## Stata 15 help for bic_note

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[R] BIC note -- Calculating and interpreting BIC

Description

This entry discusses a statistical issue that arises when using the
Bayesian information criterion (BIC) to compare models.

Stata calculates BIC, assuming N = e(N) -- we will explain -- but
sometimes it would be better if a different N were used.  Commands that
calculate BIC have an n() option, allowing you to specify the N to be
used.

In summary,

1.  If you are comparing results estimated by the same estimation
command, using the default BIC calculation is probably fine.
There is an issue, but most researchers would ignore it.

2.  If you are comparing results estimated by different estimation
commands, you need to be on your guard.

a.  If the different estimation commands share the same
definitions of observations, independence, and the like, you
are back in case 1.

b.  If they differ in these regards, you need to think about the
value of N that should be used.  For example, logit and
xtlogit differ in that the former assumes independent
observations and the latter, independent panels.

c.  If estimation commands differ in the events being used over
which the likelihood function is calculated, the information
criteria may not be comparable at all.  We say information
criteria because this would apply equally to the Akaike
information criterion (AIC), as well as to BIC.  For
instance, streg and stcox produce such incomparable results.
The events used by streg are the actual survival times,
whereas the events used by stcox are failures within risk
pools, conditional on the times at which failures occurred.

Remarks

Remarks are presented under the following headings:

Background
The problem of determining N
Example 1
Example 2
The problem of conformable likelihoods
The first problem does not arise with AIC; the second problem does
Calculating BIC correctly

Background

The AIC and the BIC are two popular measures for comparing maximum
likelihood models.  AIC and BIC are defined as

AIC = -2*ln(likelihood) + 2*k

BIC = -2*ln(likelihood) + ln(N)*k

where

k = number of parameters estimated

N = number of observations

We are going to discuss AIC along with BIC because AIC has some of the
problems that BIC has, but not all.

AIC and BIC can be viewed as measures that combine fit and complexity.
Fit is measured negatively by -2*ln(likelihood); the larger the value,
the worse the fit.  Complexity is measured positively, either by 2*k
(AIC) or ln(N)*k (BIC).

Given two models fit on the same data, the model with the smaller value
of the information criterion is considered to be better.

There is substantial literature on these measures: see Akaike (1974);
Raftery (1995); Sakamoto, Ishiguro, and Kitagawa (1986); and Schwarz
(1978).

When Stata calculates the above measures, it uses the rank of e(V) for k
and it uses e(N) for N.  e(V) and e(N) are Stata notation for results
stored by the estimation command.  e(V) is the variance-covariance matrix
of the estimated parameters, and e(N) is the number of observations in
the dataset used in calculating the result.

The problem of determining N

The difference between AIC and BIC is that AIC uses the constant 2 to
weight k, whereas BIC uses ln(N).

Determining that the value of N should be used is problematic.  Despite
appearances, the definition "N is the number of observations" is not easy
to make operational.  N does not appear in the likelihood function
itself, N is not the output of a standard statistical formula, and what
is an observation is often subjective.

Example 1

Often, what is meant by N is obvious.  Consider a simple logit model.
What is meant by N is the number of observations that are statistically
independent and that corresponds to M, the number of observations in the
dataset used in the calculation.  We will write N=M.

But now assume that the same dataset has a grouping variable and the data
are thought to be clustered within group.  To keep the problem simple,
let's pretend that there are G groups and m observations within group, so
that M = G*m.  Because you are worried about intragroup correlation, you
fit your model with xtlogit, grouping on the grouping variable.  Now you
wish to calculate BIC.  What is the N that should be used?  N=M or N=G?

That is a deep question.  If the observations really are independent,
then you should use N=M.  If the observations within group are not just
correlated but are duplicates of one another, and they had to be so, then
you should use N=G.  Between those two extremes, you should probably use
a number between M and G, but determining what that number should be from
measured correlations is difficult.  Using N=M is conservative in that,
if anything, it overweights complexity.  Conservativeness, however, is
subjective, too:  using N=G could be considered more conservative in that
fewer constraints are being placed on the data.

When the estimated correlation is high, our reaction would be that using
N=G is probably more reasonable.  Our first reaction, however, would be
that using BIC to compare models is probably a misuse of the measure.

