Recognizing Nested Statistical Models

Why Recognizing Nesting Matters

When two actuarial models are compared, one may be a restricted version of the other, or the models may simply be different alternatives. That distinction is structural: it depends on whether every distribution allowed by one model can also be represented by the other. Recognizing the relationship helps organize a likelihood-based comparison, but nesting alone does not determine how a test should be calibrated.

The key is to compare the sets of distributions permitted by the models, rather than to compare their names or the number of parameters they happen to display. A restriction such as fixing a shape parameter can create an exact nested model. A parameter limit can instead make one model part of the closure of another without including it in the original parameter space.

Definition: A model family $M_0$ is nested within a model family $M_1$ if every probability distribution allowed by $M_0$ is also allowed by $M_1$. Equivalently, for each parameter value in $M_0$, there is a parameter value in $M_1$ that gives the same distribution for the observations. The nesting is strict if $M_1$ also allows distributions that are not in $M_0$.

For continuous data, matching distributions means matching their probability laws, not merely matching a density at one observed value. For discrete data, it means matching the probability at every possible count. The observation scheme also matters: model distributions should be compared on the same data and support. If the data have been truncated, censored, or otherwise transformed, compare the distributions actually used for those observations.

A Direct Parameter-Restriction Test

A practical way to check nesting is to write down the larger model’s full parameter space and ask whether a restriction on its parameters reproduces the smaller model. Check the parameter ranges and any required parameter mapping. Then verify that the resulting distribution is identical across the full support.

Write both parameterizations.
Record the distribution and the permitted values of every parameter.

Propose a restriction or mapping.
For example, a shape parameter might be fixed at 1, or a dispersion parameter might be set to 0.

Check equality of distributions.
Substitute the restriction into the larger model’s probability function and verify equality for all possible observations.

Check whether the restriction is allowed.
If it is only approached as a parameter tends to infinity, or is excluded from the parameter space, the smaller family is not literally a subset of that parameterized family.

Justification: Suppose the smaller model has density $f_0(x;\eta)$, and the larger model has density $f_1(x;\theta)$. If a mapping $h$ satisfies $f_0(x;\eta)=f_1(x;h(\eta))$ for every possible $x$, then each distribution in $M_0$ is represented in $M_1$. For any observed sample $x_1,\ldots,x_n$, multiplying the equal densities gives the same likelihood at the mapped parameter values. This establishes containment; equality at just one observation does not.

Exponential Severity as a Gamma Restriction

Use the gamma distribution with shape $k>0$ and scale $\theta>0$. Its density for $x>0$ is

f_X(x;k,\theta)=\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}.

An exponential distribution with mean $\theta$ has density $f_X(x)=\theta^{-1}e^{-x/\theta}$, for $x>0$. Substituting $k=1$ into the gamma density gives exactly this exponential density, because $\Gamma(1)=1$ and $x^0=1$. Thus the exponential family is nested within the gamma family when the gamma model permits $k=1$, with the scale parameters matched.

Worked Example: Checking an Exponential Restriction in a Gamma Model

A hypothetical severity analysis uses a gamma model with unknown shape and scale. The restricted model is exponential with mean 2 dollars in the stated units. Determine whether the exponential model is nested, and check the resulting distributions at a loss of 3 units.

Method. Write the gamma density using shape $k$ and scale $\theta$, then impose the candidate restriction $k=1$ and set $\theta=2$. This tests equality of distributions rather than similarity of parameter names.

The restricted gamma density is

f_X(x;1,2)=\frac{x^0e^{-x/2}}{\Gamma(1)2^1} =\frac{1}{2}e^{-x/2},\qquad x>0.

The exponential model with mean 2 has rate $1/2$, so its density is $f_X(x)=(1/2)e^{-x/2}$. The two densities agree for every $x>0$, establishing that the exponential model is nested in the gamma model.

At $x=3$, both densities equal

\frac{1}{2}e^{-3/2}=0.111565\ldots\approx 0.1116.

The survival probability at 3 is also the same:

S_X(3)=e^{-3/2}=0.223130\ldots\approx 0.2231, \qquad F_X(3)=1-e^{-3/2}=0.776870\ldots\approx 0.7769.

The values are rounded to four decimals. As a check, $F_X(3)+S_X(3)=1.0000$. The common mistake is to treat a gamma scale as a rate and substitute the reciprocal incorrectly. Here the scale is 2, so the rate is $1/2$, not 2.

