Statistics Graduate Level Quiz: Likelihood Based Cis
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Likelihood Based CisQuestion 1 of 10

Independent observations follow a normal distribution with unknown mean μ\mu and unknown variance σ2\sigma^2. For a sample of size n=10n=10, the sample mean is xˉ=5\bar{x}=5 and ∑i=110(xi−xˉ)2=18.\sum_{i=1}^{10}(x_i-\bar{x})^2=18. A 95% likelihood-ratio interval for μ\mu is formed using χ1,0.952=3.84\chi^2_{1,0.95}=3.84.

Which interval results from profiling out σ2\sigma^2?

[4.08, 5.92][4.08,\ 5.92]
[4.17, 5.83][4.17,\ 5.83]
[3.99, 6.01][3.99,\ 6.01]
[4.38, 5.62][4.38,\ 5.62]
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Likelihood Based Cis

Practice Likelihood Based Cis in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Likelihood Based Cis, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Independent observations follow a normal distribution with unknown mean μ\mu and unknown variance σ2\sigma^2. For a sample of size n=10n=10, the sample mean is xˉ=5\bar{x}=5 and ∑i=110(xi−xˉ)2=18.\sum_{i=1}^{10}(x_i-\bar{x})^2=18. A 95% likelihood-ratio interval for μ\mu is formed using χ1,0.952=3.84\chi^2_{1,0.95}=3.84.

Which interval results from profiling out σ2\sigma^2?

  1. [4.08, 5.92][4.08,\ 5.92] (correct answer)
  2. [4.17, 5.83][4.17,\ 5.83]
  3. [3.99, 6.01][3.99,\ 6.01]
  4. [4.38, 5.62][4.38,\ 5.62]
Explanation: When you see a likelihood-ratio (LR) interval formed by profiling out a nuisance parameter, your goal is to find all values of the parameter of interest where the profile log-likelihood ratio statistic stays within the critical threshold. Here, the full log-likelihood for (μ,σ2)(\mu, \sigma^2) under normality yields a profile log-likelihood for μ\mu alone by plugging in the MLE of σ2\sigma^2 at each fixed μ\mu. That profile MLE is σ^2(μ)=1n[∑(xi−xˉ)2+n(xˉ−μ)2]\hat{\sigma}^2(\mu) = \frac{1}{n}\left[\sum(x_i - \bar{x})^2 + n(\bar{x}-\mu)^2\right]. The profile LR statistic simplifies to: −2log⁡R(μ)=nlog⁡(1+n(xˉ−μ)2∑(xi−xˉ)2)≤3.84-2\log R(\mu) = n\log\left(1 + \frac{n(\bar{x}-\mu)^2}{\sum(x_i-\bar{x})^2}\right) \leq 3.84 With n=10n=10, xˉ=5\bar{x}=5, and ∑(xi−xˉ)2=18\sum(x_i-\bar{x})^2=18, set the statistic equal to 3.84: 10log⁡(1+10(μ−5)218)=3.8410\log\left(1 + \frac{10(\mu-5)^2}{18}\right) = 3.84 1+10(μ−5)218=e0.384≈1.46851 + \frac{10(\mu-5)^2}{18} = e^{0.384} \approx 1.4685 10(μ−5)218=0.4685  ⟹  (μ−5)2=0.8433  ⟹  ∣μ−5∣≈0.918\frac{10(\mu-5)^2}{18} = 0.4685 \implies (\mu-5)^2 = 0.8433 \implies |\mu-5| \approx 0.918 This gives the interval [4.08, 5.92][4.08,\ 5.92], confirming A. Choice B ([4.17,5.83][4.17, 5.83]) results from incorrectly using a tt-based pivot without proper LR construction. Choice C ([3.99,6.01][3.99, 6.01]) over-widens the interval, likely from misapplying a chi-squared quantile directly to (xˉ−μ)2(\bar{x}-\mu)^2. Choice D ([4.38,5.62][4.38, 5.62]) is too narrow, corresponding to an error where σ2\sigma^2 is treated as known or the sample size is mishandled. Your study tip: always remember that profiling replaces σ2\sigma^2 with its conditional MLE at each μ\mu, converting the LR statistic into a log-ratio of observed variances — the algebra collapses neatly into the logarithmic form above.

