Statistics Graduate Level Quiz: Consistency And Efficiency

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Consistency And EfficiencyQuestion 1 of 10

Let $T_n$ be a consistent estimator of $θ$ . Define $V_n=T_n$ unless an event $A_n$ occurs, in which case $V_n=U_n$ , where $U_n$ is an arbitrary estimator that may be unbounded. Suppose $P(A_n)→0$ . Which conclusion is necessarily correct?

V_n

is consistent because the probability that it differs from

T_n

tends to zero.

V_n

is consistent only if

U_n

is itself a consistent estimator of

θ

V_n

need not be consistent because

U_n

may be unbounded on the events

A_n

V_n

is consistent only if the mean squared error of

U_n

remains uniformly bounded.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Consistency And Efficiency

Practice Consistency And Efficiency 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 Consistency And Efficiency, 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

$V_n$ is consistent because the probability that it differs from $T_n$ tends to zero. (correct answer)
$V_n$ is consistent only if $U_n$ is itself a consistent estimator of $θ$ .
$V_n$ need not be consistent because $U_n$ may be unbounded on the events $A_n$ .
$V_n$ is consistent only if the mean squared error of $U_n$ remains uniformly bounded.

Explanation: When you see a question about consistency, your first instinct should be to return to the definition:

T_n

is consistent for

\theta

T_n \xrightarrow{P} \theta

, meaning for every

\varepsilon > 0

P(|T_n - \theta| > \varepsilon) \to 0

. Now consider

V_n

. You can decompose its deviation probability using the law of total probability:

P(|V_n - \theta| > \varepsilon) = P(|V_n - \theta| > \varepsilon \mid A_n^c)\,P(A_n^c) + P(|V_n - \theta| > \varepsilon \mid A_n)\,P(A_n).

A_n^c

V_n = T_n

, so the first term is bounded above by

P(|T_n - \theta| > \varepsilon) \to 0

. The second term is bounded above by

P(A_n) \to 0

, regardless of how wild

U_n

is — even if it's unbounded. Both terms vanish, so the entire expression goes to zero, confirming A is correct:

V_n \xrightarrow{P} \theta

. B is wrong because it imposes an unnecessary condition.

U_n

never needs to be consistent on its own; it only contributes on the shrinking event

A_n

, which is probabilistically negligible. C is the most tempting distractor — it correctly notes that

U_n

may be unbounded, but unboundedness in the values of

U_n

doesn't matter here because the probability of being in

A_n

collapses to zero, killing the contribution entirely. D similarly invents a condition (bounded MSE) that the proof never requires. The key study tip: consistency is purely about probability concentrating near

\theta

. When an event has probability tending to zero, whatever happens on that event becomes irrelevant to convergence in probability — no moment conditions needed.

Question 2

Two estimators of the same scalar parameter satisfy $√n(T_n-θ)⇒N(0,4)$ and $n^{1/3}(U_n-θ)⇒N(0,1)$ . Which comparison is most appropriate?

Both are consistent, but $T_n$ has the faster convergence rate and is asymptotically more concentrated in a rate-based comparison. (correct answer)
Both are consistent, but $U_n$ is asymptotically more efficient because its displayed limiting variance is smaller.
The estimators are asymptotically equivalent because each has a nondegenerate normal limit after an appropriate rescaling.
Neither estimator is necessarily consistent because convergence in distribution does not imply convergence in probability.

Explanation: When comparing estimators asymptotically, the most important concept is convergence rate: how quickly does the estimator concentrate around the true parameter as

n \to \infty

? The normalization factor tells you everything. If

a_n(T_n - \theta) \Rightarrow N(0, \sigma^2)

, then

T_n - \theta = O_p(a_n^{-1})

, meaning the estimation error shrinks at rate

a_n^{-1}

. Here,

T_n

requires multiplying by

\sqrt{n}

to get a nondegenerate limit, so its error is

O_p(n^{-1/2})

. Meanwhile,

U_n

only needs

n^{1/3}

, meaning its error is

O_p(n^{-1/3})

