Statistics Graduate Level Quiz: Likelihood Ratio Tests And Wilks Theorem
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Likelihood Ratio Tests And Wilks TheoremQuestion 1 of 10

Counts O1,…,O6O_1,\ldots,O_6 arise from a multinomial distribution with six cells. Under the null model, the cell probabilities are pj(θ)p_j(\theta), where the scalar parameter θ\theta is estimated from the same counts. The model is regular, and every fitted probability is positive. The alternative is the saturated multinomial model, and Ej=npj(θ^)E_j=n p_j(\widehat\theta).

Which statistic and asymptotic degrees of freedom give the multinomial likelihood ratio test?

Use 2∑j=16Ejlog⁡(Ej/Oj)2\sum_{j=1}^6 E_j\log(E_j/O_j) with a χ52\chi^2_5 reference distribution.
Use 2∑j=16Ojlog⁡(Oj/Ej)2\sum_{j=1}^6 O_j\log(O_j/E_j) with a χ52\chi^2_5 reference distribution.
Use ∑j=16(Oj−Ej)2/Ej\sum_{j=1}^6 (O_j-E_j)^2/E_j with a χ42\chi^2_4 reference distribution.
Use 2∑j=16Ojlog⁡(Oj/Ej)2\sum_{j=1}^6 O_j\log(O_j/E_j) with a χ42\chi^2_4 reference distribution.
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Likelihood Ratio Tests And Wilks Theorem

Practice Likelihood Ratio Tests And Wilks Theorem in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Likelihood Ratio Tests And Wilks Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Question 1

Counts O1,…,O6O_1,\ldots,O_6 arise from a multinomial distribution with six cells. Under the null model, the cell probabilities are pj(θ)p_j(\theta), where the scalar parameter θ\theta is estimated from the same counts. The model is regular, and every fitted probability is positive. The alternative is the saturated multinomial model, and Ej=npj(θ^)E_j=n p_j(\widehat\theta).

Which statistic and asymptotic degrees of freedom give the multinomial likelihood ratio test?

  1. Use 2∑j=16Ejlog⁡(Ej/Oj)2\sum_{j=1}^6 E_j\log(E_j/O_j) with a χ52\chi^2_5 reference distribution.
  2. Use 2∑j=16Ojlog⁡(Oj/Ej)2\sum_{j=1}^6 O_j\log(O_j/E_j) with a χ52\chi^2_5 reference distribution.
  3. Use ∑j=16(Oj−Ej)2/Ej\sum_{j=1}^6 (O_j-E_j)^2/E_j with a χ42\chi^2_4 reference distribution.
  4. Use 2∑j=16Ojlog⁡(Oj/Ej)2\sum_{j=1}^6 O_j\log(O_j/E_j) with a χ42\chi^2_4 reference distribution. (correct answer)
Explanation: Whenever you see a likelihood ratio test in a multinomial setting, you need two things exactly right: the formula and the degrees of freedom. The multinomial log-likelihood ratio statistic (also called G2G^2) compares observed counts to expected counts under the null. Twice the log-likelihood ratio between the saturated model and the null model reduces to: G2=2∑j=16Ojlog⁡ ⁣(OjEj)G^2 = 2\sum_{j=1}^6 O_j \log\!\left(\frac{O_j}{E_j}\right) Note carefully that the observed counts OjO_j appear in front of the logarithm — this comes directly from the log-likelihood of the multinomial, where each cell contributes Ojlog⁡pjO_j \log p_j. The ratio inside the log compares the saturated model's fitted probabilities (Oj/nO_j/n) to the null model's fitted probabilities (Ej/nE_j/n). For degrees of freedom: with 6 cells, the saturated model has 5 free parameters (since probabilities sum to 1). The null model estimates one scalar parameter θ\theta, leaving only 4 free parameters after fitting. The asymptotic χ2\chi^2 degrees of freedom equal the difference: 5−1=45 - 1 = 4. So the correct answer is D. A is wrong because it swaps OjO_j and EjE_j in both the coefficient and the log argument — that expression has no standard likelihood interpretation. B has the correct formula for G2G^2 but uses χ52\chi^2_5, ignoring that one parameter was estimated from the data. C is Pearson's chi-squared statistic, not the likelihood ratio test, and furthermore uses χ42\chi^2_4 — mixing the wrong statistic with the right degrees of freedom. A useful rule of thumb: degrees of freedom = (cells −1- 1) −- (parameters estimated). Never forget to subtract estimated parameters, or you'll be tempted by the χ52\chi^2_5 trap in B.