Question 2

A regression model has a two-dimensional parameter of interest β=(β1,β2)\boldsymbol{\beta}=(\beta_1,\beta_2) and three nuisance parameters collected in γ\boldsymbol{\gamma}. All parameters are interior, identifiable, and subject to the usual regularity conditions. Let ℓp(β)=sup⁡γℓ(β,γ)\ell_p(\boldsymbol{\beta})=\sup_{\boldsymbol{\gamma}}\ell(\boldsymbol{\beta},\boldsymbol{\gamma}).

Which construction gives an asymptotic 95% joint profile-likelihood confidence region for β\boldsymbol{\beta}?

  1. 2{ℓ(β^,γ^)−ℓp(β)}≤χ2,0.9522\{\ell(\widehat{\boldsymbol{\beta}},\widehat{\boldsymbol{\gamma}})-\ell_p(\boldsymbol{\beta})\}\leq\chi^2_{2,0.95} (correct answer)
  2. 2{ℓ(β^,γ^)−ℓp(β)}≤χ5,0.9522\{\ell(\widehat{\boldsymbol{\beta}},\widehat{\boldsymbol{\gamma}})-\ell_p(\boldsymbol{\beta})\}\leq\chi^2_{5,0.95}
  3. 2{ℓ(β^,γ^)−ℓ(β,γ^)}≤χ2,0.9522\{\ell(\widehat{\boldsymbol{\beta}},\widehat{\boldsymbol{\gamma}})-\ell(\boldsymbol{\beta},\widehat{\boldsymbol{\gamma}})\}\leq\chi^2_{2,0.95}
  4. 2{ℓ(β^,γ^)−ℓp(β)}≤χ1,0.9522\{\ell(\widehat{\boldsymbol{\beta}},\widehat{\boldsymbol{\gamma}})-\ell_p(\boldsymbol{\beta})\}\leq\chi^2_{1,0.95}
Explanation: Whenever you see a question about profile likelihood regions, anchor yourself to Wilks' theorem: the likelihood ratio statistic 2{ℓ(β^,γ^)−ℓ(β,γ)}2\{\ell(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\gamma}}) - \ell(\boldsymbol{\beta}, \boldsymbol{\gamma})\} converges in distribution to a χ2\chi^2 whose degrees of freedom equal the dimension of the parameter being tested, not the total parameter count. The profile likelihood ℓp(β)=sup⁡γℓ(β,γ)\ell_p(\boldsymbol{\beta}) = \sup_{\boldsymbol{\gamma}} \ell(\boldsymbol{\beta}, \boldsymbol{\gamma}) plugs in the nuisance MLE at each fixed β\boldsymbol{\beta}, so the statistic 2{ℓ(β^,γ^)−ℓp(β)}2\{\ell(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\gamma}}) - \ell_p(\boldsymbol{\beta})\} is a proper profile likelihood ratio. Because β\boldsymbol{\beta} is two-dimensional, Wilks' theorem gives asymptotic distribution χ22\chi^2_2, and the 95% region is exactly 2{ℓ(β^,γ^)−ℓp(β)}≤χ2,0.9522\{\ell(\hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\gamma}}) - \ell_p(\boldsymbol{\beta})\} \leq \chi^2_{2, 0.95}. That is answer A. B uses χ52\chi^2_5, which would correspond to testing all five parameters simultaneously — a miscount that ignores how profiling eliminates the nuisance dimensions from the asymptotic distribution. C substitutes ℓ(β,γ^)\ell(\boldsymbol{\beta}, \hat{\boldsymbol{\gamma}}) instead of ℓp(β)\ell_p(\boldsymbol{\beta}); this fixes the nuisance estimates at their overall MLE values rather than re-optimizing them at each β\boldsymbol{\beta}, so it is not a profile likelihood ratio and loses its clean χ22\chi^2_2 limit. D uses χ12\chi^2_1, appropriate only for a scalar parameter of interest. The key study tip: degrees of freedom in a likelihood ratio test always count the dimension of the subspace you're testing, and profiling over nuisance parameters does not add degrees of freedom — it removes those dimensions from the problem.