. Since

n^{-1/2} \ll n^{-1/3}

for large

n

T_n

converges to

\theta

much faster. Both estimators are consistent — the errors vanish — but

T_n

is vastly superior in terms of concentration around

\theta

. This confirms A is correct. B is a classic trap: it fixates on the displayed variances (4 vs. 1) without recognizing they apply after different scalings. Comparing limiting variances is only meaningful when both estimators share the same convergence rate. Here they don't, so this comparison is meaningless —

U_n

's smaller displayed variance is irrelevant given its much slower rate. C is wrong because "nondegenerate normal limit after rescaling" does not imply equivalence. The whole point is that the rescaling itself differs, and that difference encodes the rate. D is incorrect because convergence in distribution to a point mass at zero does imply consistency, and both limits here collapse to zero once you account for the

1/a_n

factor shrinking. Study tip: Always extract the convergence rate from the normalization factor first — never compare limiting variances across estimators with different rates.

Question 3

Let $X_1,\ldots,X_n$ be independent observations from $N(θ,1)$ . For a fixed constant $λ>0$ , define the soft-threshold estimator $T_n=sign(\bar X_n)(|\bar X_n|-λ)_+$ , where $(a)_+=max(a,0)$ .

If the true parameter is $θ=2λ$ , what is the asymptotic behavior of $T_n$ ?

It converges in probability to $2λ$ because the sample mean is consistent and thresholding reduces only finite-sample variance.
It converges in probability to zero because every fixed positive penalty eventually dominates the sampling variation.
It converges in probability to $λ$ and is therefore inconsistent for $θ$ under the fixed penalty. (correct answer)
It has no probability limit because the threshold creates a discontinuity at every nonzero value of $θ$ .

Explanation: When you encounter a question about soft-thresholding estimators, the key question to ask is: what does the estimator converge to as

n \to \infty

? Start by identifying the probability limit of

\bar{X}_n

, then trace it through the thresholding operation. By the Law of Large Numbers,

\bar{X}_n \xrightarrow{p} \theta = 2\lambda

. Now apply the soft-threshold function:

T_n = \text{sign}(\bar{X}_n)(|\bar{X}_n| - \lambda)_+

. Since

g(x) = \text{sign}(x)(|x| - \lambda)_+

is a continuous function at

x = 2\lambda

(the threshold only creates a kink at zero and discontinuity of derivative at

\pm\lambda

, not a jump discontinuity there), the Continuous Mapping Theorem applies. Therefore

T_n \xrightarrow{p} g(2\lambda) = \text{sign}(2\lambda)(2\lambda - \lambda) = \lambda

. Since

\theta = 2\lambda \neq \lambda

, the estimator is inconsistent — confirming C. A is wrong because it claims

T_n \to 2\lambda

. Thresholding is not variance reduction that vanishes asymptotically; it permanently subtracts

\lambda

from the limit, introducing a fixed bias of

-\lambda

. B is wrong because the penalty

\lambda

does not dominate when

\theta > \lambda

. The threshold shrinks the estimate but never kills it entirely when

\bar{X}_n \to 2\lambda > \lambda

. D is wrong because the soft-threshold function is continuous everywhere — unlike hard thresholding — so the Continuous Mapping Theorem applies and a probability limit exists. Study tip: With regularized estimators, always check whether the penalty persists in the limit. A fixed penalty

\lambda

(not shrinking with

n

) will typically bias the probability limit, making the estimator inconsistent.

Question 4

A parametric likelihood model $f(x;θ)$ is fitted to independent observations whose true distribution is not contained in the model family. Assume standard regularity conditions for misspecified maximum likelihood estimation and that the Kullback–Leibler minimizing parameter is unique.

Which statement best characterizes consistency and efficiency of the maximum likelihood estimator?

It remains consistent for the scientific parameter intended by the analyst because likelihood maximization corrects mild model misspecification.
It converges to the Kullback–Leibler minimizer, while model-based efficiency claims may fail and sandwich covariance may be required. (correct answer)
It is inconsistent for every well-defined target because consistency is impossible whenever the assumed density family is incorrect.
It converges to the Kullback–Leibler minimizer and retains the inverse-Fisher covariance formula regardless of misspecification.

