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Likelihood Ratios As Measures Of Relative Fit

Calculate likelihood ratios consistently and explain what they do—and do not—say about the relative fit of competing models.

Intermediate 10 min read

What You'll Learn

  • Define a likelihood for discrete observations and continuous loss data.
  • Compare maximized likelihoods using a clearly stated numerator and denominator.
  • Calculate likelihood ratios for Poisson, Bernoulli, and exponential models.
  • Use log-likelihood differences to simplify comparisons.
  • Explain why a larger maximized likelihood is not, by itself, proof that a model is better.
  • Avoid confusing likelihood with event probability or a calibrated test result.

Introduction

When several probability models are being considered for the same observed losses, one basic comparison is how large each model’s likelihood can become after its parameters are fitted. A likelihood ratio summarizes this comparison. It measures relative fit to the observed data at the chosen parameter values; it does not, by itself, establish that one model is true or that the observed difference is statistically meaningful. The ideas here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Likelihood and Likelihood Ratio

Suppose the observed data are \(x=(x_1,\ldots,x_n)\), and a model assigns a joint probability mass function or density \(f(x;\theta)\), where \(\theta\) is an unknown parameter. Once the data are observed, the likelihood is the same mathematical expression viewed as a function of \(\theta\): \(L(\theta;x)=f(x;\theta)\). For independent observations, the joint likelihood is the product of the individual probability mass functions or densities.

Definition: The maximized likelihood for a model \(M\) is the largest value of its likelihood over the parameter values allowed by that model: \(\widehat L_M=\max_{\theta\in\Theta_M}L_M(\theta;x)\), when a maximum exists. For two competing models \(M_A\) and \(M_B\), define the relative-fit likelihood ratio in the stated direction as \(R_{A:B}=\widehat L_A/\widehat L_B\). A ratio greater than 1 means model \(M_A\) has the larger maximized likelihood for these data; a ratio less than 1 means model \(M_B\) does.

Always name the numerator and denominator. Reversing them changes the ratio to its reciprocal and reverses the interpretation. If the models are compared at specified parameter values rather than each being fitted, use those likelihood values in the ratio and say so; that is not a comparison of maximized likelihoods.

For discrete observations, the likelihood at a parameter value equals the probability assigned to the observed outcome. For continuous observations, it is a density at the observed data, not the probability of observing those exact values. In either case, the likelihood is treated as a function of the model parameter. It is not a probability distribution over parameter values and need not sum or integrate to 1 over \(\theta\). A density, and hence a continuous-data likelihood, can exceed 1.

Why Likelihood Ratios Are Convenient

For independent observations, multiplying many probability mass functions or densities can produce very small numbers. Taking logarithms turns the product into a sum. Because the logarithm is strictly increasing, maximizing the likelihood and maximizing its logarithm give the same fitted parameter. Also, a ratio of positive likelihoods can be found by subtracting log-likelihoods and exponentiating.

$$ \ell(\theta;x)=\log L(\theta;x) =\sum_{j=1}^{n}\log f(x_j;\theta), \qquad R_{A:B} =\exp\!\left(\widehat\ell_A-\widehat\ell_B\right). $$

This follows from \(\log(ab)=\log a+\log b\) and \(\exp(\log a-\log b)=a/b\), for positive likelihoods \(a\) and \(b\). Any factors that are identical for both models and do not depend on their parameters cancel in the ratio. They should still be handled consistently when writing the likelihoods.

Worked Likelihood-Ratio Comparisons

Worked Example: Fitting a Poisson Claim-Count Model

A hypothetical portfolio records four independent claim counts, \(1,2,3,0\), with each count assumed to follow a Poisson distribution with rate \(\lambda\) claims per policy period. Compare the maximized likelihood with the likelihood at the specified rate \(\lambda=1\). Give the ratio as maximized likelihood divided by likelihood at \(\lambda=1\).