Stata uses N=M.  An informal survey of web-based literature suggests that
N=M is the popular choice.

There is another reason, not so good, to choose N=M.  It makes
across-model comparisons more likely to be valid when performed without
thinking about the issue.  Say that you wish to compare the logit and
xtlogit results.  Thus, you need to calculate

BIC_p = -2*ln(likelihood_p) + ln(N_p)*k

BIC_x = -2*ln(likelihood_x) + ln(N_x)*k

Whatever N you use, you must use the same N in both formulas.  Stata's
choice of N=M at least meets that test.

Example 2

In the above example, using N=M is reasonable.  Now let's look at when
using N=M is wrong even if popular.

Consider a model fit by stcox.  Using N=M is certainly wrong if for no
other reason than M is not even a well-defined number.  The same data can
be represented by different datasets with different numbers of
observations.  For example, in one dataset, there might be 1 observation
per subject.  In another, the same subjects could have two records each,
the first recording the first half of the time at risk and the second
recording the remaining part.  All statistics calculated by Stata on
either dataset would be the same, but M would be different.

Deciding on the right definition, however, is difficult.  Viewed one way,
N in the Cox regression case should be the number of risk pools, R,
because the Cox regression calculation is made on the basis of the
independent risk pools.  Viewed another way, N should be the number of
subjects, N_subj, because, even though the likelihood function is based
on risk pools, the parameters estimated are at the subject level.

You can decide which argument you prefer.

For parametric survival models, in single-record data, N=M is
unambiguously correct.  For multirecord data, there is an argument for
N=M and for N=N_subj.

The problem of conformable likelihoods

The problem of conformable likelihoods does not concern N.  Researchers
sometimes use information criteria such as BIC and AIC to make
comparisons across models.  For that to be valid, the likelihoods must be
conformable; that is, the likelihoods must all measure the same thing.

It is common to think of the likelihood function as the
Pr(data|parameters), but in fact, the likelihood is

Pr(particular events in the data | parameters)

You must ensure that the events are the same.

For instance, they are not the same in the semiparametric Cox regression
and the various parametric survival models.  In Cox regression, the
events are, at each failure time, that the subjects observed to fail in
fact failed, given that failures occurred at those times.  In the
parametric models, the events are that each subject failed exactly when
the subject was observed to fail.

The formula for AIC and BIC is

measure = -2*ln(likelihood) + complexity

When you are comparing models, if the likelihoods are measuring different
events, even if the models obtain estimates of the same parameters,
differences in the information measures are irrelevant.

The first problem does not arise with AIC; the second problem does

Regardless of model, the problem of defining N never arises with AIC
because N is not used in the AIC calculation.  AIC uses a constant 2 to
weight complexity as measured by k, rather than ln(N).

For both AIC and BIC, however, the likelihood functions must be
conformable; that is, they must be measuring the same event.

Calculating BIC correctly

When using BIC to compare results, and especially when using BIC to
compare results from different models, you should think carefully about
how N should be defined.  Then specify that number by using the n()
option:

. estimates stats full sub, n(74)

----------------------------------------------------------------
Model |  Obs    ll(null)   ll(model)   df        AIC         BIC
------+---------------------------------------------------------
full |  102   -45.03321   -20.59083    4   49.18167    58.39793
sub |  102   -45.03321   -27.17516    3   60.35031    67.26251
----------------------------------------------------------------
Note:  N = 74 used in calculating BIC

Both estimates stats and estat ic allow the n() option; see [R] estimates
stats and [R] estat ic.

References

Akaike, H. 1974. A new look at the statistical model identification.
IEEE transactions on Automatic Control 19: 716-723.

Raftery, A. 1995. Bayesian model selection in social research. In Vol. 25
of Sociological Methodology, ed. P. V. Marsden, 111-163.  Oxford:
Blackwell.

Sakamoto, Y., M. Ishiguro, and G. Kitagawa. 1986.  Akaike Information
Criterion Statistics. Dordrecht, The Netherlands:  Reidel.

Schwarz, G. 1978. Estimating the dimension of a model.  Annals of
Statistics 6: 461-464.

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