Explanation: When you encounter questions about misspecified maximum likelihood estimation, anchor your thinking around one central result: the MLE converges not to the "true" parameter, but to the value that minimizes the Kullback–Leibler (KL) divergence from the true distribution to the model family. Formally,

\hat{\theta}_{MLE} \xrightarrow{p} \theta^* = \arg\min_\theta KL(f_0 \| f(\cdot;\theta))

, where

f_0

is the true data-generating distribution. This is the White (1982) misspecification framework. Choice B is correct because it captures both consequences of misspecification precisely. The MLE remains consistent for

\theta^*

, the KL minimizer — so consistency holds, but for a pseudo-true parameter, not necessarily any scientifically meaningful quantity. Critically, the standard Fisher information covariance

I(\theta^*)^{-1}

is no longer valid; the correct asymptotic covariance is the "sandwich" estimator

I(\theta^*)^{-1} J(\theta^*) I(\theta^*)^{-1}

, where

J

is the outer-product-of-scores matrix. Choice A is wrong because likelihood maximization does not correct misspecification — it simply finds the best approximation within the wrong model family, which may be far from the scientific target. Choice C overcorrects in the other direction. Inconsistency is not universal under misspecification; the MLE is perfectly consistent for

\theta^*

, just not for the true parameter if

\theta^*

doesn't correspond to it. Choice D is the subtlest trap: while convergence to the KL minimizer is correct, claiming the inverse-Fisher formula still holds is false. That formula requires the information equality, which breaks down outside the true model. Remember: misspecification = valid KL-convergence + broken information equality → always think "sandwich standard errors" when the model may be wrong.

Question 5

For observations from $N(μ,σ^2)$ with unknown $μ$ , the sample mean has asymptotic variance $σ^2/n$ , while the sample median has asymptotic variance $πσ^2/(2n)$ . Approximately how many observations must the median use to match the asymptotic variance of a mean based on $n$ observations?

The median requires about $(2/π)n$ observations, which is fewer than $n$ , because it is the more efficient estimator in the normal model.
The median requires about $n$ observations because both estimators are consistent at the root-sample-size rate and therefore have the same leading constant.
The median requires about $(π^2/4)n$ observations because one must equate asymptotic standard deviations rather than variances, introducing a squared factor.
The median requires about $(π/2)n$ observations because its asymptotic relative efficiency compared with the mean is $2/π$ . (correct answer)

Explanation: When comparing two estimators, a powerful tool is asymptotic relative efficiency (ARE), defined as the ratio of their asymptotic variances. The question is essentially asking you to use ARE to find the sample size that makes the median as precise as the mean. The mean's asymptotic variance is

\sigma^2/n

, and the median's is

\pi\sigma^2/(2n)

. The ARE of the median relative to the mean is:

\text{ARE} = \frac{\sigma^2/n}{\pi\sigma^2/(2n)} = \frac{2}{\pi} \approx 0.637

This means the median is only about 63.7% as efficient as the mean. To match the mean's precision using

n

observations, the median must use

m

observations such that:

\frac{\pi\sigma^2}{2m} = \frac{\sigma^2}{n} \implies m = \frac{\pi}{2}n \approx 1.571n

So D is correct: the median needs roughly

(\pi/2)n

observations — about 57% more than the mean requires. A inverts the logic entirely. Since

2/\pi < 1

, claiming the median needs fewer observations would mean it's more efficient — but under normality, the mean is the more efficient estimator, not the median. B confuses the rate of convergence (both are

\sqrt{n}

-consistent) with efficiency. Same rate does not mean same leading constant — that's precisely what ARE captures. C introduces a spurious squaring step. You equate asymptotic variances directly, not standard deviations, so no squaring occurs and no

\pi^2/4

factor appears. A quick study anchor: ARE tells you the inverse sample-size multiplier needed for one estimator to match another. If ARE

= 2/\pi

, the less efficient estimator needs

\pi/2

times the sample size — always take the reciprocal.

