Likelihood-Ratio Invariance Under Reparameterization

Why Parameter Scales Should Not Change a Likelihood-Ratio Test

A severity model may describe the same distribution using different parameters. An exponential model, for example, can be parameterized by its rate or its mean; a lognormal model can be parameterized by the mean of the logarithms or by the median loss. These choices affect the numerical values and interpretation of the parameters, but they should not affect a likelihood-ratio comparison if the two parameterizations describe exactly the same model and the hypotheses are translated correctly.

This invariance is useful in practice because a model’s natural parameter may not be the scale used in a question, software output, or actuarial interpretation. The essential check is not whether the parameter values look the same. It is whether the parameter transformation gives a one-to-one correspondence between the full parameter spaces and between the null parameter spaces.

Definition: Suppose a model has parameter $\theta\in\Theta$ and likelihood $L(\theta)$. A reparameterization $\phi=g(\theta)$ is one-to-one if each $\theta\in\Theta$ corresponds to exactly one $\phi\in\Phi=g(\Theta)$, and each $\phi\in\Phi$ corresponds to exactly one $\theta$. The likelihood in the new parameter is the same data likelihood written as $L_\phi(\phi)=L(g^{-1}(\phi))$. It is not multiplied by a Jacobian for changing the parameter.

For a likelihood-ratio test, write $\Theta_0\subseteq\Theta$ for the null parameter space and $\Theta$ for the full space. The likelihood-ratio statistic and its commonly reported test statistic are

\Lambda=\frac{\sup_{\theta\in\Theta_0}L(\theta)} {\sup_{\theta\in\Theta}L(\theta)}, \qquad T=-2\log\Lambda =2\left\{\sup_{\theta\in\Theta}\ell(\theta) -\sup_{\theta\in\Theta_0}\ell(\theta)\right\},

where $\ell(\theta)=\log L(\theta)$. The numerator is the largest likelihood allowed by the null, and the denominator is the largest likelihood allowed by the full model. The null space must be expressed in the new parameter as $\Phi_0=g(\Theta_0)$; changing the parameter while leaving the null condition unchanged can define a different test.

Why the Statistic Is Invariant

The result follows from the fact that a one-to-one transformation changes coordinates, not the set of distributions being compared. For every $\theta$ in the full space there is a corresponding $\phi=g(\theta)$, and the likelihood value at that point is unchanged. The same correspondence holds within the null space when the null is translated as $\Phi_0=g(\Theta_0)$.

Map the full parameter space.
As $\theta$ ranges over $\Theta$, $\phi=g(\theta)$ ranges over all of $\Phi$, so the possible likelihood values are exactly the same.

Map the null space.
As $\theta$ ranges over $\Theta_0$, $\phi$ ranges over $\Phi_0=g(\Theta_0)$, again giving exactly the same likelihood values.

Take the ratio.
Both the unrestricted maximum and the null-constrained maximum are unchanged, so their ratio and $-2\log\Lambda$ are unchanged.

In equations, the correspondence gives

\sup_{\phi\in\Phi}L_\phi(\phi) =\sup_{\theta\in\Theta}L(\theta), \qquad \sup_{\phi\in\Phi_0}L_\phi(\phi) =\sup_{\theta\in\Theta_0}L(\theta).

Dividing the second equality by the first proves that $\Lambda$, and therefore $T$, is identical in either parameterization. This argument concerns the likelihood values as the parameter varies. It does not require the transformation to be differentiable. Differentiability and related regularity conditions can matter for an asymptotic reference distribution, however, so invariance of the computed statistic does not by itself establish that a chi-square approximation is appropriate.

Theorem: A likelihood-ratio statistic is invariant under a one-to-one reparameterization when the full parameter space and null parameter space are both mapped exactly to their corresponding spaces. The maximum-likelihood estimate also maps across: if $\hat{\theta}$ maximizes $L(\theta)$, then $g(\hat{\theta})$ maximizes $L_\phi(\phi)$.

Worked Severity Comparisons

Worked Example: Exponential Rate Versus Mean

A hypothetical portfolio has three independent, fully observed positive claim severities of \$2, \$4, and \$6, modeled as exponential with a common rate $\lambda>0$. Thus $f_X(x)=\lambda e^{-\lambda x}$ for $x>0$, and the mean severity is $\theta=1/\lambda$. Test $H_0:\theta=6$ against the unrestricted exponential model. Calculate $T$ using the rate parameter and verify it using the mean parameter.

