Why BIC Balances Fit and Complexity
A more complicated regression model can often fit the observed data better than a simpler one. But improved fit alone does not settle which model to prefer: adding parameters also makes a model more flexible. The Bayesian Information Criterion, or BIC, combines a measure of fit with a penalty for the number of estimated parameters. Its penalty depends on the sample size through a logarithm.
BIC is calculated from a model’s likelihood, so it can be used for different kinds of regression models when their likelihoods are defined and their comparison is appropriate. It is not a test statistic with a null distribution, and a low BIC does not establish that a model is true. It is a criterion for comparing a specified set of fitted candidate models.
The two parts of the formula work in opposite directions. A larger maximized log-likelihood indicates a better fit to the observed data; because log-likelihoods are often negative, a larger value is less negative. Multiplying by \(-2\) turns better fit into a smaller first term. The second term, \(d\ln(n)\), is nonnegative for \(n\geq 1\) and increases as the model has more estimated parameters.
The parameter count refers to the parameters estimated by the likelihood, not just the regression slopes. For example, in a Gaussian regression model with an estimated error variance, the variance is a parameter too. If a variance or other quantity is known and fixed rather than estimated, it is not counted as an estimated parameter. The exact count depends on the model being fitted.
What Makes a Comparison Appropriate?
Before comparing BIC values, check that the candidate models are fitted to the same observations and response and use likelihoods defined on a comparable basis. If one model is fitted to a different set of observations, a change in log-likelihood can reflect the changed data as well as the model. The comparison would not isolate the fit-complexity tradeoff that BIC is meant to summarize.
The likelihood also has to match the model and the data. BIC does not replace the assumptions of the chosen regression model or make a misspecified likelihood correct. For instance, if a Gaussian regression likelihood is used, its assumptions about the response distribution and error structure need to be suitable for the setting. BIC itself is not a diagnostic check for those assumptions.
In the usual setup, \(n\) counts the independent observational units contributing to the likelihood. It is not necessarily the number of rows in a data file if rows are repeated measurements or otherwise dependent. When observations are dependent, the dependence needs to be represented appropriately in the likelihood, and the relevant sample-size convention must match that model.
Worked Example: Calculating BIC for One Gaussian Regression
Suppose a hypothetical Gaussian regression is fitted to data from \(n=48\) independent observations. Its maximized log-likelihood is \(-72.4\). The model estimates an intercept, three slopes, and an error variance, so \(d=5\). Calculate its BIC.
The formula is $$ \mathrm{BIC}=-2\ell(\hat{\theta})+d\ln(n). $$
Substitute the supplied values: $$ \mathrm{BIC}=-2(-72.4)+5\ln(48). $$
Since \(\ln(48)\approx 3.8712\), $$ \mathrm{BIC}=144.8+5(3.8712)=144.8+19.356=164.156. $$
Thus the BIC is approximately 164.2. It is a criterion value, not a quantity with the units of the response. By itself, it does not say whether the regression is adequate; it becomes useful when compared with BIC values from suitable candidate models fitted to the same data.
How the Penalty Affects a Model Comparison
The penalty for each additional estimated parameter is \(\ln(n)\). This means the cost of adding one parameter depends on the sample size. The following algebra shows how that cost enters a comparison. It does not explain why the penalty has this particular form; the focus here is how to use the defined criterion.
Suppose model A has \(d\) estimated parameters and maximized log-likelihood \(\ell_A\). Model B adds one estimated parameter, so it has \(d+1\) parameters and maximized log-likelihood \(\ell_B\). Subtract A’s BIC from B’s:
If the added parameter improves the maximized log-likelihood, then \(\ell_B-\ell_A\) is positive. B has the smaller BIC precisely when $$ -2(\ell_B-\ell_A)+\ln(n)<0, $$ which is equivalent to $$ 2(\ell_B-\ell_A)>\ln(n). $$ Thus, in this one-parameter comparison, the improvement in fit must be large enough to offset the extra penalty. This is an algebraic way to read the BIC difference, not a significance test: the inequality does not provide a p-value or show that the added parameter is scientifically important.
Worked Example: Does One Added Predictor Lower BIC?
Consider two hypothetical Gaussian regressions fitted to the same \(n=30\) observations. Model A includes an intercept and one predictor, along with an estimated error variance, so \(d_A=3\); its maximized log-likelihood is \(-51.0\). Model B adds a second predictor, so \(d_B=4\); its maximized log-likelihood is \(-48.8\). Assume both models use the same response and a suitable likelihood for the data. Calculate and compare their BIC values.