Question 3

For a scalar parameter ψ\psi, numerical maximization over nuisance parameters produces a nonconcave profile log-likelihood. Using the selected likelihood-ratio cutoff, the resulting confidence set is exactly [−2,−1]∪[0.5,1.5].[-2,-1]\cup[0.5,1.5]. The global maximum occurs at ψ^=1\widehat\psi=1, while the negative interval surrounds a lower local maximum that also remains above the cutoff.

How should this likelihood-ratio confidence set ordinarily be reported?

  1. Report [0.5,1.5][0.5,1.5], retaining only the global-mode component.
  2. Report [−2,1.5][-2,1.5], filling the gap to create one interval.
  3. Report [−2,−1]∪[0.5,1.5][-2,-1]\cup[0.5,1.5] as a disconnected confidence set. (correct answer)
  4. Report [−2,−1][-2,-1], retaining only the secondary-mode component.
Explanation: When a profile log-likelihood is nonconcave, the likelihood-ratio confidence set is constructed mechanically: collect all parameter values whose profile log-likelihood falls within the specified cutoff of the global maximum. If that set happens to be disconnected, you report it as-is — the math dictates the shape, not your aesthetic preference. Here, the cutoff captures two disjoint intervals, [−2,−1][-2,-1] and [0.5,1.5][0.5,1.5], because both regions satisfy the likelihood-ratio criterion. The correct answer is C: report [−2,−1]∪[0.5,1.5][-2,-1]\cup[0.5,1.5] as a disconnected confidence set. This is statistically honest — it communicates that the data are consistent with two separate regions of plausible parameter values, a meaningful inferential message about the likelihood surface. A is wrong because discarding [−2,−1][-2,-1] arbitrarily ignores values that legitimately satisfy the confidence criterion. The global mode being at ψ^=1\widehat\psi=1 doesn't disqualify the secondary region if it clears the cutoff — the rule is about the threshold, not uniqueness of the mode. B is wrong because filling the gap with [−2,1.5][-2,1.5] includes values in (−1,0.5)(−1,0.5) that fail the likelihood-ratio criterion. Merging intervals to appear "cleaner" inflates the confidence set and misrepresents the likelihood surface — this is a conservative distortion, not a valid simplification. D is wrong for the mirror reason as A: keeping only the secondary interval while dropping the global-mode interval [0.5,1.5][0.5,1.5] is indefensible. Study tip: On any question involving nonconcave likelihoods, remember that the confidence set is defined by a threshold condition applied globally — shape follows function. Never prune or pad a valid confidence set for cosmetic reasons.

Question 4

Suppose XX and YY are independent, with X∼Poisson⁡(λ1),Y∼Poisson⁡(λ2).X\sim\operatorname{Poisson}(\lambda_1),\qquad Y\sim\operatorname{Poisson}(\lambda_2). The parameter of interest is the rate ratio ψ=λ1/λ2\psi=\lambda_1/\lambda_2, and the observed counts are x=12x=12 and y=8y=8.

After expressing λ1=ψλ2\lambda_1=\psi\lambda_2 and profiling over λ2\lambda_2, which expression is the profile log-likelihood for ψ\psi, up to an additive constant independent of ψ\psi?

