CIs from Asymptotic Normality - Statistics Graduate Level

1. Under the stated fixed-parameter asymptotics, which comparison of the two intervals is most accurate?
   Answer: Both have first-order correct coverage, although their endpoints generally differ at smaller asymptotic orders Whenever you see a question mixing delta method transformations with Wald intervals, the core concept to invoke is asymptotic equivalence under smooth transformations. Both procedures are valid to first order, but they differ in higher-order behavior. Here's why C is correct. Starting from $\sqrt{n}(\hat{\theta}-\theta)\xrightarrow{d}N(0,V)$, the delta method tells you that for any smooth function $g$ with nonzero derivative, $\sqrt{n}(g(\hat{\theta})-g(\theta))\xrightarrow{d}N(0,[g'(\theta)]^2 V)$. Applying $g(x)=\log(x)$ gives $\sqrt{n}(\log\hat{\theta}-\log\theta)\xrightarrow{d}N(0,V/\theta^2)$. A Wald interval on the log scale then exponentiates back. Both the direct and log-transformed intervals achieve the same nominal coverage asymptotically — both are first-order correct — but their finite-sample endpoints differ because they linearize around different scales. Neither dominates in general at this order; differences appear only at smaller asymptotic orders (e.g., $O(n^{-1})$ corrections). A is wrong because exponentiation does not destroy asymptotic normality on the log scale; the delta method guarantees asymptotic normality is preserved through smooth, invertible transformations. B is wrong because direct Wald intervals are first-order correct — the claim that they lack transformation invariance confuses a practical limitation with a theoretical failure. Both intervals are first-order valid. D is wrong because the two intervals are algebraically distinct constructions. Even with the same variance estimate, exponentiating a symmetric log-scale interval produces an asymmetric interval on the original scale. Study tip: Remember that the delta method preserves first-order coverage — any smooth, invertible reparametrization yields an asymptotically equivalent interval. Differences between such intervals only show up in second-order ($O(n^{-1})$) terms.
2. Ignoring small-sample degrees-of-freedom corrections, which approximate $95\%$ confidence interval follows from the stated cluster asymptotics?
   Answer: $(0.808,\ 1.592)$, scaling the sandwich variance by the number of independent clusters When working with cluster-robust inference, the key is tracking exactly what the sandwich estimator is estimating and how it scales with the number of clusters $G$. The asymptotic result $\sqrt{G}(\widehat{\beta}-\beta)\xrightarrow{d}N(0,\Omega)$ tells you that the variance of $\sqrt{G}\,\widehat{\beta}$ converges to $\Omega$. To get the variance of $\widehat{\beta}$ itself, you divide: $\text{Var}(\widehat{\beta})\approx\Omega/G$. The sandwich estimate gives you $\widehat{\Omega}=2.56$, so the estimated variance of $\widehat{\beta}$ is $2.56/64=0.04$, yielding a standard error of $\sqrt{0.04}=0.20$. The 95% CI is then $1.20\pm1.96(0.20)\approx(0.808,\ 1.592)$, confirming A is correct. Each wrong answer reflects a specific scaling mistake. B treats $\widehat{\Omega}=2.56$ directly as a variance (not dividing by $G$), then takes $\sqrt{2.56}=1.6$ as the standard error — but $2.56$ estimates $\Omega$, the variance of the scaled quantity $\sqrt{G}\,\widehat{\beta}$, not of $\widehat{\beta}$ itself. C goes further in the wrong direction, using $\sqrt{2.56}=1.6$ without any division, producing an absurdly wide interval that ignores the $1/\sqrt{G}$ shrinkage. D substitutes the total observation count for $G$; unless clusters are of size one, this conflates the number of independent units with the total sample size, which is the very problem cluster-robust methods exist to address. Your study tip: always ask yourself what quantity the sandwich estimate is approximating the variance of. Trace the scaling carefully — divide by $G$ (not total $n$) to recover the variance of $\widehat{\beta}$.