For model A, $$ \mathrm{BIC}_A=-2(-51.0)+3\ln(30). $$ Since \(\ln(30)\approx 3.4012\), $$ \mathrm{BIC}_A=102.0+3(3.4012)=102.0+10.2036=112.2036. $$
For model B, $$ \mathrm{BIC}_B=-2(-48.8)+4\ln(30) =97.6+4(3.4012). $$ Therefore, $$ \mathrm{BIC}_B=97.6+13.6048=111.2048. $$
Model B’s BIC is lower by approximately \(112.2036-111.2048=0.9988\), or about 1.0. The log-likelihood gain is \(-48.8-(-51.0)=2.2\), so twice that gain is \(4.4\). The added-parameter penalty is \(\ln(30)\approx3.4012\), and \(4.4>3.4012\). The fit improvement therefore outweighs the additional penalty in this BIC comparison, making B the preferred model of these two by BIC. This does not establish that its new predictor has a causal effect or that B is the true model.
Reading BIC Values Across Several Candidates
When there are several candidate models, calculate each BIC using the same convention and compare the resulting values. The candidate with the smallest BIC is the one preferred by this criterion among the models considered. This ranking does not imply that every model with a larger BIC is useless, or that the lowest-BIC model captures every important feature of the data.
The next hypothetical calculation illustrates how a modest improvement in log-likelihood can be outweighed by the cost of additional parameters. Each model is fitted to the same 80 independent observations, using a suitable likelihood for the response.
Worked Example: Ranking Three Candidate Regressions
A hypothetical analysis considers three regression models for the same response and the same \(n=80\) observations. Model A has \(d=3\) estimated parameters and maximized log-likelihood \(-120.0\). Model B has \(d=4\) and log-likelihood \(-117.5\). Model C has \(d=5\) and log-likelihood \(-116.9\). Calculate their BIC values and identify the smallest.
Here, \(\ln(80)\approx4.3820\). For Model A, $$ \mathrm{BIC}_A=-2(-120.0)+3\ln(80) =240.0+3(4.3820)=253.146\approx253.1. $$ For Model B, $$ \mathrm{BIC}_B=-2(-117.5)+4\ln(80) =235.0+4(4.3820)=252.528\approx252.5. $$ For Model C, $$ \mathrm{BIC}_C=-2(-116.9)+5\ln(80) =233.8+5(4.3820)=255.710\approx255.7. $$
Model B has the smallest BIC, approximately 252.5, so it is preferred among these three candidates by BIC. Model C fits the observed data better in the log-likelihood sense than Model B, but its extra complexity raises its BIC above Model B’s. The ranking is conditional on this candidate set and these data; it is not a guarantee that Model B is scientifically correct or will be best for every purpose.
What BIC Does—and Does Not—Tell You
A BIC value is useful because it places model fit and parameter count on one scale for a specified comparison. A smaller value means the combination of fit and penalty is smaller under the formula. It is not a direct measure of prediction error in the response’s units, a proportion of variation explained, or a probability that the model is true.
BIC values are most informative as relative quantities within a well-defined comparison. The absolute value, considered alone, has no universal threshold for declaring a model “good.” A preferred model by BIC may still have poor residual behavior, omit important structure, or fail to answer the scientific question. Candidate models should be chosen for defensible reasons, and their assumptions and purpose still matter.
Common Mistakes
- Counting only slope coefficients. That can undercount parameters in a likelihood. Include every parameter that is estimated in that likelihood, such as an estimated Gaussian error variance, as applicable.
- Using the wrong sign for fit. The formula uses \(-2\ell(\hat{\theta})\). If one model has a higher maximized log-likelihood, its fit term is smaller, all else equal.
- Forgetting the logarithm. The penalty is \(d\ln(n)\), not \(dn\) and not \(d\ln(d)\).
- Comparing models fitted to different data without checking comparability. Different observations or likelihood scales can make the values unsuitable for a direct ranking.
- Treating the smallest BIC as proof. A lower value selects a preferred candidate under this criterion; it does not prove the model is true, establish causation, or test a null hypothesis.
Check Your Understanding
Use the definition and comparison logic to answer these questions.
- A model has maximized log-likelihood \(-40.0\), \(n=20\), and \(d=3\). Write the expression for its BIC before evaluating it.
- In a Gaussian regression with an estimated error variance, should that variance be counted among the parameters estimated by the likelihood? Explain.
- Two models use the same observations, and one has a larger maximized log-likelihood but also more estimated parameters. Why can its BIC still be larger?
- For a one-parameter extension, express the condition for the more complex model to have a lower BIC in terms of the log-likelihood gain and \(n\).
- Why is it incorrect to say that the model with the lowest BIC has been proved to be true?