Statistics Graduate Level Quiz: Cis From Asymptotic Normality

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Cis From Asymptotic NormalityQuestion 1 of 10

An estimator of $\mu=(\mu_X,\mu_Y)^\mathsf{T}$ satisfies $\sqrt{n}(\widehat{\mu}-\mu)\xrightarrow{d}N\left(0,\begin{pmatrix}4&1\\1&1\end{pmatrix}\right)$ . In a sample with $n=100$ , the estimates are $\widehat{\mu}_X=2$ and $\widehat{\mu}_Y=1$ . The parameter of interest is $\rho=\mu_X/\mu_Y$ , with $\mu_Y\ne0$ .

Using a first-order delta-method Wald interval, which approximate $95\%$ confidence interval should be reported for $\rho$ ?

(1.412,\ 2.588)

, using the sum of the two marginal asymptotic variances

(0.040,\ 3.960)

, using the root-scaled standard deviation without sample-size adjustment

(1.661,\ 2.339)

, treating the numerator and denominator estimates as independent

(1.608,\ 2.392)

, using the gradient of the ratio and the covariance term

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Cis From Asymptotic Normality

Practice Cis From Asymptotic Normality 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 Cis From Asymptotic Normality, 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.

All questions

Question 1

Using a first-order delta-method Wald interval, which approximate $95\%$ confidence interval should be reported for $\rho$ ?

$(1.412,\ 2.588)$ , using the sum of the two marginal asymptotic variances
$(0.040,\ 3.960)$ , using the root-scaled standard deviation without sample-size adjustment
$(1.661,\ 2.339)$ , treating the numerator and denominator estimates as independent
$(1.608,\ 2.392)$ , using the gradient of the ratio and the covariance term (correct answer)

Explanation: 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}

(\hat\mu_X, \hat\mu_Y) = (2,1)

. The asymptotic variance of

\sqrt{n}(\hat\rho - \rho)

(\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.

Question 2

For a fixed parameter value $\theta>0$ , suppose $\sqrt{n}(\widehat{\theta}-\theta)\xrightarrow{d}N(0,V)$ and a consistent estimator of $V$ is available. One analyst forms a direct Wald interval for $\theta$ . Another applies the delta method to $\log(\theta)$ , forms a Wald interval on the log scale, and exponentiates its endpoints.

Under the stated fixed-parameter asymptotics, which comparison of the two intervals is most accurate?

Only the direct interval has first-order correct coverage because exponentiation destroys asymptotic normality
Only the log-scale interval has first-order correct coverage because direct Wald intervals are not transformation invariant
Both have first-order correct coverage, although their endpoints generally differ at smaller asymptotic orders (correct answer)
The intervals have identical endpoints for every sample whenever the same variance estimate is used

Explanation: 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.

Question 3

A continuous population has median $m$ and density $f(m)>0$ . The sample median satisfies $\sqrt{n}(\widehat{m}-m)\xrightarrow{d}N\left(0,\frac{1}{4f(m)^2}\right)$ . For $n=400$ , the sample median is $10.00$ and a consistent density estimate at the median is $\widehat{f}(\widehat{m})=0.10$ .

Which is the approximate $95\%$ confidence interval for the population median?

$(9.510,\ 10.490)$ , using the density in the denominator of the quantile standard error (correct answer)
$(9.951,\ 10.049)$ , using the estimated density as a multiplicative standard-error factor
$(5.100,\ 14.900)$ , omitting the sample-size scaling from the quantile standard error
$(9.755,\ 10.245)$ , using $1/(4f(m))$ as the asymptotic standard deviation

Explanation: 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}

\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.

Question 4

An estimator has a nonnegligible first-order asymptotic bias: $\sqrt{n}(\widehat{\theta}_n-\theta)\xrightarrow{d}N(b,V)$ , where $b$ and $V$ are known constants with $V>0$ .

Which interval has asymptotic coverage $1-\alpha$ for $\theta$ under the stated limit?

$\widehat{\theta}_n\pm z_{1-\alpha/2}\sqrt{V/n}$ , because the bias vanishes without root scaling
$\widehat{\theta}_n-b\pm z_{1-\alpha/2}\sqrt{V/n}$ , because the limiting mean is subtracted directly
$\widehat{\theta}_n-b/\sqrt{n}\pm z_{1-\alpha/2}\sqrt{V/n}$ , using a first-order bias correction (correct answer)
$\widehat{\theta}_n+b/\sqrt{n}\pm z_{1-\alpha/2}\sqrt{V/n}$ , using the limiting mean as a centering adjustment

Explanation: 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

b/\sqrt{n}

— a common source of confusion on exam problems involving non-standard limits.

Question 5

A positive parameter is estimated by $\widehat{\theta}$ , and the estimator satisfies $\sqrt{n}(\widehat{\theta}-\theta)\xrightarrow{d}N(0,4\theta^2)$ . In a sample of size $n=100$ , the estimate is $\widehat{\theta}=0.80$ .

