Deriving Akaike Weights From AIC Differences

From AIC Differences to Weights

AIC gives a score for comparing fitted models, balancing fit against a penalty for model complexity. When several candidate models are under consideration, the smallest AIC identifies the model with the lowest score among them. Akaike weights extend that comparison: they assign each candidate a normalized share of relative support, with the shares summing to one.

The calculation has two essential parts. First, measure each candidate’s AIC difference from the minimum. Second, convert those differences to relative likelihoods and divide each by their total. The factor of one-half in the exponent is important: omitting it changes the resulting weights.

Definition: For candidate model $i$, let $\mathrm{AIC}_i$ be its Akaike information criterion and let $\mathrm{AIC}_{\min}$ be the smallest AIC in the candidate set. Define $\Delta_i=\mathrm{AIC}_i-\mathrm{AIC}_{\min}$. The relative likelihood is $L_i=\exp(-\Delta_i/2)$, and the Akaike weight is $$ w_i=\frac{\exp(-\Delta_i/2)}{\displaystyle\sum_{j=1}^{M}\exp(-\Delta_j/2)}, $$ where $M$ is the number of models in the stated candidate set. Thus $w_i\geq 0$ and $\sum_{i=1}^{M}w_i=1$.

AIC is calculated as $\mathrm{AIC}_i=-2\ell_i(\hat{\theta}_i)+2k_i$, where $\ell_i(\hat{\theta}_i)$ is the maximized log likelihood and $k_i$ is the number of estimated free parameters. The weights use differences in the resulting AIC scores, not differences in log likelihood alone. In particular, the penalty for the number of parameters is already reflected in AIC.

Why the Negative Half-Difference Appears

The exponent follows directly from the definition of AIC. The AIC difference compares each candidate’s score with the best score in the set. Exponentiating minus one-half of that difference gives a relative likelihood on a scale where the best candidate has value one.

Write the difference.
For model $i$, $\Delta_i=\mathrm{AIC}_i-\mathrm{AIC}_{\min}$.

Substitute the AIC definition.
Since $\mathrm{AIC}_i=-2\ell_i(\hat{\theta}_i)+2k_i$, the difference is the difference in those penalized scores.

Exponentiate on the likelihood scale.
Applying $\exp(-\Delta_i/2)$ makes the relative quantity proportional to $\exp(\ell_i(\hat{\theta}_i)-k_i)$. The factor of one-half cancels the factor of two multiplying the log likelihood and the parameter penalty in AIC.

Normalize across candidates.
Divide each relative likelihood by the sum of all candidates’ relative likelihoods. This makes the weights add to one.

For completeness, the key algebra in Step 3 is

\exp\left(-\frac{\mathrm{AIC}_i-\mathrm{AIC}_{\min}}{2}\right) =\frac{\exp(-\mathrm{AIC}_i/2)}{\exp(-\mathrm{AIC}_{\min}/2)} \propto \exp\left(\ell_i(\hat{\theta}_i)-k_i\right).

The proportionality is across the candidate set: the denominator involving the minimum AIC is common to every model. The weight formula then divides each relative quantity by their sum. This derivation explains both the negative sign and the factor of one-half; neither is an optional convention.

Worked Example: Two Claim-Frequency Models

A hypothetical insurer compares two fitted models for claim counts from the same portfolio and observation period. Model P is Poisson with fitted mean $\hat{\lambda}=1.25$, and Model N is negative binomial with fitted mean $\hat{\mu}=1.20$ and fitted size parameter $\hat{r}=2.0$. Suppose the fit reports AIC values 80 and 82, respectively. Calculate the Akaike weights within these two candidates.

Method. Subtract the minimum AIC and convert each difference to a relative likelihood. The minimum is 80, so

\Delta_P=80-80=0,\qquad \Delta_N=82-80=2.

The relative likelihoods are

L_P=\exp(-0/2)=1,\qquad L_N=\exp(-2/2)=\exp(-1)\approx 0.367879.

Their total is $1+0.367879=1.367879$. Therefore,

w_P=\frac{1}{1.367879}\approx 0.7311,\qquad w_N=\frac{0.367879}{1.367879}\approx 0.2689.

To four decimal places, the weights are 0.7311 and 0.2689. They are nonnegative and sum to 1.0000 after rounding. The method avoids the common error of using $\exp(-\Delta_i)$ instead of $\exp(-\Delta_i/2)$.

Applying the Formula to Several Candidates

With more than two models, the procedure is unchanged. The denominator includes the relative likelihood of every candidate in the stated set. A model’s weight is not obtained by comparing it only with the best model; it must be normalized against the whole set.

Worked Example: Three Hypothetical Severity Models

For the same hypothetical sample of positive claim severities, consider a gamma model with fitted shape 2.4 and scale 500, a lognormal model with fitted log-mean 6.1 and log-standard-deviation 0.8, and a Weibull model with fitted shape 1.3 and scale 900. Suppose their AIC values are 120, 122, and 124, respectively. Use the three models as the candidate set and calculate the weights.

Step 1: Find differences from the minimum. The minimum AIC is 120. In the order gamma, lognormal, Weibull, the differences are $0$, $2$, and $4$.

Step 2: Compute relative likelihoods.

L_{\mathrm{Gamma}}=\exp(0)=1, \quad L_{\mathrm{Lognormal}}=\exp(-2/2)=\exp(-1)\approx 0.367879, \quad L_{\mathrm{Weibull}}=\exp(-4/2)=\exp(-2)\approx 0.135335.

Step 3: Add the relative likelihoods. Their sum is $1+0.367879+0.135335=1.503214$.