Method and assumptions. The observations are independent and identically distributed, with no truncation or censoring, and the exponential model is assumed to hold. The rate $\lambda>0$ and mean $\theta>0$ are in one-to-one correspondence. The observed sum is $S=2+4+6=12$, and the sample size is $n=3$.

Ignoring no parameter-dependent terms, the likelihood and log likelihood in the rate parameter are

L(\lambda)=\lambda^3e^{-12\lambda}, \qquad \ell(\lambda)=3\log\lambda-12\lambda.

Differentiating gives $\ell'(\lambda)=3/\lambda-12$. Setting this equal to zero gives $\hat{\lambda}=3/12=0.25$. The second derivative is $-3/\lambda^2<0$, so this is the maximum. Under the null, $\theta=6$ means $\lambda_0=1/6$.

The unrestricted minus null log likelihood is

\ell(0.25)-\ell(1/6) =3\log\left(\frac{0.25}{1/6}\right)-12\left(0.25-\frac16\right) =3\log(1.5)-1.

Since $\log(1.5)=0.4054651$, this difference is $1.2163953-1=0.2163953$, and $T=2(0.2163953)=0.4327906$. To four decimal places, $T=0.4328$.

Now use the mean parameter. The likelihood is $L(\theta)=\theta^{-3}e^{-12/\theta}$, so $\ell(\theta)=-3\log\theta-12/\theta$. Its maximum is at $\hat{\theta}=12/3=4$, and the null value is $\theta_0=6$. Thus

\ell(4)-\ell(6) =3\log\left(\frac64\right)+12\left(\frac16-\frac14\right) =3\log(1.5)-1 =0.2163953.

This gives the same $T=0.4328$, as required. Common mistake avoided: the null mean of 6 corresponds to rate $1/6$, not rate 6. Substituting 6 as the rate would test a different severity model.

Worked Example: Gamma Scale Versus Log-Scale

Suppose two hypothetical independent claim severities are \$2 and \$6. Model each as gamma with known shape $k=2$ and unknown scale $\theta>0$, using density $f_X(x)=x^{k-1}e^{-x/\theta}/(\Gamma(k)\theta^k)$ for $x>0$. Test $H_0:\theta=1$ against the unrestricted model, first using $\theta$ and then using $\eta=\log\theta$.

Method and assumptions. The shape is fixed at 2, the scale is positive, and both observations are complete and independent. Since $\eta=\log\theta$ maps $(0,\infty)$ one-to-one onto the real line, the corresponding null is $\eta=0$.

Let $S=2+6=8$ and $n=2$. The terms in the log likelihood that depend on $\theta$ are $-nk\log\theta-S/\theta$. Differentiating gives $-nk/\theta+S/\theta^2$; setting this to zero yields $\hat{\theta}=S/(nk)=8/4=2$. The second derivative at this value is negative, so it is a maximum. The log-likelihood difference between the estimate and the null is

\ell(2)-\ell(1) =4\log\left(\frac12\right)+8\left(1-\frac12\right) =-4\log 2+4.

Using $\log 2=0.6931472$, the difference is $4-2.7725887=1.2274113$. Therefore $T=2(1.2274113)=2.4548226$, or $2.4548$ to four decimal places.

In the log-scale parameter, $\hat{\eta}=\log 2$ and $\eta_0=0$. Substituting $\theta=e^\eta$, the parameter-dependent log likelihood becomes $-4\eta-8e^{-\eta}$. Its difference at $\log2$ and 0 is

\{-4\log2-8e^{-\log2}\}-\{-4(0)-8e^0\} =-4\log2-4+8 =4-4\log2 =1.2274113.

The resulting statistic is again $2.4548$. Common mistake avoided: the unrestricted estimate on the log scale is $\log 2$, not 2. Also, the null $\theta=1$ must become $\eta=0$; retaining $\eta=1$ would not preserve the null hypothesis.

Worked Example: Lognormal Mean-Log Versus Median

A hypothetical set of three complete claim severities is $\$1$, $\$e$, and $\$e^2$. Assume independent lognormal observations with known log-scale standard deviation $\sigma=1$ and unknown $\mu$, so $\log X$ is normal with mean $\mu$ and variance 1. Test whether the median severity is \$1 against the unrestricted model. Compare the parameterizations $\mu$ and $m=e^\mu$, where $m$ is the median.

Method and assumptions. The lognormal model and known value $\sigma=1$ are assumed, and all three positive observations are fully observed. The mapping $m=e^\mu$ is one-to-one from the real line to positive medians. The null $m=1$ is therefore $\mu=\log1=0$.