Method and assumptions. Use the Poisson probability mass function \(P(X=x)=e^{-\lambda}\lambda^x/x!\), independently for the four counts. The total count is \(1+2+3+0=6\), and the product of the factorial terms is \(1!\,2!\,3!\,0!=12\). Thus

$$ L(\lambda)=\frac{\lambda^6e^{-4\lambda}}{12}, \qquad \ell(\lambda)=6\log\lambda-4\lambda-\log 12. $$

Differentiate the log-likelihood: \(\ell'(\lambda)=6/\lambda-4\). Setting this equal to zero gives \(\widehat\lambda=6/4=1.5\). The second derivative is \(\ell''(\lambda)=-6/\lambda^2<0\) for \(\lambda>0\), so this is a maximum.

At the fitted rate, \(L(1.5)=1.5^6e^{-6}/12\). Since \(1.5^6=11.390625\), this is \(0.94921875e^{-6}\approx0.0023529\). At \(\lambda=1\), \(L(1)=e^{-4}/12\approx0.0015263\). Therefore

$$ R_{\widehat\lambda:1} =\frac{L(1.5)}{L(1)} =1.5^6e^{-2} \approx1.5416. $$

The maximized likelihood is about 1.5416 times the likelihood at rate 1 for these data. Common mistake avoided: the sample mean is 1.5, not 1. The ratio comparing the fitted rate with itself would be exactly 1; it is not the comparison requested here.

Worked Example: Comparing a Fitted Bernoulli Rate With a Fixed Rate

In a hypothetical independent sample of ten policies, eight have at least one claim during a period. Assume each policy has the same probability \(p\) of at least one claim. Compare the maximum likelihood over \(0\leq p\leq1\) with the likelihood at \(p=0.5\).

Method and assumptions. For the observed sequence with eight successes and two failures, the likelihood is \(L(p)=p^8(1-p)^2\). If the data are recorded only as a success count, the binomial coefficient \(\binom{10}{8}\) appears in both likelihoods and cancels in their ratio.

The log-likelihood is \(8\log p+2\log(1-p)\). Its derivative is \(8/p-2/(1-p)\), which is zero when \(8(1-p)=2p\), giving \(\widehat p=0.8\). The log-likelihood is concave for \(0<p<1\), so this is its maximum. The two likelihoods are

$$ L(0.8)=0.8^8(0.2)^2 =0.16777216(0.04)=0.0067108864, \qquad L(0.5)=0.5^{10}=0.0009765625. $$

Hence \(R_{\widehat p:0.5}=0.0067108864/0.0009765625=6.8719476736\), or \(6.8719\) to four decimal places. The fitted probability gives a substantially larger likelihood for this sample than the fixed probability. Common mistake avoided: comparing just \(0.8\) with \(0.5\) ignores that the likelihood concerns the entire observed sample, not one policy.

Worked Example: Continuous Losses and an Exponential Rate

A hypothetical insurer observes two independent loss amounts, \(1\) and \(2\), measured in thousands of dollars. Assume each loss has an exponential distribution with rate \(\lambda\) per thousand dollars, so its density is \(f(x;\lambda)=\lambda e^{-\lambda x}\) for \(x\geq0\) and \(\lambda>0\). Compare the maximum likelihood with the likelihood at \(\lambda=1\).

Method and assumptions. With total observed loss \(1+2=3\), the joint density, viewed as a likelihood, is \(L(\lambda)=\lambda^2e^{-3\lambda}\). The log-likelihood is \(2\log\lambda-3\lambda\). Its derivative \(2/\lambda-3\) vanishes at \(\widehat\lambda=2/3\); its second derivative \(-2/\lambda^2<0\) confirms a maximum.