Poisson Frequency and Negative Binomial Parameterizations

A useful negative binomial parameterization has mean $\mu>0$ and dispersion parameter $\alpha>0$. For count $k=0,1,2,\ldots$, its probability mass function is

P(X=k)= \frac{\Gamma(k+1/\alpha)} {\Gamma(1/\alpha)k!} \left(\frac{1}{1+\alpha\mu}\right)^{1/\alpha} \left(\frac{\alpha\mu}{1+\alpha\mu}\right)^k.

Under this parameterization, the variance is $\mu+\alpha\mu^2$, so positive $\alpha$ represents overdispersion relative to a Poisson count with the same mean. The Poisson distribution is obtained in the limit as $\alpha$ decreases to 0. Whether to call it nested depends on the stated parameter space: if the negative binomial family is extended to include $\alpha=0$ by defining the limiting distribution there, Poisson is included at that boundary. If the parameter space requires $\alpha>0$, Poisson is only in the closure of the family, not an element of it.

Derivation: Put $r=1/\alpha$. For fixed $k$, the negative binomial probability can be written as $$ P(X=k)=\frac{1}{k!}\left(\prod_{j=0}^{k-1}(r+j)\right) \left(\frac{r}{r+\mu}\right)^r \left(\frac{\mu}{r+\mu}\right)^k. $$ As $\alpha$ decreases to 0, $r$ increases without bound. The product divided by $r^k$ tends to 1, $(r/(r+\mu))^r$ tends to $e^{-\mu}$, and $r^k(\mu/(r+\mu))^k$ tends to $\mu^k$. Therefore $P(X=k)$ tends to $e^{-\mu}\mu^k/k!$, the Poisson probability. This is a limit argument, not a finite substitution for $r$.

Worked Example: Is a Poisson Count Included in a Negative Binomial Model?

A hypothetical portfolio’s annual claim count is modeled with mean $\mu=2$. Compare a negative binomial model with $\alpha=0.5$ to a Poisson model with the same mean, and state precisely how the Poisson model relates to the parameterized negative binomial family.

Method. Use the mean-dispersion parameterization. Calculate a negative binomial probability and its variance, then compare with the Poisson model. Do not assume that any finite positive dispersion value gives a Poisson distribution.

For $\alpha=0.5$, $1/\alpha=2$, and

\frac{1}{1+\alpha\mu}=\frac{1}{1+0.5(2)}=\frac12, \qquad \frac{\alpha\mu}{1+\alpha\mu}=\frac{1}{2}.

At $k=3$, the negative binomial probability is

P(X=3)= \frac{\Gamma(5)}{\Gamma(2)3!} \left(\frac12\right)^2\left(\frac12\right)^3 =\frac{24}{1\cdot 6}\cdot\frac{1}{32} =\frac{4}{32}=0.1250.

Its variance is $2+0.5(2^2)=4$. For a Poisson count with mean 2, the variance is 2 and

P(X=3)=e^{-2}\frac{2^3}{3!} =e^{-2}\frac{8}{6} =0.180447\ldots\approx 0.1804.

The probabilities differ, so this positive-dispersion negative binomial is not the Poisson distribution. The Poisson family is included only if the negative binomial parameter space is extended to $\alpha=0$; otherwise it is a limiting family, not an exactly nested member. The calculations are rounded to four decimals. A common error is to set the negative binomial size $r$ equal to 1 and call that Poisson. Here $r=1$ gives a geometric distribution, not a Poisson distribution; the Poisson limit requires $r$ to increase without bound while the mean stays fixed.

Overlap Does Not Mean Nesting

Two model families can share distributions without either containing the other. Gamma and Weibull severity models illustrate this distinction. Let a Weibull model have shape $c>0$ and scale $\lambda>0$, with density

f_X(x;c,\lambda)=\frac{c}{\lambda} \left(\frac{x}{\lambda}\right)^{c-1} e^{-(x/\lambda)^c},\qquad x>0.