Question 6

Let $X_1,\ldots,X_n$ be independent observations from $N(θ,1)$ . Define the Hodges estimator $H_n=0$ if $|\bar X_n|≤n^{-1/4}$ and $H_n=\bar X_n$ otherwise.

Which statement best explains the estimator's apparent efficiency advantage?

It uniformly dominates the sample mean because its asymptotic variance is no larger at every fixed or local parameter sequence.
It is superefficient at $θ=0$ and equivalent to the sample mean at fixed nonzero $θ$ , but the improvement is not uniform locally. (correct answer)
It is inconsistent at $θ=0$ because setting the estimate exactly to zero prevents a nondegenerate limiting distribution.
It is less efficient at every fixed $θ$ because the threshold $n^{-1/4}$ converges to zero more slowly than $n^{-1/2}$ .

Explanation: Whenever you see a question about superefficiency or the Hodges estimator, think carefully about the distinction between pointwise asymptotic behavior and uniform asymptotic behavior — this is the core tension being tested. The Hodges estimator exploits the fact that when

\theta = 0

, the sample mean

\bar{X}_n

falls inside

[-n^{-1/4}, n^{-1/4}]

with probability approaching 1. By shrinking to zero in that region,

H_n

achieves asymptotic variance 0 at

\theta = 0

, beating the Cramér-Rao bound of 1. For any fixed

\theta \neq 0

, eventually

|\bar{X}_n| > n^{-1/4}

, so

H_n = \bar{X}_n

and efficiency matches the sample mean. This is exactly what B describes: superefficiency at one point, equivalence elsewhere — but no uniform improvement. The critical flaw the estimator hides is revealed through local sequences

\theta_n = c/\sqrt{n}

. Along such sequences, the threshold region captures the mean with non-negligible probability, inflating the asymptotic risk dramatically. The Le Cam–Hájek theory formalizes this: no estimator can be superefficient on a set of positive measure without paying a price locally. A is wrong because the improvement is definitively not uniform — local sequences expose catastrophic risk inflation, so

H_n

does not uniformly dominate

\bar{X}_n

. C is wrong because

H_n

is actually consistent at

\theta = 0

(it equals 0 exactly, which is the true value), and the limiting distribution is degenerate at 0 — which is a feature, not a failure. D is wrong because the slower threshold is precisely why the estimator achieves superefficiency, not reduced efficiency, at fixed

\theta

. Your takeaway: superefficiency is always local and never free — when you see it, immediately ask what happens along

\theta_n = c/\sqrt{n}

sequences.

Question 7

Let $T_n$ be an unbiased estimator of $θ$ such that $Var(T_n)→0$ . Let $S_n$ be sufficient for $θ$ , and define the Rao–Blackwellized estimator $R_n=E(T_n\mid S_n)$ . Which conclusion follows without requiring completeness of $S_n$ ?

$R_n$ is consistent only if it has exactly the same variance as $T_n$ for every sample size.
$R_n$ is unbiased, has no larger mean squared error than $T_n$ , and is consistent. (correct answer)
$R_n$ is unbiased but may fail to be consistent because conditional expectation need not preserve convergence in probability.
$R_n$ is the unique minimum-variance unbiased estimator because sufficiency alone guarantees uniqueness.

Explanation: Whenever you see a question combining sufficiency, conditional expectation, and consistency, you should immediately reach for the Rao–Blackwell Theorem and the basic properties of conditional expectation — no completeness required. The Rao–Blackwell Theorem guarantees three things using only sufficiency. First, unbiasedness is preserved:

E(R_n) = E[E(T_n \mid S_n)] = E(T_n) = \theta

. Second, MSE cannot increase: by the conditional variance formula,

Var(T_n) = Var(R_n) + E[Var(T_n \mid S_n)]

, so

Var(R_n) \leq Var(T_n)