Question 2

A regular parametric model has parameter θ=(θ1,θ2,θ3,θ4)\theta=(\theta_1,\theta_2,\theta_3,\theta_4). The null hypothesis is specified by the three equations θ1+θ2=0\theta_1+\theta_2=0, 2θ1+2θ2=02\theta_1+2\theta_2=0, and θ3θ4=1\theta_3\theta_4=1. At the true null value, θ3\theta_3 and θ4\theta_4 are nonzero, and all other regularity conditions for Wilks' theorem hold.

What is the asymptotic null distribution of the likelihood ratio statistic?

  1. It is χ12\chi^2_1 because only the nonlinear restriction contributes locally.
  2. It is χ22\chi^2_2 because the restriction Jacobian has rank 22. (correct answer)
  3. It is χ32\chi^2_3 because the null is written using three equations.
  4. Wilks' theorem is inapplicable because one restriction is nonlinear.
Explanation: Whenever you see a Wilks' theorem question involving multiple constraints, your first job is to determine the effective degrees of freedom — not by counting equations, but by computing the rank of the constraint Jacobian evaluated at the null. Wilks' theorem states that the likelihood ratio statistic is asymptotically χr2\chi^2_r, where rr is the rank of the matrix of partial derivatives of the constraints with respect to θ\theta, evaluated at the true null value. Here the three constraints are g1=θ1+θ2g_1 = \theta_1+\theta_2, g2=2θ1+2θ2g_2 = 2\theta_1+2\theta_2, and g3=θ3θ4−1g_3 = \theta_3\theta_4-1. The Jacobian is: Row 2 is exactly twice row 1, so rows 1 and 2 are linearly dependent — they encode the same restriction. Row 3 is independent (since θ3,θ4≠0\theta_3, \theta_4 \neq 0). The rank is therefore 2, giving an asymptotic χ22\chi^2_2 distribution, confirming answer B. Choice C is the classic trap: naively counting three written equations. Redundant constraints don't add degrees of freedom. Choice A is wrong because the linear restriction θ1+θ2=0\theta_1+\theta_2=0 does contribute one independent restriction — it's not rendered irrelevant by the nonlinear one. Choice D is wrong because Wilks' theorem extends to smooth nonlinear constraints, provided the Jacobian has full rank locally and θ3,θ4≠0\theta_3, \theta_4 \neq 0 ensures smoothness here. The key study habit: always compute the Jacobian rank, not the equation count. Redundancy collapses degrees of freedom silently.

Question 3

In a random-intercept model, an investigator tests whether the random-intercept variance is zero. Write the variance component as τ2\tau^2, so the hypotheses are H0:τ2=0H_0:\tau^2=0 and H1:τ2>0H_1:\tau^2>0. Assume the remaining parameters are regular and that the standard one-boundary-parameter asymptotic result applies.

Which null distribution and approximate level-0.050.05 critical value should be used for the likelihood ratio statistic?