When $c=1$, this is exponential with mean $\lambda$. The gamma model also becomes exponential when its shape is 1 and its scale is $\lambda$. The two families therefore overlap, but neither contains the other. For example, a Weibull model with $c=2$ has a quadratic term in the exponential tail, $e^{-(x/\lambda)^2}$, unlike the gamma model’s exponential-in-$x$ tail. In the other direction, a gamma model with shape 2 has density proportional to $x e^{-x/\theta}$; a Weibull density matching its exponential tail would need $c=1$, whose density has no factor $x$. Thus the shared exponential case does not make the two full families nested.

Worked Example: Distinguishing Shared Exponential Models From Nested Families

A hypothetical severity analyst compares gamma and Weibull models. Determine whether either full family contains the other, and verify their shared exponential member at $x=2$ when the scale or mean is 1.

Method. First check the common restriction, then show that each family also has members outside the other. Equality at one loss value alone cannot establish family containment.

For gamma shape $k=1$ and scale $\theta=1$, the density at $x=2$ is

f_X(2;1,1)=e^{-2}=0.135335\ldots\approx 0.1353.

For Weibull shape $c=1$ and scale $\lambda=1$, the density is also $e^{-2}$, so the two families share this exponential distribution. Now take gamma shape $k=2$, scale $\theta=1$, and Weibull shape $c=2$, scale $\lambda=1$. At $x=1$, their densities are respectively

f_{\mathrm{gamma}}(1)=1\cdot e^{-1}=0.3679, \qquad f_{\mathrm{Weibull}}(1)=2\cdot 1\cdot e^{-1}=0.7358.

These values show that those particular parameter choices do not give the same distribution; by themselves, they do not prove nonnesting. The full check is the tail form: a Weibull shape of 2 has a quadratic exponent in $x$, which no gamma density’s linear exponent can match. Conversely, gamma shape 2 has the factor $x$; matching its linear exponential tail with a Weibull requires shape 1, whose density lacks that factor. Neither full family contains the other. The density values are rounded to four decimals.

Common Mistakes and Exam Traps

Comparing names instead of distributions. Two models with different names may share a restricted distribution, as gamma and exponential do. Conversely, sharing one distribution does not establish that either full family is nested within the other.
Ignoring the permitted parameter space. A limiting distribution is not automatically an included member. State whether the endpoint or limit is part of the model’s parameter space.
Confusing parameterizations. A gamma scale and gamma rate are reciprocals. A negative binomial size parameter and dispersion parameter are also reciprocals in the parameterization used here.
Checking only one value. Equal densities at one loss amount, or equal probabilities at one count, do not establish identical distributions. Verify equality over the full support.
Assuming nesting settles the test calibration. Nesting describes the relationship between model families. It does not by itself establish the assumptions or reference distribution for a likelihood-based test.

Key takeaway: A model is nested only when the smaller family’s distributions can all be reproduced by the larger family under allowed parameter restrictions. Identify the mapping, verify equality across the support, and label boundary limits separately from exact inclusion.

Check Your Understanding

For each problem, identify the relevant restriction or support argument before deciding whether the models are nested.

A gamma severity model has shape $k>0$ and scale $\theta>0$. An exponential model has mean 3. State the restriction and matching scale that reproduce the exponential model. (Answer: Set $k=1$ and $\theta=3$; the exponential family is nested in the gamma family.)
A negative binomial count model uses mean $\mu$ and dispersion $\alpha$. At $\mu=4$ and $\alpha=0.125$, calculate the variance and state the restriction for a Poisson model in the boundary-extended parameterization. (Answer: Variance $=4+0.125(16)=6$; the Poisson restriction is $\alpha=0$.)
A Weibull severity model has shape $c$ and scale $\lambda$. For $c=1,\lambda=3$, identify the distribution and calculate its survival probability at $x=3$. (Answer: Exponential with mean 3; $S_X(3)=e^{-1}\approx 0.3679$.)
Compare a binomial count with fixed $n=12$ and $p=0.25$ to a Poisson model with mean 3. Use support to decide whether these are the same distribution or whether either family contains the other. (Answer: Neither contains the other; the binomial has zero probability above 12, while the Poisson assigns positive probability there.)
A zero-inflated Poisson model assigns extra probability to zero, with mixing parameter $\omega\in[0,1)$; with probability $1-\omega$, the count is Poisson with mean $\mu$. Identify the Poisson restriction and whether it is an interior or boundary value. (Answer: $\omega=0$ gives the Poisson model, and it is a boundary value of the stated parameter space.)

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