  1. ℓp(ψ)=12log⁡ψ−20log⁡(1+ψ)\ell_p(\psi)=12\log\psi-20\log(1+\psi) (correct answer)
  2. ℓp(ψ)=8log⁡ψ−20log⁡(1+ψ)\ell_p(\psi)=8\log\psi-20\log(1+\psi)
  3. ℓp(ψ)=12log⁡ψ−12log⁡(1+ψ)\ell_p(\psi)=12\log\psi-12\log(1+\psi)
  4. ℓp(ψ)=20log⁡ψ−20log⁡(1+ψ)\ell_p(\psi)=20\log\psi-20\log(1+\psi)
Explanation: When you encounter a profile likelihood problem, your goal is to eliminate nuisance parameters by substituting their MLEs (expressed as functions of the parameter of interest) back into the full log-likelihood. Start with the joint log-likelihood. Since X∼Poisson(λ1)X \sim \text{Poisson}(\lambda_1) and Y∼Poisson(λ2)Y \sim \text{Poisson}(\lambda_2) independently, the full log-likelihood (dropping constants not involving parameters) is ℓ(λ1,λ2)=xlog⁡λ1−λ1+ylog⁡λ2−λ2.\ell(\lambda_1, \lambda_2) = x\log\lambda_1 - \lambda_1 + y\log\lambda_2 - \lambda_2. Substituting λ1=ψλ2\lambda_1 = \psi\lambda_2 gives ℓ(ψ,λ2)=xlog⁡(ψλ2)−ψλ2+ylog⁡λ2−λ2=xlog⁡ψ+(x+y)log⁡λ2−(1+ψ)λ2.\ell(\psi, \lambda_2) = x\log(\psi\lambda_2) - \psi\lambda_2 + y\log\lambda_2 - \lambda_2 = x\log\psi + (x+y)\log\lambda_2 - (1+\psi)\lambda_2. To profile, maximize over λ2\lambda_2: setting the derivative to zero yields λ^2(ψ)=x+y1+ψ.\hat{\lambda}_2(\psi) = \frac{x+y}{1+\psi}. Substituting back and using x+y=20x+y = 20, x=12x = 12: ℓp(ψ)=12log⁡ψ+20log⁡ ⁣(201+ψ)−20=12log⁡ψ−20log⁡(1+ψ)+const,\ell_p(\psi) = 12\log\psi + 20\log\!\left(\frac{20}{1+\psi}\right) - 20 = 12\log\psi - 20\log(1+\psi) + \text{const}, confirming answer A. Answer B uses y=8y = 8 instead of x=12x = 12 in the first term — this would arise from confusing which count multiplies log⁡ψ\log\psi. Answer C correctly identifies the 12log⁡ψ12\log\psi term but uses xx alone (12) instead of x+yx+y (20) in the second term, forgetting that both observations contribute to the total count constraining λ2\lambda_2. Answer D replaces xlog⁡ψx\log\psi with (x+y)log⁡ψ(x+y)\log\psi, incorrectly attributing the full count to ψ\psi rather than recognizing that only xx directly links to ψλ2\psi\lambda_2. A reliable strategy: always track which terms carry ψ\psi versus λ2\lambda_2 before profiling. The coefficient of log⁡ψ\log\psi comes solely from xlog⁡λ1=xlog⁡ψ+xlog⁡λ2x\log\lambda_1 = x\log\psi + x\log\lambda_2, while x+yx + y appears only after maximizing over λ2\lambda_2.

Question 5

In a signal-detection model, the signal amplitude aa is the parameter of interest and the signal frequency ω\omega is a nuisance parameter. Under the null hypothesis a=0a=0, the data distribution does not depend on ω\omega, so ω\omega is not identifiable. A profile likelihood-ratio statistic maximizes over ω\omega under the alternative.

Which approach is most defensible for calibrating a likelihood-based confidence set for aa?