3. Which is the approximate $95\%$ confidence interval for the population median?
   Answer: $(9.510,\ 10.490)$, using the density in the denominator of the quantile standard error Whenever you encounter asymptotic confidence intervals for sample quantiles, your anchor should be the delta-method result: $\sqrt{n}(\widehat{m}-m)\xrightarrow{d}N\!\left(0,\frac{1}{4f(m)^2}\right)$. This tells you the asymptotic standard deviation of $\widehat{m}$ is $\frac{1}{2f(m)\sqrt{n}}$, where the density appears in the denominator — a fact that trips up many students. Plugging in the given values, the estimated standard error is $\frac{1}{2\,\widehat{f}(\widehat{m})\sqrt{n}} = \frac{1}{2(0.10)(20)} = \frac{1}{4} = 0.25$. The 95% interval is then $10.00 \pm 1.96(0.25) = (9.510,\ 10.490)$, confirming A is correct. Choice B produces an absurdly tight interval $(9.951, 10.049)$ by misusing the density as a multiplicative factor rather than placing it in the denominator — effectively computing $\widehat{f}\cdot\frac{1}{\sqrt{n}}$ instead of $\frac{1}{2\widehat{f}\sqrt{n}}$. Choice C yields a wildly wide interval by forgetting the $\sqrt{n}$ scaling entirely, using $\frac{1}{2f(m)}=5$ as the standard error rather than $\frac{1}{2f(m)\sqrt{n}}$. Choice D uses $\frac{1}{4f(m)}=2.5$ as the standard deviation — this confuses the asymptotic variance $\frac{1}{4f(m)^2}$ with the standard deviation, and also ignores the $\sqrt{n}$ denominator. A useful memory anchor: the sample median's standard error looks like $\frac{1}{2f(m)\sqrt{n}}$ — think of $2f(m)$ as the "effective information" at the median, analogous to how larger density means less spread in estimation.
4. Which condition is sufficient to justify the plug-in Wald interval $\widehat{\theta}_n\pm z_{1-\alpha/2}\sqrt{V(\widehat{\eta}_n)/n}$?
   Answer: $\widehat{\eta}_n\xrightarrow{p}\eta$; no independence from $\widehat{\theta}_n$ or root-rate convergence is required When building a Wald confidence interval with an estimated variance, the core question is: *what do we need from $\widehat{\eta}_n$ so that $V(\widehat{\eta}_n)$ behaves like $V(\eta)$ asymptotically?* The answer comes directly from the continuous mapping theorem and Slutsky's theorem. If $\widehat{\eta}_n \xrightarrow{p} \eta$ and $V(\cdot)$ is continuous at $\eta$, then by the continuous mapping theorem, $V(\widehat{\eta}_n) \xrightarrow{p} V(\eta)$. Slutsky's theorem then lets you combine this with $\sqrt{n}(\widehat{\theta}_n - \theta) \xrightarrow{d} N(0, V(\eta))$ to conclude that the studentized statistic $\frac{\sqrt{n}(\widehat{\theta}_n - \theta)}{\sqrt{V(\widehat{\eta}_n)}} \xrightarrow{d} N(0,1)$, validating the plug-in Wald interval. No independence assumption or root-$n$ rate is required — Slutsky's theorem handles everything. A is correct. Choice B is wrong because it imposes unnecessary conditions. Slutsky's theorem requires only convergence in probability for $V(\widehat{\eta}_n)$, not asymptotic independence or a root-$n$ rate for $\widehat{\eta}_n$ itself. Choice C is wrong because unbiasedness in finite samples does not imply consistency, and without $V(\widehat{\eta}_n) \xrightarrow{p} V(\eta)$, the interval's asymptotic validity breaks down entirely. Choice D is wrong because almost-sure equality $V(\widehat{\eta}_n) = V(\eta)$ a.s. is an extraordinarily strong (and essentially unrealistic) condition, and the claim that convergence in probability is "not sufficient" directly contradicts Slutsky's theorem. The strategy here: whenever you see a plug-in estimator replacing a nuisance parameter, immediately think continuous mapping + Slutsky. Consistency ($\xrightarrow{p}$) plus continuity is the minimal sufficient toolkit — nothing more is needed.