  1. Use χ12\chi^2_1 with critical value approximately 3.843.84.
  2. Use 12χ02+12χ12\tfrac12\chi^2_0+\tfrac12\chi^2_1 with critical value approximately 2.712.71. (correct answer)
  3. Use χ22\chi^2_2 with critical value approximately 5.995.99.
  4. Use 12χ12+12χ22\tfrac12\chi^2_1+\tfrac12\chi^2_2 with critical value approximately 4.234.23.
Explanation: Whenever you test a variance component like τ2≥0\tau^2 \geq 0, the null hypothesis places the parameter on the boundary of its admissible space (variance cannot be negative). Standard likelihood ratio theory assumes the null value lies in the interior of the parameter space, so the usual χ12\chi^2_1 result breaks down here. The correct framework comes from the one-boundary-parameter result: when the true parameter is on the boundary, the likelihood ratio statistic under H0H_0 converges in distribution to an equal mixture of a point mass at zero and a χ12\chi^2_1, written as 12χ02+12χ12\tfrac{1}{2}\chi^2_0 + \tfrac{1}{2}\chi^2_1. Intuitively, in half of large samples the restricted and unrestricted MLEs coincide (giving LR = 0), and in the other half the unrestricted estimate moves into the interior (giving a χ12\chi^2_1-like contribution). For a level-0.050.05 test, you need the 95th percentile of this mixture. Since P(χ12>c)=0.05P(\chi^2_1 > c) = 0.05 gives c≈3.84c \approx 3.84, the mixture satisfies P(LR>c)=12P(χ12>c)=0.05P(\text{LR} > c) = \tfrac{1}{2}P(\chi^2_1 > c) = 0.05 when c≈2.71c \approx 2.71 (the 90th percentile of χ12\chi^2_1). This confirms answer B. Choice A uses χ12\chi^2_1 with critical value 3.843.84, ignoring the boundary constraint — this is the most common trap and yields a conservative test. Choice C uses χ22\chi^2_2, which would apply if two interior parameters were being tested simultaneously — incorrect here. Choice D uses 12χ12+12χ22\tfrac{1}{2}\chi^2_1 + \tfrac{1}{2}\chi^2_2, which arises when testing two boundary parameters, not one. As a study tip: anytime you test H0:σ2=0H_0: \sigma^2 = 0 or any variance component, immediately flag it as a boundary-parameter problem and shift from χk2\chi^2_k to the 12χk−12+12χk2\tfrac{1}{2}\chi^2_{k-1} + \tfrac{1}{2}\chi^2_k mixture distribution.

Question 4

A researcher fits a single-component Gaussian model and a two-component Gaussian mixture model with unknown component means, variances, and mixing proportion. The researcher proposes testing one component against two components by comparing twice the maximized log-likelihood difference with a chi-square distribution whose degrees of freedom equal the nominal difference in parameter counts.

Which assessment of the proposed calibration is most accurate?

  1. It is valid because both models are fitted by maximum likelihood and one is nested in the other.
  2. It is valid after subtracting one degree of freedom for the mixing probabilities summing to one.
  3. It is generally invalid because mixture parameters are unidentified under the null and regular Wilks conditions fail. (correct answer)
  4. It is generally invalid only when the two estimated component means happen to be numerically equal.
Explanation: Whenever you see a likelihood ratio test (LRT) applied to mixture models, your first instinct should be to check whether Wilks' theorem conditions are satisfied — because mixture models are a classic setting where they silently fail. Wilks' theorem guarantees that 2(ℓ1−ℓ0)→dχk22(\ell_1 - \ell_0) \xrightarrow{d} \chi^2_k under the null when the null hypothesis corresponds to an interior point of the parameter space and the model is identifiable under the null. In a one-vs-two-component Gaussian test, both conditions collapse simultaneously. Under H0H_0 (one component), the mixing proportion π\pi is unidentified — you can set π=0\pi = 0 or π=1\pi = 1, or make the two means equal, all achieving the same likelihood. This means the null lies on the boundary of the parameter space, and multiple parameter configurations are observationally equivalent. The regularity conditions Wilks requires simply do not hold, so the χ2\chi^2 reference distribution is incorrect, and the test is miscalibrated in general. C is correct for exactly this reason: the unidentifiability of mixture parameters under the null is a fundamental violation, not a fixable technicality. A is wrong because nesting alone is insufficient — Wilks' theorem requires regularity conditions beyond nesting, which fail here. B is wrong because adjusting degrees of freedom is a red herring; the problem isn't parameter counting, it's that the asymptotic distribution is not chi-square at all. D is wrong because the invalidity is structural and general, not confined to the edge case where estimated means coincide numerically. As a study strategy: memorize that mixture model order testing is a canonical counterexample to naive LRT application — expect exam questions to probe whether you know the chi-square calibration fails and why.