  1. Use a χ12\chi^2_1 cutoff because only aa is the parameter of interest.
  2. Use a χ22\chi^2_2 cutoff because both aa and ω\omega are optimized.
  3. Fix ω\omega at its alternative estimate and use a χ12\chi^2_1 cutoff.
  4. Calibrate the profiled statistic under the null, such as by parametric bootstrap. (correct answer)
Explanation: When a nuisance parameter is unidentifiable under the null hypothesis, standard asymptotic theory for likelihood-ratio statistics breaks down entirely. This is the classic Davies problem: because ω\omega disappears from the likelihood when a=0a=0, the profile likelihood-ratio statistic no longer follows the usual χ2\chi^2 reference distribution, regardless of how many parameters you count. Recognizing this setup should immediately signal that off-the-shelf calibration is unreliable. The most defensible approach is D — calibrating the profiled statistic empirically under the null, for example via parametric bootstrap. You simulate data under a=0a=0, compute the profiled statistic (maximizing over ω\omega) on each simulated dataset, and build the reference distribution directly. This respects the actual null distribution, which can have heavier tails than any χ2\chi^2. Choice A applies standard Wilks' theorem for a one-dimensional parameter of interest, but Wilks' theorem requires the model to be regular at the null — identifiability of all parameters is part of that regularity. It fails here. Choice B tries to account for the extra optimization over ω\omega by adding a degree of freedom, but this is equally unfounded: the χ22\chi^2_2 approximation also relies on regularity conditions that are violated. Adjusting the degrees of freedom does not fix a fundamentally non-standard distribution. Choice C eliminates the optimization by conditioning on the alternative's ω^\hat{\omega}, but this ignores the uncertainty in ω\omega and produces a statistic whose null distribution is still non-standard and poorly characterized. As a study strategy: whenever you see a nuisance parameter that vanishes under H0H_0, immediately set aside Wilks' theorem and think simulation-based calibration.

Question 6

For an interior scalar parameter ψ\psi in a regular model, the signed root profile likelihood-ratio statistic is approximately standard normal. An analyst wants a one-sided 95% upper confidence bound and will determine its endpoint on the upper side of ψ^\widehat\psi.

To which value should the ordinary profile likelihood-ratio statistic 2{ℓp(ψ^)−ℓp(ψ)}2\{\ell_p(\widehat\psi)-\ell_p(\psi)\} be equated at the upper endpoint?

  1. 3.84=(1.960)23.84=(1.960)^2
  2. 2.71=(1.645)22.71=(1.645)^2 (correct answer)
  3. 1.6451.645 without squaring
  4. 5.995.99 from two degrees of freedom
Explanation: Whenever you see a question mixing profile likelihood-ratio statistics with confidence bounds, the key is carefully tracking the relationship between the signed root statistic and its square. The signed root profile likelihood-ratio statistic is rp(ψ)=sgn(ψ^−ψ)2{ℓp(ψ^)−ℓp(ψ)}r_p(\psi) = \text{sgn}(\widehat{\psi}-\psi)\sqrt{2\{\ell_p(\widehat\psi)-\ell_p(\psi)\}}, which is approximately N(0,1)N(0,1). For a one-sided 95% upper bound, you want the value ψU\psi_U such that P(ψ^≤ψU)=0.95P(\widehat\psi \leq \psi_U) = 0.95. On the upper side of ψ^\widehat\psi, the signed root is negative (since ψ^−ψ<0\widehat\psi - \psi < 0), so you set rp(ψU)=−1.645r_p(\psi_U) = -1.645. Squaring both sides gives 2{ℓp(ψ^)−ℓp(ψU)}=(1.645)2=2.712\{\ell_p(\widehat\psi)-\ell_p(\psi_U)\} = (1.645)^2 = 2.71. That confirms B is correct. Choice A is the trap most students fall into. The value 3.84=(1.960)23.84 = (1.960)^2 corresponds to a two-sided 95% interval, where you split the 5% across both tails, each carrying z0.025=1.960z_{0.025}=1.960. For a one-sided bound, you use the full 5% in one tail, giving z0.05=1.645z_{0.05}=1.645, not 1.9601.960. Choice C correctly identifies 1.6451.645 but forgets the squaring step — the profile likelihood-ratio statistic equals rp2r_p^2, not rpr_p itself. Choice D, 5.995.99, comes from a chi-squared distribution with two degrees of freedom, which is irrelevant here since ψ\psi is a scalar (one degree of freedom). A useful pattern: one-sided 95% bounds always square the one-tailed z0.05=1.645z_{0.05}=1.645, giving 2.712.71, while two-sided 95% intervals square z0.025=1.960z_{0.025}=1.960, giving 3.843.84. Memorize this pairing — confusing them is among the most common errors on likelihood-based inference questions.