5. Which interval has asymptotic coverage $1-\alpha$ for $\theta$ under the stated limit?
   Answer: $\widehat{\theta}_n-b/\sqrt{n}\pm z_{1-\alpha/2}\sqrt{V/n}$, using a first-order bias correction Whenever you encounter a question about confidence intervals for a biased estimator, your first move should be to translate the asymptotic distributional statement back into a statement about $\widehat{\theta}_n$ itself, then solve for $\theta$. The given limit tells you $\sqrt{n}(\widehat{\theta}_n - \theta) \xrightarrow{d} N(b, V)$, which means approximately $\widehat{\theta}_n \approx \theta + b/\sqrt{n} + \sqrt{V/n}\cdot Z$ where $Z \sim N(0,1)$. Rearranging for $\theta$: $\theta \approx \widehat{\theta}_n - b/\sqrt{n} - \sqrt{V/n}\cdot Z$. So the bias-corrected pivot is $\sqrt{n}(\widehat{\theta}_n - b/\sqrt{n} - \theta) \xrightarrow{d} N(0,V)$, and the natural $1-\alpha$ interval is $\widehat{\theta}_n - b/\sqrt{n} \pm z_{1-\alpha/2}\sqrt{V/n}$. That's exactly C. A is wrong because ignoring the bias $b$ shifts the interval's center by $b/\sqrt{n}$, producing asymptotic coverage strictly less than $1-\alpha$ — the bias does not vanish at the $1/\sqrt{n}$ scale relevant here. B subtracts the full constant $b$ rather than the rescaled bias $b/\sqrt{n}$. This overcorrects by a factor of $\sqrt{n}$, displacing the interval wildly from $\theta$ and giving zero asymptotic coverage. D adds $b/\sqrt{n}$ instead of subtracting it, moving the center in the wrong direction and inflating the bias rather than removing it. The key strategy: always convert the root-$n$ CLT statement into a statement about $\widehat{\theta}_n - \theta$ directly. The asymptotic mean $b$ lives on the $\sqrt{n}$ scale, so the actual bias in $\widehat{\theta}_n$ is $b/\sqrt{n}$ — a common source of confusion on exam problems involving non-standard limits.
6. Which statement about an asymptotic confidence interval for one regression coefficient is correct?
   Answer: A heteroskedasticity-consistent sandwich standard error gives first-order valid coverage for the projection coefficient When you see a question involving OLS under heteroskedasticity, the key distinction to hold in mind is between model validity and inference validity — and specifically, what target parameter you're actually estimating. OLS consistently estimates the population linear projection coefficient $\beta = (E[XX'])^{-1}E[XY]$ under standard regularity conditions, regardless of whether the conditional mean is correctly specified or homoskedasticity holds. By a central limit theorem applied to the score $X_i u_i$, the OLS estimator is asymptotically normal. The sandwich ("heteroskedasticity-consistent") variance estimator consistently estimates the true asymptotic variance $V = (E[XX'])^{-1} E[X_i X_i' u_i^2] (E[XX'])^{-1}$, yielding confidence intervals with first-order correct coverage. This makes B correct. A is wrong because asymptotic normality of OLS alone is not sufficient for valid inference — you also need a consistent variance estimator. The homoskedastic model-based standard error assumes $E[u_i^2 | X_i] = \sigma^2$, which fails under heteroskedasticity, biasing the estimated variance and invalidating coverage. C is wrong because heteroskedasticity does not prevent a CLT from applying. The score $X_i u_i$ satisfies the conditions of a standard CLT (finite variance, independence) even when error variances differ across observations. Heteroskedasticity affects the form of the asymptotic variance, not whether normality holds. D is wrong because the sandwich estimator's validity is purely large-sample and requires no distributional assumption on the errors — conditional normality is irrelevant here. Your study tip: whenever a question mixes OLS, heteroskedasticity, and inference, ask yourself two things separately — consistency of the point estimator, and consistency of the variance estimator. They can fail independently.