Step 4: Divide each value by the sum.

w_{\mathrm{Gamma}}=\frac{1}{1.503214}\approx 0.6652,\quad w_{\mathrm{Lognormal}}=\frac{0.367879}{1.503214}\approx 0.2447,\quad w_{\mathrm{Weibull}}=\frac{0.135335}{1.503214}\approx 0.0900.

To four decimal places, the weights are 0.6652, 0.2447, and 0.0900. Their rounded sum is 0.9999; the difference from one is due to rounding, and the unrounded weights sum exactly to one. This check can catch an omitted candidate or a denominator error.

The weight calculation requires only the AIC values once they have been obtained. It does not require the severity distributions to have the same parameterization or number of parameters; their respective complexity penalties are already included in their AIC scores. The candidates must, however, be compared on a common basis, including the same observations and a consistent likelihood basis.

Worked Example: Calculate AIC First, Then Weights

A hypothetical portfolio’s annual claim counts are fitted with three candidate distributions. The Poisson model has one estimated parameter, $\hat{\lambda}=1.6$, and maximized log likelihood $-50.1$. A negative binomial model has two estimated parameters, $\hat{\mu}=1.6$ and $\hat{r}=2.5$, and maximized log likelihood $-49.6$. A geometric model on counts $0,1,2,\ldots$ has one estimated parameter, fitted success probability $\hat{p}=0.40$, and maximized log likelihood $-50.7$. These are hypothetical fitted outputs for the same count observations. Calculate the AIC weights.

Step 1: Calculate each AIC. Use $\mathrm{AIC}=-2\ell(\hat{\theta})+2k$.

\mathrm{AIC}_{P}=-2(-50.1)+2(1)=100.2+2=102.2,

\mathrm{AIC}_{NB}=-2(-49.6)+2(2)=99.2+4=103.2,

\mathrm{AIC}_{G}=-2(-50.7)+2(1)=101.4+2=103.4.

The minimum is 102.2, so the AIC differences are $0$, $1.0$, and $1.2$, in that order.

Step 2: Convert differences to relative likelihoods.

L_P=1,\qquad L_{NB}=\exp(-1/2)\approx 0.606531,\qquad L_G=\exp(-1.2/2)=\exp(-0.6)\approx 0.548812.

Their sum is $1+0.606531+0.548812=2.155343$. Therefore,

w_P=\frac{1}{2.155343}\approx 0.4640,\quad w_{NB}=\frac{0.606531}{2.155343}\approx 0.2814,\quad w_G=\frac{0.548812}{2.155343}\approx 0.2546.

To four decimal places, the weights are 0.4640, 0.2814, and 0.2546, and they sum to 1.0000. The calculation avoids a second common error: using the unpenalized log likelihoods directly and thereby failing to include the parameter-count penalty already reflected in AIC.

What the Weights Do—and Do Not—Say

Akaike weights distribute relative support over the candidate models included in the calculation. They are conditional on that stated set: adding, removing, or changing a candidate changes the denominator, and can change every weight even when the AIC values of the retained models do not change. The AIC differences among retained models remain the same if a constant is added to all their AIC values, but the normalized weights depend on which models are included.

The weights are not probabilities that the models are true, nor are they Bayesian posterior probabilities. They are normalized relative likelihoods derived from AIC differences. Their interpretation is limited to the candidates and comparison basis used. A weight of 0.7 does not mean there is a 70% probability that a model is the data-generating process.

This distinction matters in actuarial work. A selected severity or frequency model may affect estimates of expected losses, tail quantities, or premiums, but the weights alone do not validate any candidate or settle whether its assumptions are appropriate. They summarize one particular AIC-based comparison. Model diagnostics and the decision context still matter.

Common Exam Trap: Dropping the One-Half

Suppose two models have AIC values 80 and 82. A tempting but incorrect calculation is to use $\exp(-\Delta_i)$. It gives relative values $1$ and $\exp(-2)\approx 0.1353$, and therefore weights $1/(1+0.1353)\approx 0.8808$ and $0.1353/(1+0.1353)\approx 0.1192$. This is wrong because AIC differences must be divided by two in the exponent.

The correct relative values are $1$ and $\exp(-2/2)=\exp(-1)\approx 0.3679$. Normalizing gives weights 0.7311 and 0.2689. Always write $\exp(-\Delta_i/2)$ before substituting numbers.

A second useful check is to verify that the minimum-AIC model has $\Delta=0$ and hence relative likelihood 1. All other differences must be nonnegative. Finally, divide by the sum over the complete candidate set and check that the unrounded weights sum to one.

Key takeaway: Compute each AIC difference from the minimum, exponentiate its negative half, and normalize over every candidate in the stated set. Akaike weights sum to one, but they are relative support measures—not probabilities that candidate models are true.

Check Your Understanding

For each problem, calculate the weights over exactly the candidate set stated. Give final weights to four decimal places.

Two hypothetical fitted claim-count models have AIC values 50 and 54. Calculate their Akaike weights. (Answer: 0.8808, 0.1192)
Three candidate models for hypothetical repair costs have AIC values 210, 211, and 213. Calculate the weights in that order. (Answer: 0.5466, 0.3315, 0.1219)
Three hypothetical severity models have AIC values 95, 95, and 97. Calculate the weights in that order. (Answer: 0.4223, 0.4223, 0.1554)
For three hypothetical frequency models, the fitted log likelihoods and parameter counts are: Model A, $-31.4$ and $k=2$; Model B, $-30.8$ and $k=3$; Model C, $-32.0$ and $k=1$. Calculate each AIC and then the weights in A, B, C order. (Answer: AIC values 66.8, 67.6, 66.0; weights 0.3162, 0.2120, 0.4718)
Three models have AIC values 100, 101, and 102. Calculate the weights first over all three models, then over the reduced candidate set containing only the first two. (Answer: all three: 0.5065, 0.3072, 0.1863; first two only: 0.6225, 0.3775)

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