Question 5

Independent observations are sampled from a N(μ,σ2)N(\mu,\sigma^2) distribution, with both parameters initially unknown. For a sample of size 2020, the summaries are ∑i=120(Xi−Xˉ)2=40\sum_{i=1}^{20}(X_i-\bar X)^2=40 and ∑i=120Xi2=50\sum_{i=1}^{20}X_i^2=50. Consider the likelihood ratio test of H0:μ=0H_0:\mu=0 against H1:μ≠0H_1:\mu\ne0.

Using Wilks' theorem, which conclusion is correct at significance level 0.050.05?

  1. The statistic is about 8.938.93; compare it with χ12\chi^2_1 and reject H0H_0.
  2. The statistic is about 4.464.46; compare it with χ22\chi^2_2 and do not reject H0H_0.
  3. The statistic is about 4.464.46; compare it with χ12\chi^2_1 and reject H0H_0. (correct answer)
  4. The statistic is about −4.46-4.46; compare it with χ12\chi^2_1 and do not reject H0H_0.
Explanation: When you see a likelihood ratio test (LRT) for a normal distribution with unknown parameters, your first instinct should be to identify the degrees of freedom carefully — this is where most errors occur. Under H0:μ=0H_0: \mu = 0, you estimate only σ2\sigma^2 (1 free parameter). Under H1H_1, you estimate both μ\mu and σ2\sigma^2 (2 free parameters). By Wilks' theorem, the LRT statistic −2log⁡Λ-2\log\Lambda follows a χ2\chi^2 distribution with degrees of freedom equal to the difference in parameters: 2−1=12 - 1 = 1. Now compute the MLEs. Under H1H_1: μ^=Xˉ\hat\mu = \bar X and σ^12=1n∑(Xi−Xˉ)2=4020=2\hat\sigma^2_1 = \frac{1}{n}\sum(X_i - \bar X)^2 = \frac{40}{20} = 2. Under H0H_0 (μ=0\mu = 0): σ^02=1n∑Xi2=5020=2.5\hat\sigma^2_0 = \frac{1}{n}\sum X_i^2 = \frac{50}{20} = 2.5. The LRT statistic is: −2log⁡Λ=nlog⁡ ⁣(σ^02σ^12)=20log⁡ ⁣(2.52)≈20(0.2231)≈4.46-2\log\Lambda = n\log\!\left(\frac{\hat\sigma^2_0}{\hat\sigma^2_1}\right) = 20\log\!\left(\frac{2.5}{2}\right) \approx 20(0.2231) \approx 4.46 This exceeds the χ12\chi^2_1 critical value of 3.8413.841 at α=0.05\alpha = 0.05, so you reject H0H_0 — confirming C is correct. Choice A doubles the statistic to 8.938.93, which has no valid derivation here. Choice B gets the statistic right but uses χ22\chi^2_2, incorrectly counting the total parameters under H1H_1 rather than the difference in parameter counts. Choice D produces a negative statistic, which is impossible since −2log⁡Λ≥0-2\log\Lambda \geq 0 always. Study tip: Always remember that the degrees of freedom for Wilks' theorem equals the difference in the number of free parameters between the two hypotheses — not the total number estimated under either model alone.

Question 6

A Gaussian linear regression is fitted by maximum likelihood. The unrestricted model contains an intercept, two slope coefficients, and an unknown error variance. Under the null hypothesis, both slopes are zero, while the intercept and variance remain unrestricted. The maximized log-likelihoods are −121.7-121.7 under the unrestricted model and −125.4-125.4 under the null model.

Which likelihood ratio statistic and Wilks reference distribution are appropriate?