Using the delta method for $\log(\widehat{\theta})$ and then transforming back, which approximate $95\%$ confidence interval for $\theta$ is obtained?

$(0.486,\ 1.114)$ , obtained from a symmetric Wald interval on the original scale
$(0.541,\ 1.184)$ , obtained by exponentiating the log-scale Wald endpoints (correct answer)
$(0.593,\ 1.079)$ , obtained by treating the log-scale variance as $4\widehat{\theta}^2/n$
$(0.016,\ 39.65)$ , obtained without dividing the log-scale standard error by $\sqrt{n}$

Explanation: 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}

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})

(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.

Question 6

Suppose $\sqrt{n}(\widehat{\theta}_n-\theta)\xrightarrow{d}N(0,V(\eta))$ , where $\eta$ is an unknown nuisance parameter. An estimator $\widehat{\eta}_n$ is available, and $V(\cdot)$ is continuous at the true value of $\eta$ .

Which condition is sufficient to justify the plug-in Wald interval $\widehat{\theta}_n\pm z_{1-\alpha/2}\sqrt{V(\widehat{\eta}_n)/n}$ ?

$\widehat{\eta}_n\xrightarrow{p}\eta$ ; no independence from $\widehat{\theta}_n$ or root-rate convergence is required (correct answer)
$\sqrt{n}(\widehat{\eta}_n-\eta)=O_p(1)$ and asymptotic independence from $\widehat{\theta}_n$ are both required
$E(\widehat{\eta}_n)=\eta$ for every $n$ ; consistency is unnecessary when the estimator is unbiased
$V(\widehat{\eta}_n)=V(\eta)$ with probability one; convergence in probability is not sufficiently strong

Explanation: 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.

Question 7

A nonnegative variance component $\tau$ is estimated under the constraint $\tau\ge0$ . When the true value is $\tau=0$ , the estimator has the nonregular limit $\sqrt{n}\widehat{\tau}\xrightarrow{d}\max(0,Z)$ , where $Z\sim N(0,\sigma^2)$ .

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?

Truncation changes only finite-sample interval length, so the interval retains exact first-order coverage at the boundary
The variance component is unidentified at zero, so no asymptotically valid confidence procedure can be constructed
The estimator converges at a rate slower than $\sqrt{n}$ , so every root-rate interval is asymptotically too narrow
The limiting law is not centered normal, so ordinary normal critical values need not produce nominal boundary coverage (correct answer)

Explanation: 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.

Question 8

Two estimators computed from the same observations satisfy $\sqrt{n}\left((\widehat{\theta}_1,\widehat{\theta}_2)^\mathsf{T}-(\theta_1,\theta_2)^\mathsf{T}\right)\xrightarrow{d}N\left(0,\begin{pmatrix}9&3\\3&4\end{pmatrix}\right)$ . For $n=400$ , the observed difference is $\widehat{\theta}_1-\widehat{\theta}_2=0.40$ .

Which is the approximate $95\%$ confidence interval for $\theta_1-\theta_2$ ?

$(0.047,\ 0.753)$ , obtained by omitting the covariance between the estimators
$(0.141,\ 0.659)$ , obtained by incorporating their positive covariance (correct answer)
$(-0.027,\ 0.827)$ , obtained by adding twice the covariance term
$(0.387,\ 0.413)$ , obtained by dividing the asymptotic variance by $n$ twice

Explanation: 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

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.

Question 9

Data consist of $G=64$ independent clusters, with potentially dependent observations within each cluster. An estimating-equation estimator satisfies $\sqrt{G}(\widehat{\beta}-\beta)\xrightarrow{d}N(0,\Omega)$ . For one coefficient, $\widehat{\beta}=1.20$ and the cluster-level sandwich estimate of the corresponding component of $\Omega$ is $2.56$ .

Ignoring small-sample degrees-of-freedom corrections, which approximate $95\%$ confidence interval follows from the stated cluster asymptotics?

$(0.808,\ 1.592)$ , scaling the sandwich variance by the number of independent clusters (correct answer)
$(1.161,\ 1.239)$ , treating the estimated variance component itself as a standard error
$(-1.936,\ 4.336)$ , using the root-scaled standard deviation without dividing by $\sqrt{G}$
$(1.151,\ 1.249)$ , replacing the cluster count by the total number of observations

Explanation: 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}

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}

Question 10

Independent observations follow a heteroskedastic regression process. Ordinary least squares is used, and the inferential target is the coefficient of the population linear projection, not necessarily a correctly specified conditional-mean model. Standard regularity conditions hold.

Which statement about an asymptotic confidence interval for one regression coefficient is correct?

The model-based homoskedastic standard error remains valid because ordinary least squares is asymptotically normal
A heteroskedasticity-consistent sandwich standard error gives first-order valid coverage for the projection coefficient (correct answer)
Neither standard error is valid because heteroskedasticity prevents a central limit theorem for ordinary least squares
The sandwich standard error is valid only if the regression errors are normally distributed conditional on the covariates

Explanation: 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.