The logged observations are $y_1=0$, $y_2=1$, and $y_3=2$. Apart from terms not depending on $\mu$, the log likelihood is $-\tfrac12\sum_{j=1}^3(y_j-\mu)^2$. It is maximized at the sample mean $\hat{\mu}=(0+1+2)/3=1$, corresponding to $\hat{m}=e$. The sum of squared deviations at the unrestricted estimate is $(0-1)^2+(1-1)^2+(2-1)^2=1+0+1=2$. Under the null $\mu=0$, it is $0^2+1^2+2^2=5$. Hence

\ell(\hat{\mu})-\ell(0) =\frac12(5-2)=1.5, \qquad T=2(1.5)=3.0000.

Using the median parameter gives $\hat{m}=e$ and $m_0=1$, which correspond exactly to $\hat{\mu}=1$ and $\mu_0=0$. The same likelihood values and $T=3.0000$ follow. Common mistake avoided: the median is $e^\mu$, not $\mu$ itself. Testing $m=1$ means testing $\mu=0$, rather than $\mu=1$.

What Invariance Does—and Does Not—Guarantee

The result applies to equivalent parameterizations of the same probability model, not to arbitrary changes in the model. A parameter transformation must preserve the full set of distributions and the set allowed under the null. If either set changes, the maximized likelihoods may change and the comparison is no longer the same test.

A parameter transformation is also distinct from transforming the data. If observed severities are converted from dollars to thousands of dollars, the density with respect to the new data scale has a data-transformation factor. When the same data transformation is used under both hypotheses, that common factor cancels in the likelihood ratio. But it is not the reason parameter reparameterization is invariant: parameter changes do not require adding a Jacobian to the likelihood.

The following distinctions are useful when checking a calculation:

Situation	What to verify	Consequence
One-to-one parameter change	Map both full and null parameter spaces exactly	The likelihood-ratio statistic is unchanged
Null value stated on a new scale	Translate the null using the parameter map	The same distributions remain under comparison
Many-to-one parameter map	Check whether distinct parameter values represent distinct distributions	It may not be a valid reparameterization of the original model
Changed model or changed null set	Compare the sets of distributions, not just parameter labels	The statistic may differ because the hypotheses differ

For example, $\phi=\theta^2$ is one-to-one if the original parameter space is $\theta>0$, but it is not one-to-one if $\theta$ can be any real number. In the latter case, $\theta$ and $-\theta$ map to the same $\phi$. One cannot claim invariance by treating $\phi$ as an ordinary one-to-one coordinate for that original space without checking whether the duplicate values correspond to the same distribution and whether the null and full model have been represented correctly.

In exam calculations, a reliable procedure is to write the parameter map, transform the unrestricted estimate, translate the null, and compare the two maximized log likelihoods. Agreement between scales is a useful check on algebra and interpretation. A disagreement usually indicates a mistaken null translation, an incorrect estimate, or a likelihood expression that does not describe the same model.

Key takeaway: A likelihood-ratio statistic depends on the likelihood values available under the null and full models, not on the names or scales used for their parameters. Preserve those model spaces exactly, and the statistic is invariant.

Check Your Understanding

For each problem, identify the parameter correspondence and translate the null before calculating or judging the likelihood-ratio comparison.

Two independent hypothetical severities, \$2 and \$6, follow an exponential distribution with unknown mean $\theta>0$. Test $H_0:\theta=2$ against the unrestricted model. Find $T$, and state the equivalent null in rate parameter $\lambda=1/\theta$. (Answer: $T=1.2274$; the rate null is $\lambda=0.5$.)
Two independent hypothetical severities, \$3 and \$9, follow a gamma distribution with known shape $k=1.5$ and unknown scale $\theta>0$. Test $H_0:\theta=3$. Re-express the comparison using $\eta=\log\theta$. (Answer: $T=0.2739$; the null is $\eta=\log3$, and the statistic is unchanged.)
Three hypothetical severities have logarithms $1,2,3$. Assume a lognormal model with known log-scale standard deviation $\sigma=1$; test whether the median is $e^2$. Calculate $T$ using $\mu$, and state the estimate and null value on the median scale. (Answer: $T=0.0000$; $\hat{\mu}=2$, $\hat{m}=e^2$, and the null is $\mu=2$.)
A model uses $\theta>0$, and a colleague proposes $\phi=\theta^2$ with null $\theta=3$. Is this a one-to-one reparameterization on the stated parameter space, and what is the corresponding null value? (Answer: Yes; $\phi=9$, and the likelihood-ratio statistic is unchanged if the full and null spaces are both mapped.)

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