  1. The statistic is 7.47.4 and is compared with a χ22\chi^2_2 distribution. (correct answer)
  2. The statistic is 3.73.7 and is compared with a χ22\chi^2_2 distribution.
  3. The statistic is 7.47.4 and is compared with a χ42\chi^2_4 distribution.
  4. The statistic is −7.4-7.4 and is compared with a χ22\chi^2_2 distribution.
Explanation: Whenever you see a likelihood ratio test question, anchor yourself to two things: the formula for the test statistic and the degrees of freedom, which equals the number of restrictions imposed by the null hypothesis. The likelihood ratio (LR) statistic is defined as Λ=−2(ℓ0−ℓ1)\Lambda = -2(\ell_0 - \ell_1), where ℓ0\ell_0 is the maximized log-likelihood under the null and ℓ1\ell_1 is the maximized log-likelihood under the unrestricted model. Here, ℓ1=−121.7\ell_1 = -121.7 and ℓ0=−125.4\ell_0 = -125.4, so: Λ=−2(−125.4−(−121.7))=−2(−3.7)=7.4\Lambda = -2(-125.4 - (-121.7)) = -2(-3.7) = 7.4 By Wilks' theorem, this statistic follows a χ2\chi^2 distribution with degrees of freedom equal to the number of restrictions. The null hypothesis sets both slopes to zero — that's exactly 2 restrictions. The intercept and variance are free under both models, so they don't contribute. Therefore, the reference distribution is χ22\chi^2_2, confirming answer A is correct. Answer B uses 3.73.7, which is the raw log-likelihood difference ℓ1−ℓ0\ell_1 - \ell_0 — it forgets the factor of −2-2 that scales the statistic to match the chi-squared distribution. Answer C correctly computes 7.47.4 but uses χ42\chi^2_4, mistakenly counting all four parameters in the unrestricted model rather than only the two parameters being constrained. Answer D uses a negative value, which is impossible — because ℓ1≥ℓ0\ell_1 \geq \ell_0 always (the unrestricted model can only do at least as well), the statistic is always non-negative. Your memory anchor: LR statistic = −2×Δℓ-2 \times \Delta\ell, and degrees of freedom = number of constraints, not total parameters.

Question 7

Two nested logistic regression models are fitted to the same binary-response data. The reduced model has residual deviance 132.6132.6. The full model adds three slope coefficients and has residual deviance 124.1124.1. Assume the models are regular and the sample size is sufficiently large.

What is the likelihood ratio test result for the three added predictors?

  1. The statistic is 8.58.5 on 33 degrees of freedom, giving a p-value between 0.0250.025 and 0.050.05. (correct answer)
  2. The statistic is 8.58.5 on 11 degree of freedom, giving a p-value below 0.010.01.
  3. The statistic is 4.254.25 on 33 degrees of freedom, giving a p-value above 0.100.10.
  4. The statistic is 256.7256.7 on 33 degrees of freedom, giving a p-value effectively equal to zero.
Explanation: When comparing nested generalized linear models, your go-to tool is the likelihood ratio test (LRT). The key insight: deviance is defined as −2log⁡L-2\log\mathcal{L}, so the difference in deviances between a reduced and full model directly gives the LRT statistic, which follows a χ2\chi^2 distribution under the null hypothesis that the added predictors have zero effect. Here, the test statistic is simply 132.6−124.1=8.5132.6 - 124.1 = 8.5. The degrees of freedom equal the number of additional parameters in the full model, which is 3 (one for each added slope coefficient). Consulting a χ32\chi^2_3 table, the critical values are approximately 9.359.35 (p=0.025p = 0.025) and 7.817.81 (p=0.05p = 0.05). Since 8.58.5 falls between these, the p-value lies between 0.0250.025 and 0.050.05, confirming answer A is correct. Answer B misidentifies the degrees of freedom as 1 instead of 3. This would apply only if a single parameter were added — confusing the number of constraints with something else. Answer C halves the statistic to 4.254.25, which would only make sense if you were converting to an FF-statistic or mistakenly divided by something; on 3 df, 4.254.25 yields p>0.10p > 0.10, a completely different conclusion. Answer D appears to confuse the LRT statistic with an exponentiated likelihood ratio (i.e., computing e8.5/2e^{8.5/2} or similar), producing a nonsensical value of 256.7256.7. Study tip: Always remember the LRT statistic = difference in deviances, and df = difference in number of parameters. Memorize a few χ2\chi^2 critical values (3 df: 7.817.81 at 0.050.05, 11.3411.34 at 0.010.01) to quickly bracket p-values on exams.

Question 8

In a regular parametric model, a likelihood ratio test imposes rr smooth, locally independent restrictions. Consider alternatives that approach the null at the rate n−1/2n^{-1/2}, as well as fixed alternatives that remain separated from the null as the sample size grows.

