Consistency & Efficiency - Statistics Graduate Level

1. Which statement is most accurate?
   Answer: Both estimators are consistent, but their relative asymptotic efficiency depends on the weights and the heteroskedastic variance pattern. 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.
2. 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 $S_n$?
   Answer: $R_n$ is unbiased, has no larger mean squared error than $T_n$, and is consistent. 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.
3. 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?
   Answer: The median requires about $(π/2)n$ observations because its asymptotic relative efficiency compared with the mean is $2/π$. 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.
4. Which statement correctly describes the consistency of this plug-in estimator?
   Answer: It is consistent when $θ≠0$ but inconsistent at $θ=0$, where it remains Bernoulli with success probability $1/2$. Consistency questions hinge on a precise definition: an estimator $\hat{\theta}_n$ is consistent for $g(\theta)$ if $\hat{\theta}_n \xrightarrow{P} g(\theta)$ as $n \to \infty$. 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 $g$ at $\theta$, 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.
5. Which statement best explains the estimator's apparent efficiency advantage?
   Answer: It is superefficient at $θ=0$ and equivalent to the sample mean at fixed nonzero $θ$, but the improvement is not uniform locally. 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.
6. 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?
   Answer: Both are consistent, but $T_n$ has the faster convergence rate and is asymptotically more concentrated in a rate-based comparison. 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.
7. If the true parameter is $θ=2λ$, what is the asymptotic behavior of $T_n$?
   Answer: It converges in probability to $λ$ and is therefore inconsistent for $θ$ under the fixed penalty. 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.
8. Which statement best characterizes consistency and efficiency of the maximum likelihood estimator?
   Answer: It converges to the Kullback–Leibler minimizer, while model-based efficiency claims may fail and sandwich covariance may be required. 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.
9. 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?
   Answer: It is consistent even though its bias equals $1$ and its mean squared error equals $n$. 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.
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?
    Answer: $V_n$ is consistent because the probability that it differs from $T_n$ tends to zero. When you see a question about consistency, your first instinct should be to return to the definition: $T_n$ is consistent for $\theta$ if $T_n \xrightarrow{P} \theta$, 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: $P(|V_n - \theta| > \varepsilon) = P(|V_n - \theta| > \varepsilon \mid A_n^c)\,P(A_n^c) + P(|V_n - \theta| > \varepsilon \mid A_n)\,P(A_n).$ On $A_n^c$, $V_n = T_n$, 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.

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