Why the Scale of the Likelihood Matters
A loss model may be fitted using transformed observations, such as \(Z=\ln X\), because the transformed data are easier to describe with a familiar distribution. But a likelihood calculated from the density of \(Z\) is a likelihood for \(Z\), not directly for the original loss \(X\). To compare candidate models consistently, express their likelihoods for the same observed quantity.
This distinction matters when using information criteria. AIC and BIC depend on the maximized log-likelihood as well as the number of estimated parameters. If the models use the same fixed transformation, its Jacobian adjustment is common to all their log-likelihoods: the adjustment changes the criterion values but not their ranking. If candidates use different transformations, omitting their different adjustments can change the ranking, even when the models describe identical distributions for losses.
For independent, uncensored observations \(x_1,\ldots,x_n\), a model for \(Z=g(X)\) with parameter vector \(\theta\) therefore gives
Here the transformation is fixed and does not depend on \(\theta\). Thus \(J\) is constant with respect to the parameters, so maximizing \(\ell_Z(\theta)\) also maximizes \(\ell_X(\theta)\). The maximized likelihood values are nevertheless different. If the transformation depends on an estimated parameter, its Jacobian can depend on that parameter; it must then be included in the likelihood before maximizing.
The Log Transformation and Information Criteria
For positive losses, the natural-log transformation is \(Z=\ln X\), with inverse \(X=e^Z\). Since \(d(\ln x)/dx=1/x\), the original-scale density and log-likelihood are
The same principle applies to another fixed one-to-one transformation, but its derivative must be used. For example, if \(Z=\log_{10}X\), then \(g'(x)=1/(x\ln 10)\), so the adjustment is \(-\sum_j\ln(x_j\ln 10)\), not merely \(-\sum_j\ln x_j\).
A density specified directly for \(X\) is already on the original loss scale and does not receive a further transformation adjustment. Conversely, if candidate models are all fitted to the same transformed observations using the same transformation, \(J\) is identical across candidates. In that case, omitting it leaves their AIC and BIC differences unchanged, although the reported criterion values are not the original-scale values.
Worked Comparisons
Worked Example: Comparing Two Models on the Log-Loss Scale
A hypothetical portfolio has two independent positive losses, \(x_1=1\) and \(x_2=e\), in a consistent monetary unit. Compare a normal model for \(Z=\ln X\) with an exponential model for \(Z\), using AIC. In both cases convert the maximized likelihood to the original \(X\) scale.
Method and transformation. The transformed observations are \(z_1=0\) and \(z_2=1\). For either model, \(J=-\sum_j\ln x_j=-(0+1)=-1\). The sample size is \(n=2\).
Normal model. The maximum-likelihood estimates are the sample mean and the variance with divisor \(n\): \(\hat{\mu}=(0+1)/2=0.5\) and \(\hat{\sigma}^2=((0-0.5)^2+(1-0.5)^2)/2=0.25\). Therefore, \(\hat{\ell}_Z=-\frac{2}{2}\{\ln(2\pi(0.25))+1\}=-\ln(\pi/2)-1=-1.4516\). The original-scale value is \(\hat{\ell}_X=-1.4516-1=-2.4516\). With \(k=2\), \(\mathrm{AIC}_X=2(2)-2(-2.4516)=8.9032\), or \(8.90\) to two decimals.
Exponential model. For an exponential distribution on \(z\geq0\) with rate \(\lambda\), the log-likelihood is \(2\ln\lambda-\lambda(0+1)\). Its maximum occurs at \(\hat{\lambda}=2/(0+1)=2\), giving \(\hat{\ell}_Z=2\ln2-2=-0.6137\). The original-scale value is \(\hat{\ell}_X=-0.6137-1=-1.6137\). With \(k=1\), \(\mathrm{AIC}_X=2-2(-1.6137)=5.2274\), or \(5.23\) to two decimals.
The exponential-on-\(Z\) candidate has the smaller AIC, \(5.23\) versus \(8.90\). Its density on the original scale is \(f_X(x)=2x^{-3}\) for \(x\geq1\), as follows from \(f_Z(\ln x)/x\). The common adjustment of \(+2\) to the two AIC values does not affect this ranking.
Common mistake avoided. Comparing the transformed-scale AIC values alone would give \(6.90\) for the normal model and \(3.23\) for the exponential model. That happens to give the same ranking because both models use the same transformation, but those are not their original-scale AIC values.
Worked Example: Different Transformations Can Give a False Ranking
Use the same hypothetical losses \(1\) and \(e\). Candidate A models \(Z=\ln X\) as normal. Candidate B models \(W=2\ln X\) as normal. Show why comparing their transformed-scale AIC values without Jacobians is misleading.
Candidate A. As calculated above, the transformed observations are \(0,1\), the fitted normal parameters are \(\hat{\mu}_Z=0.5\) and \(\hat{\sigma}_Z^2=0.25\), and \(\hat{\ell}_Z=-1.4516\). Its Jacobian is \(J_A=-\sum\ln x_j=-1\). Thus \(\hat{\ell}_{X,A}=-2.4516\), and with \(k=2\), \(\mathrm{AIC}_{X,A}=8.9032\).
Candidate B. The observations are \(w_1=0\) and \(w_2=2\). The fitted normal parameters are \(\hat{\mu}_W=(0+2)/2=1\) and \(\hat{\sigma}_W^2=((0-1)^2+(2-1)^2)/2=1\). Hence \(\hat{\ell}_W=-\frac{2}{2}\{\ln(2\pi(1))+1\}=-\ln(2\pi)-1=-2.8379\). For \(g(x)=2\ln x\), \(g'(x)=2/x\), so \(J_B=\ln(2/1)+\ln(2/e)=2\ln2-1=0.3863\). Consequently \(\hat{\ell}_{X,B}=-2.8379+0.3863=-2.4516\), and \(\mathrm{AIC}_{X,B}=4-2(-2.4516)=8.9032\).