7. Which is the approximate $95\%$ confidence interval for $\theta_1-\theta_2$?
   Answer: $(0.141,\ 0.659)$, obtained by incorporating their positive covariance When you see joint asymptotic normality of two estimators, the key question for any linear combination is: what is the asymptotic variance of that combination? For $\widehat{\theta}_1 - \widehat{\theta}_2$, you apply the delta method (or simple variance rules) to the joint limiting distribution. Given $\sqrt{n}(\widehat{\theta}_1 - \widehat{\theta}_2 - (\theta_1-\theta_2)) \xrightarrow{d} N(0, \sigma^2)$, the asymptotic variance is $\sigma^2 = \text{Var}_1 + \text{Var}_2 - 2\text{Cov} = 9 + 4 - 2(3) = 7$. So the variance of $\widehat{\theta}_1 - \widehat{\theta}_2$ is $7/n = 7/400 = 0.0175$, giving standard error $\sqrt{0.0175} \approx 0.1323$. The 95% CI is $0.40 \pm 1.96(0.1323) \approx (0.141, 0.659)$, confirming B is correct. A is wrong because it ignores the covariance entirely, using $\sigma^2 = 9 + 4 = 13$, which gives $\text{SE} \approx 0.180$ and a wider interval $(0.047, 0.753)$. Omitting covariance inflates the standard error when the covariance is positive. C adds twice the covariance rather than subtracting it, using $9 + 4 + 2(3) = 19$. This is the variance formula for a sum $\widehat{\theta}_1 + \widehat{\theta}_2$, not a difference — a classic sign error. D divides by $n$ twice (i.e., uses $7/n^2$), drastically shrinking the interval. The asymptotic variance $\sigma^2/n$ already accounts for one factor of $n$; dividing again is a fundamental misapplication. Your study tip: always write out the full variance formula $\text{Var}(aX + bY) = a^2\sigma_X^2 + b^2\sigma_Y^2 + 2ab\,\text{Cov}(X,Y)$ before plugging in numbers — sign errors on the covariance term are the most common trap on these problems.
8. What is the principal problem with constructing a nominal $95\%$ interval as $\widehat{\tau}\pm1.96\widehat{\sigma}/\sqrt{n}$ and truncating its lower endpoint at zero?
   Answer: The limiting law is not centered normal, so ordinary normal critical values need not produce nominal boundary coverage Whenever you encounter questions about boundary estimation and constrained parameters, your first instinct should be to ask: does standard asymptotic theory still apply? Classical confidence intervals rely on the estimator being asymptotically normal and centered at the true value. At the boundary $\tau = 0$, that assumption breaks down. The passage tells you exactly what goes wrong: the limit distribution is $\max(0, Z)$ rather than $Z \sim N(0, \sigma^2)$. This is a half-normal distribution — it places all probability mass on the nonnegative half-line and has a point mass at zero. It is not symmetric and not centered at zero. When you construct $\widehat{\tau} \pm 1.96\widehat{\sigma}/\sqrt{n}$, those critical values come from the symmetric standard normal, which no longer matches the actual sampling distribution at the boundary. As a result, the interval's true coverage at $\tau = 0$ is not 95%, regardless of truncation. This is precisely why D is correct: the nonstandard limiting law invalidates normal critical values, breaking nominal boundary coverage. A is wrong because truncation does more than alter finite-sample length — it fundamentally changes the interval's asymptotic coverage properties at the boundary. Claiming "exact first-order coverage" is retained is false. B is wrong because $\tau$ is perfectly identified at zero; the issue is non-regularity (boundary behavior), not lack of identification. Valid procedures (e.g., likelihood-ratio inversion) can be constructed. C is wrong because $\widehat{\tau}$ still converges at rate $\sqrt{n}$ — the problem is the shape of the limit, not the speed of convergence. Study tip: On boundary inference problems, always check the shape of the limit distribution, not just its rate. A half-normal limit immediately signals that symmetric normal critical values will fail — look for answers referencing the non-centered or nonstandard limiting law.
9. Using the delta method for $\log(\widehat{\theta})$ and then transforming back, which approximate $95\%$ confidence interval for $\theta$ is obtained?