. Third, consistency follows because

Var(R_n) \leq Var(T_n) \to 0

, and since

R_n

is unbiased with variance shrinking to zero, Chebyshev's inequality gives

R_n \xrightarrow{p} \theta

. All three conclusions in B follow cleanly — this is the correct answer. A is wrong because it inverts the variance relationship. Rao–Blackwell guarantees

Var(R_n) \leq Var(T_n)

, not equality. Consistency requires variance going to zero, which is satisfied since

Var(R_n)

is sandwiched between 0 and

Var(T_n) \to 0

. C is wrong because it confuses two different types of convergence. Consistency here is established via Chebyshev — unbiasedness plus vanishing variance is sufficient to prove convergence in probability, regardless of any subtleties in conditional expectation. D is the classic completeness trap. Uniqueness of the UMVUE requires the sufficiency to also be complete. Sufficiency alone only gives you a estimator at least as good as

T_n

, not the unique best one. Remember: Rao–Blackwell gives you unbiasedness + variance reduction for free; you need completeness only when claiming uniqueness of the UMVUE.

Question 8

Let $X_1,\ldots,X_n$ be independent observations from $N(θ,1)$ . The parameter of interest is $g(θ)=1\{θ>0\}$ , and an analyst uses the plug-in estimator $g(\bar X_n)=1\{\bar X_n>0\}$ .

Which statement correctly describes the consistency of this plug-in estimator?

It is consistent for every $θ$ because consistency is always preserved under measurable transformations.
It is inconsistent for every $θ$ because indicator transformations eliminate root-sample-size convergence.
It is consistent when $θ≠0$ but inconsistent at $θ=0$ , where it remains Bernoulli with success probability $1/2$ . (correct answer)
It is consistent at $θ=0$ but inconsistent for sufficiently small nonzero values because the sign is difficult to estimate.

Explanation: Consistency questions hinge on a precise definition: an estimator

\hat{\theta}_n

is consistent for

g(\theta)

\hat{\theta}_n \xrightarrow{P} g(\theta)

n \to \infty

. When you see a plug-in estimator applied to a transformed parameter, your first instinct should be to check whether the transformation is continuous at the true parameter value — because the continuous mapping theorem guarantees consistency only at continuity points. Here,

\bar{X}_n \xrightarrow{P} \theta

by the Law of Large Numbers. The indicator function

g(x) = 1\{x > 0\}

is continuous everywhere except at

x = 0

. When

\theta \neq 0

, the continuous mapping theorem applies directly:

1\{\bar{X}_n > 0\} \xrightarrow{P} 1\{\theta > 0\}

, so the estimator is consistent. But when

\theta = 0

\bar{X}_n \sim N(0, 1/n)

, which means

P(\bar{X}_n > 0) = 1/2

for every finite

n

. The estimator never settles — it remains

\text{Bernoulli}(1/2)

forever, not converging in probability to

g(0) = 0

. This confirms answer C. Choice A is wrong because the continuous mapping theorem requires continuity of

g

\theta

, which fails at

\theta = 0

. Consistency is not universally preserved under all measurable transformations. Choice B overcorrects: the estimator works perfectly well for

\theta \neq 0

, so blanket inconsistency is false. Choice D inverts the story — small nonzero

\theta

values still allow consistency (just with slower convergence in practice), while the true failure point is exactly

\theta = 0

. The key study tip: whenever you see a plug-in estimator applied to a discontinuous function, immediately locate the discontinuity and ask whether the true parameter sits there — that's precisely where consistency can break down.

Question 9

Consider the fixed-design regression model $Y_i=βx_i+ε_i$ , where the errors are independent, satisfy $E(ε_i\mid x_i)=0$ , and may be heteroskedastic. Assume the usual design conditions for consistency. An analyst compares ordinary least squares with weighted least squares using deterministic positive weights bounded away from zero and infinity. The chosen weights are not known to be proportional to the inverse error variances.

Which statement is most accurate?