Question 7

A profile-likelihood analysis gives the 95% confidence interval [2,5][2,5] for a positive scalar parameter ψ\psi, whose maximum-likelihood estimate is 33. A researcher instead wishes to report inference for ϕ=1/ψ\phi=1/\psi.

Without recomputing the likelihood optimization, which interval is the corresponding profile-likelihood interval for ϕ\phi?

  1. [2.00, 5.00][2.00,\ 5.00]
  2. [0.17, 0.50][0.17,\ 0.50]
  3. [0.20, 0.33][0.20,\ 0.33]
  4. [0.20, 0.50][0.20,\ 0.50] (correct answer)
Explanation: Whenever you encounter profile-likelihood inference under a parameter transformation, the key principle is equivariance: the profile-likelihood confidence interval transforms exactly as the parameter does, with no re-optimization needed. Here's why this works. The profile likelihood for ψ\psi defines a set of "plausible" values {ψ:ℓp(ψ)≥ℓp(ψ^)−c}\{\psi : \ell_p(\psi) \geq \ell_p(\hat\psi) - c\}, which produces the interval [2,5][2, 5]. Because ϕ=1/ψ\phi = 1/\psi is a monotone (strictly decreasing) function, the set of plausible ϕ\phi values is simply the image of [2,5][2, 5] under this map. You apply ϕ=1/ψ\phi = 1/\psi to each endpoint: 1/5=0.201/5 = 0.20 and 1/2=0.501/2 = 0.50. Since the transformation is decreasing, the endpoints swap, giving [0.20,0.50][0.20, 0.50]. That's answer D. Now, why are the others wrong? A simply copies [2,5][2, 5] unchanged — a clear error, since ϕ\phi and ψ\psi live on different scales and the transformation was never applied. B applies 1/ψ1/\psi to the MLE (1/3≈0.331/3 \approx 0.33) and seems to confuse endpoint mapping, yielding [0.17,0.50][0.17, 0.50] — the lower bound 0.170.17 doesn't correspond to either endpoint of the original interval. C gives [0.20,0.33][0.20, 0.33], which uses 1/51/5 correctly but replaces the upper bound with the transformed MLE 1/31/3 instead of 1/21/2 — a tempting trap if you conflate the MLE with a confidence boundary. Study tip: For any monotone transformation gg, just apply gg to both endpoints and re-order if necessary. Never substitute the MLE into an interval endpoint — the MLE transforms correctly on its own, but it is not a boundary of the confidence interval.

Question 8

For a scalar parameter of interest ψ\psi and nuisance parameter λ\lambda, the log-likelihood near its maximum is exactly ℓ(ψ,λ)=ℓ(2,1)−12{4(ψ−2)2+4(ψ−2)(λ−1)+2(λ−1)2}.\ell(\psi,\lambda)=\ell(2,1)-\frac{1}{2}\left\{4(\psi-2)^2+4(\psi-2)(\lambda-1)+2(\lambda-1)^2\right\}. A 95% profile-likelihood confidence interval is constructed using the cutoff χ1,0.952=3.84\chi^2_{1,0.95}=3.84.

Which interval is obtained for ψ\psi?