   Answer: $(0.541,\ 1.184)$, obtained by exponentiating the log-scale Wald endpoints Whenever you apply the delta method to construct a confidence interval, the key insight is that transforming to a stabilized scale, building the interval there, and then back-transforming often produces asymmetric intervals with better coverage for bounded or skewed parameters. Here, $\sqrt{n}(\widehat{\theta}-\theta)\xrightarrow{d}N(0,4\theta^2)$, so the asymptotic variance of $\widehat{\theta}$ is $4\theta^2/n$. Applying the delta method to $g(\theta)=\log(\theta)$ with $g'(\theta)=1/\theta$, the asymptotic variance of $\log(\widehat{\theta})$ is $(1/\theta)^2\cdot(4\theta^2/n)=4/n$. This is variance-stabilizing — the log-scale variance is constant (no $\theta$ dependence). With $n=100$, the log-scale standard error is $\sqrt{4/100}=0.2$. The 95% log-scale interval is $\log(0.80)\pm 1.96(0.2)$, giving $(-0.6132\pm 0.392)$, i.e., $(-1.005,\ -0.221)$. Exponentiating: $(e^{-1.005},\ e^{-0.221})\approx(0.541,\ 1.184)$, confirming answer B. A is wrong because it builds a symmetric Wald interval directly on the original scale using $\widehat{\theta}\pm 1.96\cdot(2\widehat{\theta}/\sqrt{n})$. This ignores the skewness advantage of the log transformation and yields a symmetric interval that can behave poorly for positive parameters. C mistakenly uses $4\widehat{\theta}^2/n$ as the log-scale variance rather than $4/n$. The whole point of the delta method here is that the log transformation removes $\theta$ from the variance. D forgets to divide by $\sqrt{n}$ entirely, producing a wildly inflated standard error and a nonsensical interval. Study tip: When a variance has the form $c\cdot\theta^2$, immediately consider $\log(\theta)$ as a stabilizing transformation — the delta method will cancel out $\theta$ and leave a constant variance, making the back-transformed interval your best CI strategy.
10. Using a first-order delta-method Wald interval, which approximate $95\%$ confidence interval should be reported for $\rho$?
    Answer: $(1.608,\ 2.392)$, using the gradient of the ratio and the covariance term Whenever you need a confidence interval for a nonlinear function of a vector parameter, the delta method is your tool. For $\rho = g(\mu_X, \mu_Y) = \mu_X/\mu_Y$, compute the gradient evaluated at the estimates: $\nabla g = \left(\frac{1}{\mu_Y},\ -\frac{\mu_X}{\mu_Y^2}\right)^\mathsf{T} = (1,\ -2)^\mathsf{T}$ at $(\hat\mu_X, \hat\mu_Y) = (2,1)$. The asymptotic variance of $\sqrt{n}(\hat\rho - \rho)$ is $(\nabla g)^\mathsf{T} \Sigma (\nabla g)$, where $\Sigma = \begin{pmatrix}4&1\\1&1\end{pmatrix}$. This gives $(1,{-2})\begin{pmatrix}4&1\\1&1\end{pmatrix}\begin{pmatrix}1\\-2\end{pmatrix} = (1,-2)\begin{pmatrix}2\\-1\end{pmatrix} = 4$. The variance of $\hat\rho$ is therefore $4/n = 4/100 = 0.04$, so $\text{SE} = 0.2$. The 95% interval is $2 \pm 1.96(0.2) \approx (1.608,\ 2.392)$, confirming D is correct. Choice A simply adds the two diagonal variances $(4+1)/100$, ignoring the gradient weights and the covariance term entirely — a fundamental misapplication of error propagation. Choice B uses $\sqrt{4}/\sqrt{100}$ as the standard error, grabbing only $\sigma_X$ and forgetting to apply the delta-method gradient, so the denominator's uncertainty and the covariance are both ignored. Choice C correctly applies the gradient but sets the covariance to zero, computing $1^2(4) + (-2)^2(1) = 8$ instead of the correct $4$ — this overestimates uncertainty whenever variables are positively correlated. The key study tip: always write out the full quadratic form $\nabla g^\mathsf{T} \Sigma \nabla g$ — the cross terms from the covariance matrix frequently change the answer dramatically and are the most common source of error on delta-method problems.

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