The two candidates are equivalent descriptions of the same family of distributions for \(X\): multiplying a normally distributed log-loss by two simply rescales its normal mean and standard deviation. Their correctly calculated original-scale likelihoods and AIC values therefore agree. If the Jacobians were omitted, the AIC values would instead be \(6.9032\) for A and \(9.6758\) for B, falsely favoring A. This is a case where different transformations produce different omitted constants, so the apparent comparison is invalid.
Worked Example: Recomputing an Original-Scale AIC
For a hypothetical candidate fitted to positive losses, suppose the sum of the natural logarithms of the observed losses is \(\ln 10\). Its maximized log-likelihood evaluated as a density for \(Z=\ln X\) is \(-6.000\), and the model has \(k=2\) estimated parameters. Calculate its original-scale log-likelihood and AIC.
Method. For \(Z=\ln X\), the Jacobian adjustment is \(J=-\sum\ln x_j=-\ln10=-2.302585\). Therefore, \(\hat{\ell}_X=\hat{\ell}_Z+J=-6.000-2.302585=-8.302585\), or \(-8.303\) to three decimals.
The original-scale AIC is \(\mathrm{AIC}_X=2k-2\hat{\ell}_X=2(2)-2(-8.302585)=4+16.605170=20.605170\), or \(20.61\) to two decimals. As a check, the transformed-scale AIC is \(4-2(-6)=16\), and the adjustment adds \(-2J=4.605170\), giving \(16+4.605170=20.605170\).
Common mistake avoided. The adjustment is \(-\ln10\), so the corrected log-likelihood is \(-8.303\), not \(-9.003\). Substituting the correct value gives AIC \(20.61\), not \(22.01\).
Why the Jacobian Adjustment Holds
The adjustment follows from preserving probability over corresponding intervals. For a small interval around \(x\), the transformed interval around \(z=g(x)\) has width approximately \(|g'(x)|\,dx\). The probability assigned to either interval must be the same: \(f_X(x)\,dx\approx f_Z(g(x))\,|g'(x)|\,dx\). Dividing by \(dx\) gives \(f_X(x)=f_Z(g(x))|g'(x)|\). Taking logarithms and summing over independent observations produces the log-likelihood formula.
This argument requires a differentiable, one-to-one transformation over the relevant support and continuous densities. If a transformation is not one-to-one, the transformed density generally sums contributions from all inverse branches; using only one derivative is then insufficient. If observations are discrete, probability masses—not continuous-density Jacobians—are used. For censored or truncated observations, the likelihood must also reflect the censoring or sampling mechanism; the transformation adjustment does not replace those likelihood terms.
Practical Checks and Exam Traps
Before comparing information criteria, identify the random quantity whose density each likelihood represents. A lognormal model described by a normal distribution for \(\ln X\) requires the factor \(1/x\) to obtain the density for \(X\). A model stated directly as a density for \(X\) does not require that factor. Mixing these likelihood scales makes the criterion comparison uninterpretable.
- Check the derivative. For \(Z=\ln X\), the adjustment is \(-\sum\ln x_j\). For \(Z=\log_{10}X\), the derivative also includes \(1/\ln10\).
- Check whether the adjustment is shared. A fixed transformation applied to every candidate creates a common adjustment. Different transformations need not.
- Keep parameter counts consistent. A fixed one-to-one transformation does not change the number of estimated parameters, but a model reparameterization or constraint must be counted according to the parameters actually estimated.
- Distinguish a ranking from a reported criterion. A shared constant leaves AIC and BIC differences unchanged, but the adjusted values are needed when reporting criteria on the original observation scale or comparing against another likelihood scale.
Check Your Understanding
For each problem, state the relevant scale and include any required Jacobian adjustment before calculating or interpreting the criterion.
- A candidate model for \(Z=\ln X\) has maximized transformed-scale log-likelihood \(-4.000\), \(k=2\), and the observed losses satisfy \(\sum_j\ln x_j=2\). Find its original-scale log-likelihood and AIC. (Answer: \(\hat{\ell}_X=-6.000\), AIC \(=16.00\).)
- For losses \(1\) and \(e\), an exponential model for \(Z=\ln X\) is fitted by maximum likelihood. Find the fitted rate, original-scale log-likelihood, and original-scale AIC. (Answer: \(\hat{\lambda}=2\), \(\hat{\ell}_X=2\ln2-3=-1.6137\), AIC \(=5.23\).)
- A fitted model has \(n=4\), \(k=3\), transformed-scale maximized log-likelihood \(-7.000\), and \(\sum_j\ln x_j=\ln16\), where the transformation is \(Z=\ln X\). Find its original-scale log-likelihood and AIC. (Answer: \(\hat{\ell}_X=-9.7726\), AIC \(=25.55\).)
- Two models use the same fixed transformation on the same observations. Their transformed-scale maximized log-likelihoods are \(-8\) and \(-9\), and their parameter counts are 2 and 3. Explain whether a common Jacobian can reverse their AIC ranking. (Answer: No; the first has the smaller AIC by 4 before adjustment, and a shared Jacobian adds the same amount to both.)
- Two candidates use different transformations and have identical original-scale likelihoods and parameter counts. Their unadjusted transformed-scale log-likelihoods are \(-5\) and \(-6\). Can the unadjusted values establish which candidate has smaller original-scale AIC? (Answer: No; the separate Jacobian adjustments are required, and the unadjusted likelihoods alone do not establish the original-scale ranking.)