimately lead to differences in substantive conclusions.

Appendix: Proof of Vuong test finite sample bias

Here we derive the inequality given in Eq. 3 of the main text. Suppose $\varvec{y}$ is a sample of $n$ independent observations from a normal distribution with zero mean and variance $\tau ^2$. Also, let $g$ be a normal probability density function with the mean estimated as the sample mean of $\varvec{y}$, and the variance fixed at $\sigma ^2$. Then the expected value of the observed log-likelihood is

$$\begin{aligned} E[ll_o]&= E_{\varvec{y}}\left[ \frac{1}{n} \ln \left( \prod _{i=1}^n \frac{1}{\sqrt{2\pi \sigma ^2}} \exp \left[ \frac{-1}{2\sigma ^2} \left( y_i - \frac{1}{n}\sum _{j=1}^n y_j \right) ^2\right] \right) \right] \nonumber \\&= -\ln \left( \sqrt{2 \pi \sigma ^2} \right) - \frac{1}{2\sigma ^2n} E_{\varvec{y}} \left[ \sum _{i=1}^n \left( y_i - \frac{1}{n}\sum _{j=1}^n y_j \right) ^2 \right] \nonumber \\&= -\ln \left( \sqrt{2 \pi \sigma ^2} \right) - \frac{\tau ^2(n-1)}{2\sigma ^2n}. \end{aligned}$$

(8)

The expected value of the expected log-likelihood is¹⁶

$$\begin{aligned} \begin{aligned} E[ll_e]&= E_{\bar{y}}\left[ E_{\varvec{y}}\left( \frac{1}{n} \sum _{i=1}^n \ln \frac{1}{\sqrt{2 \pi \sigma ^2}} - \frac{(y_i - \bar{y})^2}{2\sigma ^2} \right) \right] \\&= -\ln \left( \sqrt{2 \pi \sigma ^2} \right) - \frac{1}{2\sigma ^2} E_{\bar{y}}\left[ E_{\varvec{y}}\left( \frac{1}{n} \sum _{i=1}^n y_i^2 -2y_i\bar{y}+\bar{y}^2 \right) \right] \\&= -\ln \left( \sqrt{2 \pi \sigma ^2} \right) - \frac{1}{2\sigma ^2} E_{\bar{y}}\left[ \tau ^2 + \bar{y}^2 \right] \\&= -\ln \left( \sqrt{2 \pi \sigma ^2} \right) - \frac{\tau ^2+\frac{\tau ^2}{n}}{2\sigma ^2} \end{aligned} \end{aligned}$$

(9)

Now, considering two different values of $\sigma ^2$, $\sigma _1^2$ and $\sigma _2^2$ with $\sigma _1^2 < \sigma _2^2$, $E_1[ll_o] > E_2[ll_o]$ iff

$$\begin{aligned} -\ln \left( \sqrt{2 \pi \sigma _1^2} \right) - \frac{\tau ^2(n-1)}{2\sigma _1^2n}&> -\ln \left( \sqrt{2 \pi \sigma _2^2} \right) - \frac{\tau ^2(n-1)}{2\sigma _2^2n}\nonumber \\ \tau ^2&< \frac{\ln \left( \sigma ^2_2 \right) -\ln \left( \sigma _1^2 \right) }{\frac{n-1}{n}\left( \frac{1}{\sigma _1^2}-\frac{1}{\sigma _2^2}\right) }, \end{aligned}$$

(10)

and $E_1[ll_e] < E_2[ll_e]$ iff

$$\begin{aligned} -\ln \left( \sqrt{2 \pi \sigma _1^2} \right) - \frac{\tau ^2+\frac{\tau ^2}{n}}{2\sigma _1^2}&< -\ln \left( \sqrt{2 \pi \sigma _2^2} \right) - \frac{\tau ^2+\frac{\tau ^2}{n}}{2\sigma _2^2}\nonumber \\ \frac{\ln \left( \sigma ^2_2 \right) -\ln \left( \sigma _1^2 \right) }{\left[ 1+\frac{1}{n}\right] \left( \frac{1}{\sigma _1^2} -\frac{1}{\sigma _2^2}\right) }&< \tau ^2. \end{aligned}$$

(11)

Combining these two conditions gives the interval from Eq. 3.