Which statement best describes the large-sample behavior of the likelihood ratio statistic?

  1. Under local alternatives it remains central χr2\chi^2_r, while under fixed alternatives it has a bounded noncentral distribution.
  2. Under local alternatives it converges to a noncentral χr2\chi^2_r, while under fixed alternatives it typically diverges. (correct answer)
  3. Under local alternatives it converges to a normal distribution, while under fixed alternatives it becomes central χr2\chi^2_r.
  4. Under local alternatives it has degrees of freedom equal to the full parameter dimension, while under fixed alternatives it converges to zero.
Explanation: Whenever you encounter questions about the likelihood ratio test (LRT), anchor your thinking to two regimes: local alternatives (shrinking toward the null at rate n−1/2n^{-1/2}) and fixed alternatives (bounded away from the null regardless of nn). Under the null, the classical result gives 2log⁡Λn→dχr22\log\Lambda_n \xrightarrow{d} \chi^2_r. The deeper insight is what happens when you shift away. Under local alternatives of the form θn=θ0+δ/n\theta_n = \theta_0 + \delta/\sqrt{n}, a Le Cam-style contiguity argument shows the LRT statistic converges in distribution to a noncentral χr2(λ)\chi^2_r(\lambda), where the noncentrality parameter λ\lambda depends on δ\delta and the Fisher information. The statistic stays stochastically bounded but shifts its mass rightward. Under fixed alternatives, the log-likelihood ratio grows at rate nn (each observation contributes a fixed positive expected log-likelihood difference), so the statistic diverges to infinity — which is exactly why the LRT is consistent against fixed alternatives. This confirms B as correct. A is wrong on both counts: local alternatives do not preserve the central χr2\chi^2_r, and fixed alternatives produce divergence, not a bounded noncentral distribution. C is wrong because local alternatives produce a noncentral χr2\chi^2_r, not a normal distribution (the normality applies to the score statistic, not the LRT itself). D is wrong because degrees of freedom equal rr (the number of restrictions), not the full parameter dimension, and the statistic diverges rather than converges to zero under fixed alternatives. Study tip: Pair each asymptotic regime with its outcome — null →\to central χr2\chi^2_r, local alternative →\to noncentral χr2\chi^2_r, fixed alternative →\to divergence. This three-way mapping appears repeatedly in large-sample theory.

Question 9

A model has parameter of interest ψ=(ψ1,ψ2)\psi=(\psi_1,\psi_2) and a three-dimensional nuisance parameter λ\lambda. For each proposed value of ψ\psi, the nuisance parameter is maximized out, producing the profile log-likelihood ℓp(ψ)\ell_p(\psi). All regularity conditions for Wilks' theorem hold.

Which asymptotic 100(1−α)%100(1-\alpha)\% likelihood ratio confidence region for ψ\psi is appropriate?