  1. [0.61, 3.39][0.61,\ 3.39] (correct answer)
  2. [1.02, 2.98][1.02,\ 2.98]
  3. [0.04, 3.96][0.04,\ 3.96]
  4. [0.27, 3.73][0.27,\ 3.73]
Explanation: Whenever you see a profile likelihood problem, your goal is to eliminate the nuisance parameter λ\lambda by maximizing the log-likelihood over it for each fixed ψ\psi, then find where the profile log-likelihood drops by half the critical value. Finding the profile likelihood. For fixed ψ\psi, maximize over λ\lambda. Taking the derivative of ℓ(ψ,λ)\ell(\psi,\lambda) with respect to (λ−1)(\lambda-1) and setting it to zero: −12[4(ψ−2)+4(λ−1)]=0  ⟹  λ−1=−(ψ−2).-\frac{1}{2}\left[4(\psi-2) + 4(\lambda-1)\right] = 0 \implies \lambda-1 = -(\psi-2). Substituting back, the profile log-likelihood becomes: ℓp(ψ)=ℓ(2,1)−12{4(ψ−2)2+4(ψ−2)(−(ψ−2))+2(ψ−2)2}=ℓ(2,1)−(ψ−2)2.\ell_p(\psi) = \ell(2,1) - \frac{1}{2}\left\{4(\psi-2)^2 + 4(\psi-2)(-(\psi-2)) + 2(\psi-2)^2\right\} = \ell(2,1) - (\psi-2)^2. Constructing the interval. The profile likelihood ratio statistic is 2[ℓ(2,1)−ℓp(ψ)]=2(ψ−2)22[\ell(2,1) - \ell_p(\psi)] = 2(\psi-2)^2. Setting this equal to 3.843.84: 2(ψ−2)2=3.84  ⟹  (ψ−2)2=1.92  ⟹  ∣ψ−2∣=1.92≈1.386.2(\psi-2)^2 = 3.84 \implies (\psi-2)^2 = 1.92 \implies |\psi-2| = \sqrt{1.92} \approx 1.386. This gives the interval [2−1.386, 2+1.386]≈[0.61, 3.39][2 - 1.386,\ 2 + 1.386] \approx [0.61,\ 3.39], confirming A. The wrong answers reflect common mistakes. B results from forgetting to profile out λ\lambda entirely and using the observed curvature in ψ\psi directly (4(ψ−2)2=3.844(\psi-2)^2 = 3.84). C comes from using the full χ22\chi^2_2 cutoff instead of χ12\chi^2_1. D arises from an algebraic error during the substitution step that leaves a residual cross-term. Study tip: Always complete the profiling step fully before applying the cutoff — the most common trap is using the raw curvature in ψ\psi without accounting for how λ\lambda adjusts, which artificially narrows the interval.

Question 9

A variance-component parameter τ\tau is constrained by τ≥0\tau\geq0. Its maximum-likelihood estimate is τ^=0\widehat\tau=0. Under the null boundary value, the asymptotic distribution of the profile likelihood-ratio statistic is 12χ02+12χ12,\frac{1}{2}\chi^2_0+\frac{1}{2}\chi^2_1, where χ02\chi^2_0 denotes a point mass at zero.

Which approximate critical value should be used to obtain a 95% likelihood-ratio confidence set by inversion at this boundary?

  1. 5.995.99, the 95th percentile of χ22\chi^2_2
  2. 3.843.84, the 95th percentile of χ12\chi^2_1
  3. 2.712.71, the 90th percentile of χ12\chi^2_1 (correct answer)
  4. 1.641.64, the unsquared 95th normal percentile
Explanation: Whenever you see a likelihood-ratio test at a boundary of the parameter space, the standard χ12\chi^2_1 critical value no longer applies — you must account for the non-standard asymptotic distribution. Here, because τ≥0\tau \geq 0 and τ^=0\widehat{\tau} = 0, the likelihood-ratio statistic has the asymptotic null distribution 12χ02+12χ12\frac{1}{2}\chi^2_0 + \frac{1}{2}\chi^2_1. To invert this test and form a 95% confidence set, you need the 95th percentile of this mixture. The mixture CDF satisfies P(T≤t)=12⋅1(t≥0)+12⋅Fχ12(t)P(T \leq t) = \frac{1}{2}\cdot\mathbf{1}(t \geq 0) + \frac{1}{2}\cdot F_{\chi^2_1}(t). Setting this equal to 0.95 gives 12+12Fχ12(t)=0.95\frac{1}{2} + \frac{1}{2}F_{\chi^2_1}(t) = 0.95, so Fχ12(t)=0.90F_{\chi^2_1}(t) = 0.90. This means you need the 90th percentile of χ12\chi^2_1, which is 2.712.71 — confirming answer C. Answer A (5.99) is the 95th percentile of χ22\chi^2_2, which would apply if you were testing two unconstrained parameters simultaneously — a completely different setting. Answer B (3.84) is the standard 95th percentile of χ12\chi^2_1, which you would use if τ\tau were an interior parameter with no boundary constraint — ignoring the point mass at zero entirely. Answer D (1.64) conflates the likelihood-ratio statistic with a normal-based test statistic; the LR statistic is already squared, so you never take an unsquared normal percentile here. The key study tip: boundary parameters cut the asymptotic distribution in half, effectively shifting you one quantile level — a 95% confidence set uses the 90th percentile of χ12\chi^2_1, not the 95th.