At the fitted rate, \(L(2/3)=(4/9)e^{-2}\approx0.0601490\). At rate 1, \(L(1)=e^{-3}\approx0.0497871\). Dividing gives

$$ R_{\widehat\lambda:1} =\frac{(4/9)e^{-2}}{e^{-3}} =\frac{4e}{9} \approx1.2081. $$

Thus, the maximum likelihood is about 1.2081 times the likelihood at rate 1. These likelihoods are densities for the two observed loss amounts, not probabilities of observing those exact amounts. Their common measurement units cancel in the ratio. Common mistake avoided: treating the continuous-data density as an event probability.

What a Larger Maximized Likelihood Does Not Establish

A ratio greater than 1 describes a relative likelihood for the observed data under the particular models and parameter spaces being compared. It does not give the probability that one model is correct, a posterior odds ratio, a p-value, or a guarantee that the observed difference would be unusual under a specified null model. Those interpretations require additional assumptions and analysis.

A model with more freedom to adjust parameters can often achieve a likelihood at least as large as a more restricted alternative. Consequently, a larger maximum alone does not establish that a model is preferable after accounting for its flexibility. Likelihood ratios provide a comparison of fit at the observed data, not an automatic adjustment for model complexity or a decision rule.

The ratio is most interpretable when both likelihoods are calculated for the same observations and the same data representation. For continuous data, a change of measurement units changes density values by a common Jacobian factor; when the same transformation is applied to both models, that common factor cancels. Comparing likelihood values calculated from different data, different subsets, or inconsistent likelihood conventions does not give a meaningful relative-fit ratio.

Common Exam Traps

  • Reversing the direction. Write “model A divided by model B” before calculating. A result of 2 in the opposite direction would be 0.5.
  • Using a fitted value on both sides. If both numerator and denominator are the same maximized likelihood, the ratio is 1. State the competing parameter value or model clearly.
  • Calling likelihood a probability of a model. Likelihood evaluates how well parameter values or models account for observed data; it does not assign probabilities to models.
  • Interpreting a ratio as a test result. A likelihood ratio by itself is not a p-value or a calibrated measure of statistical significance.
  • Ignoring the data structure. For independent observations, use the joint likelihood, or its log. Comparing only one observation’s contribution can change the conclusion.
Key takeaway: State the data, model assumptions, parameter values allowed, and ratio direction. A likelihood ratio greater than 1 means only that the numerator has the larger likelihood for the observed data under the comparison as defined; it does not alone establish better overall evidence.

Check Your Understanding

For each problem, keep the ratio direction explicit and interpret it only as a relative likelihood comparison.

  1. A hypothetical sample has five Bernoulli successes and three failures. Compare the maximum likelihood over \(p\) with the likelihood at \(p=0.5\). (Answer: \(\widehat p=5/8\), and \(R_{\widehat p:0.5}=84375/65536\approx1.2875\).)
  2. Four independent hypothetical Poisson counts are \(0,1,1,2\). Compare the maximum likelihood with the likelihood at \(\lambda=0.5\). (Answer: \(\widehat\lambda=1\), and \(R_{\widehat\lambda:0.5}=16e^{-2}\approx2.1654\).)
  3. Three independent hypothetical exponential losses, measured in thousands of dollars, are \(0.5,1.5,2\). Compare the maximum likelihood for the rate \(\lambda\) with the likelihood at \(\lambda=1\). (Answer: \(\widehat\lambda=3/4\) per thousand dollars, and \(R_{\widehat\lambda:1}=(27/64)e\approx1.1468\).)
  4. Two candidate models give maximized log-likelihoods \(-12.4\) and \(-13.1\), respectively, for the same data. Calculate the ratio of the first model’s maximized likelihood to the second’s and state what it means. (Answer: \(e^{0.7}\approx2.0138\); the first model has about 2.0138 times the maximized likelihood for these data, which alone does not establish that it is preferable.)
  5. A hypothetical sample has seven Bernoulli successes and three failures. Compare the likelihood at the fitted probability with the likelihood at \(p=0.5\). (Answer: \(\widehat p=0.7\), and \(R_{\widehat p:0.5}=0.7^7(0.3)^3/0.5^{10}\approx2.2769\).)

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