Ordinary least squares is consistent and asymptotically efficient because conditional mean zero is sufficient for the Gauss–Markov conclusion.
Weighted least squares is generally inconsistent because weights not proportional to inverse variances distort the conditional regression mean.
Both estimators are consistent and necessarily have the same asymptotic variance because the weights are uniformly bounded.
Both estimators are consistent, but their relative asymptotic efficiency depends on the weights and the heteroskedastic variance pattern. (correct answer)

Explanation: Whenever you see a question pairing OLS and WLS under heteroskedasticity, your first instinct should be to separate two distinct properties: consistency and efficiency. These can come apart, and conflating them is exactly what the distractors exploit. Both OLS and WLS with deterministic, bounded weights are consistent for

\beta

under

E(\varepsilon_i \mid x_i) = 0

and standard design conditions. Consistency requires only that the estimating equations converge to the true parameter — and since any fixed positive weighting scheme preserves the unbiasedness of the score (the weights don't distort the conditional mean), both estimators converge to

\beta

. However, their asymptotic variances generally differ. The OLS asymptotic variance involves

\frac{\sum x_i^2 \sigma_i^2}{(\sum x_i^2)^2}

, while WLS replaces

x_i

with

w_i x_i

, yielding a different sandwich expression. Unless the weights happen to be proportional to

1/\sigma_i^2

, neither estimator is BLUE, and which one is more efficient depends on the specific interplay between

w_i

and

\sigma_i^2

. This makes D correct. A is wrong because Gauss–Markov's efficiency conclusion requires homoskedasticity (or correctly specified weights) — conditional mean zero alone guarantees unbiasedness, not efficiency. B is wrong because misspecified weights do not induce inconsistency; they affect variance, not the limiting value of the estimator. C is wrong because bounded weights preserve consistency but do not equalize asymptotic variances — the two sandwich expressions are generally different. A useful rule: consistency lives in the first moment (zero bias in the limit); efficiency lives in the second moment (variance). Heteroskedasticity attacks efficiency, not consistency, when the mean is correctly specified.

Question 10

For each $n$ , suppose an estimator satisfies $T_n-θ=n$ with probability $1/n$ and $T_n-θ=0$ otherwise. Which statement correctly describes this estimator sequence?

It is inconsistent because its mean squared error diverges to infinity at rate $n$ .
It is consistent and asymptotically unbiased because the nonzero error occurs with vanishing probability.
It is consistent even though its bias equals $1$ and its mean squared error equals $n$ . (correct answer)
It is inconsistent because its bias remains equal to $1$ rather than converging to zero.

Explanation: Whenever you see a question about consistency, your first instinct should be to reach for the definition:

T_n \xrightarrow{p} \theta

, meaning

P(|T_n - \theta| > \epsilon) \to 0

for every

\epsilon > 0

. Consistency is purely a statement about probability, not about moments like bias or MSE. Here,

P(T_n - \theta = n) = 1/n

and

P(T_n - \theta = 0) = 1 - 1/n

. For any fixed

\epsilon > 0

P(|T_n - \theta| > \epsilon) = P(T_n - \theta = n) = 1/n \to 0

. So by definition, $T_n$ is consistent. Now compute the moments:

\text{Bias}(T_n) = E[T_n - \theta] = n \cdot (1/n) + 0 \cdot (1 - 1/n) = 1

, and

\text{MSE}(T_n) = E[(T_n-\theta)^2] = n^2 \cdot (1/n) = n

. The bias stays at 1 and MSE diverges — yet the estimator is still consistent. Answer C captures this perfectly. A is wrong because diverging MSE does not imply inconsistency. MSE convergence to zero is sufficient but not necessary for consistency; this example is the canonical counterexample. B is wrong in a subtle way: the estimator is consistent, but it is not asymptotically unbiased — the bias equals 1 for every

n

and never vanishes. D is wrong for the same reason as A and B: a nonzero limiting bias does not preclude consistency, since consistency concerns probability concentration, not expected value. The key study takeaway: consistency ≠ small bias or small MSE. Always return to the probability definition

P(|T_n - \theta| > \epsilon) \to 0

, and remember that bias and MSE criteria are only sufficient conditions, not necessary ones.