Question 10

For four candidate values of a scalar parameter, the relative profile likelihood Rp(ψ)=Lp(ψ)Lp(ψ^)R_p(\psi)=\frac{L_p(\psi)}{L_p(\widehat\psi)} has values 0.200.20, 0.150.15, 0.140.14, and 0.020.02 at ψ1\psi_1, ψ2\psi_2, ψ3\psi_3, and ψ4\psi_4, respectively. A 95% profile-likelihood confidence set is based on χ1,0.952=3.84\chi^2_{1,0.95}=3.84.

Which candidate values belong to the confidence set?

  1. The values ψ1\psi_1, ψ2\psi_2, and ψ3\psi_3 belong.
  2. Only ψ1\psi_1 and ψ2\psi_2 belong to the set. (correct answer)
  3. Only ψ1\psi_1 belongs to the confidence set.
  4. All four candidate values belong to the confidence set.
Explanation: Whenever you see a question involving profile likelihood confidence sets, your core task is converting the relative profile likelihood into a log-likelihood ratio statistic and comparing it to a chi-squared cutoff. The profile likelihood confidence set at level 95% includes all values of ψ\psi satisfying: −2log⁡Rp(ψ)≤χ1,0.952=3.84-2\log R_p(\psi) \leq \chi^2_{1,0.95} = 3.84 Since Rp(ψ)=Lp(ψ)/Lp(ψ^)R_p(\psi) = L_p(\psi)/L_p(\hat{\psi}), you compute −2log⁡Rp(ψ)-2\log R_p(\psi) for each candidate:
  • ψ1\psi_1: −2log⁡(0.20)=−2(−1.609)=3.22≤3.84-2\log(0.20) = -2(-1.609) = 3.22 \leq 3.84 ✓
  • ψ2\psi_2: −2log⁡(0.15)=−2(−1.897)=3.79≤3.84-2\log(0.15) = -2(-1.897) = 3.79 \leq 3.84 ✓
  • ψ3\psi_3: −2log⁡(0.14)=−2(−1.966)=3.93>3.84-2\log(0.14) = -2(-1.966) = 3.93 > 3.84 ✗
  • ψ4\psi_4: −2log⁡(0.02)=−2(−3.912)=7.82>3.84-2\log(0.02) = -2(-3.912) = 7.82 > 3.84 ✗
So only ψ1\psi_1 and ψ2\psi_2 fall inside the confidence set, confirming answer B. Answer A incorrectly includes ψ3\psi_3, which fails the cutoff by a small but decisive margin (3.93 > 3.84). This is the most tempting trap — ψ3\psi_3's relative likelihood of 0.14 looks close to ψ2\psi_2's 0.15, but the log transformation amplifies small differences near the boundary. Answer C excludes ψ2\psi_2, likely from failing to perform the log conversion at all and instead using some raw threshold on RpR_p. Answer D includes all four values, ignoring the test entirely. Study tip: Always convert Rp(ψ)R_p(\psi) to the log-likelihood ratio scale before comparing to the chi-squared cutoff — never apply the 3.84 threshold directly to the relative likelihood values themselves.