  1. Include values satisfying 2[ℓp(ψ^)−ℓp(ψ)]≤χ2,1−α22[\ell_p(\widehat\psi)-\ell_p(\psi)]\le\chi^2_{2,1-\alpha}. (correct answer)
  2. Include values satisfying 2[ℓp(ψ^)−ℓp(ψ)]≤χ3,1−α22[\ell_p(\widehat\psi)-\ell_p(\psi)]\le\chi^2_{3,1-\alpha}.
  3. Include values satisfying 2[ℓp(ψ^)−ℓp(ψ)]≤χ5,1−α22[\ell_p(\widehat\psi)-\ell_p(\psi)]\le\chi^2_{5,1-\alpha}.
  4. Include values satisfying ℓp(ψ^)−ℓp(ψ)≤χ1,1−α2\ell_p(\widehat\psi)-\ell_p(\psi)\le\chi^2_{1,1-\alpha}.
Explanation: Whenever you see a likelihood ratio confidence region, the key question is: how many parameters are you making inference about? Wilks' theorem tells us that twice the log-likelihood ratio statistic converges in distribution to a chi-squared random variable whose degrees of freedom equal the dimension of the parameter of interest — not the nuisance parameter. Here, ψ=(ψ1,ψ2)\psi = (\psi_1, \psi_2) is two-dimensional, so the profile likelihood ratio statistic 2[ℓp(ψ^)−ℓp(ψ)]2[\ell_p(\hat{\psi}) - \ell_p(\psi)] is asymptotically χ22\chi^2_2 under the true ψ\psi. The correct confidence region therefore includes all ψ\psi satisfying 2[ℓp(ψ^)−ℓp(ψ)]≤χ2,1−α22[\ell_p(\hat{\psi}) - \ell_p(\psi)] \le \chi^2_{2,1-\alpha}, which is answer A. Answer B uses χ32\chi^2_3, which corresponds to the dimension of the nuisance parameter λ\lambda. This is a classic trap: the nuisance parameter is already profiled out and plays no role in the degrees of freedom of the resulting pivot. Answer C uses χ52\chi^2_5, apparently adding the dimensions of ψ\psi and λ\lambda together (2 + 3 = 5). This double-counts the nuisance parameter entirely — once it is maximized out to form the profile likelihood, it no longer contributes degrees of freedom. Answer D divides by 2, using ℓp(ψ^)−ℓp(ψ)≤χ1,1−α2\ell_p(\hat{\psi}) - \ell_p(\psi) \le \chi^2_{1,1-\alpha} without the factor of 2. The factor of 2 is essential to the Wilks result, and using χ12\chi^2_1 ignores the two-dimensional nature of ψ\psi entirely. Study tip: Always anchor degrees of freedom to the dimension of the parameter of interest. Nuisance parameters that are profiled out disappear from the chi-squared degrees of freedom entirely.

Question 10

For a regular model with scale parameter σ>0\sigma>0, one analyst tests H0:σ=1H_0:\sigma=1 using the parameterization σ\sigma. A second analyst uses the smooth one-to-one parameterization ϕ=log⁡σ\phi=\log\sigma and tests the equivalent null hypothesis H0:ϕ=0H_0:\phi=0. Both maximize the likelihood correctly under the null and alternative.

How should their likelihood ratio tests compare?

  1. They agree only asymptotically because finite-sample likelihood ratios depend on parameterization.
  2. They have different statistics, but both have the same asymptotic χ12\chi^2_1 calibration.
  3. They have identical statistics, but the logarithmic parameterization requires a χ22\chi^2_2 calibration.
  4. They have identical likelihood ratio statistics and the same asymptotic χ12\chi^2_1 calibration. (correct answer)
Explanation: When you see a question about likelihood ratio tests under different parameterizations, the key insight to reach for is invariance: the likelihood function itself is defined in terms of the data and the model, not the labeling of parameters. The likelihood ratio statistic is Λ=2[ℓ(θ^)−ℓ(θ^0)]\Lambda = 2[\ell(\hat{\theta}) - \ell(\hat{\theta}_0)], where you subtract the maximized log-likelihood under H0H_0 from the unrestricted maximum. Notice that this quantity depends entirely on likelihood values at specific parameter points — not on how you name those points. If ϕ=log⁡σ\phi = \log\sigma, then σ=1\sigma = 1 and ϕ=0\phi = 0 describe the exact same null constraint. The MLE under the alternative satisfies ϕ^=log⁡σ^\hat{\phi} = \log\hat{\sigma}, so both analysts evaluate the likelihood at the same model configurations. The numerical value of Λ\Lambda is therefore identical — not approximately, but exactly. Furthermore, both are testing a single scalar constraint in a regular model, so by Wilks' theorem both statistics follow an asymptotic χ12\chi^2_1 distribution. This confirms D is correct. A is wrong because it introduces a false finite-sample dependence on parameterization. The LRT statistic is algebraically identical regardless of parameterization at every sample size. B is wrong for the same reason — claiming the statistics differ contradicts the invariance property. C correctly notes the statistics are identical but then incorrectly inflates the degrees of freedom to χ22\chi^2_2; the calibration depends on the number of constrained dimensions (one here), not on any feature of the reparameterization. As a general study tip: the LRT is parameterization-invariant by construction — anytime a question hints otherwise, that's the distractor. Score degrees of freedom by counting independent constraints, not by counting transformations applied.