Question bank — Properties of estimators — unbiasedness, consistency, efficiency

False. $\hat{\theta }=X_1$ (just the first data point) is unbiased for $\mu$ but its variance stays ${\sigma }^2$ for all $n$, so it never homes in — unbiasedness is a fixed-$n$ property, consistency is a limit property.

False. The divide-by-$n$ sample variance is biased low for every finite $n$, yet its bias $\propto 1/n\to 0$, so it still converges to ${\sigma }^2$ and is consistent.

False. That is only asymptotic unbiasedness (the centre approaches $\theta$). You also need the variance to vanish; a distribution can be centred correctly yet stay spread out forever.

True — but only because for unbiased estimators $\text{MSE}=\text{Var}+0^2=\text{Var}$, so the comparison reduces to variance. See Bias–Variance Tradeoff.

False on two counts. It bounds unbiased estimators only (a biased constant $\hat{\theta }=0$ has zero variance), and it holds only under regularity conditions — differentiable likelihood and a support not depending on $\theta$. See Fisher Information.

False. The divide-by-$n$ estimator is biased but has smaller variance, and for normal data it actually has smaller MSE — unbiasedness is not the same as MSE-optimal.

False. Only linear $g$ preserve unbiasedness. By Jensen's inequality $\mathbb{E}[{\hat{\theta }}^2]=\text{Var}+{\theta }^2>{\theta }^2$, so ${\hat{\theta }}^2$ over-estimates ${\theta }^2$.

False. The bound is a floor; sometimes no unbiased estimator reaches it. The MVUE (minimum-variance unbiased estimator) is then the best available, but it may still sit strictly above $1/I(\theta )$.

True. $\text{Var}(\bar{X})={\sigma }^2/n$, so $n\to 2n$ gives ${\sigma }^2/2n$ — exactly half. This shrinking-to-zero variance is why $\bar{X}$ is consistent (a Law of Large Numbers statement).

False. The MLE for Uniform$(0,\theta )$ is ${\max }_i{X}_i$, which is biased low ($\mathbb{E}[\max ]=\frac{n}{n+1}\theta <\theta$). MLEs are only asymptotically unbiased and consistent — see Maximum Likelihood Estimation.

The deviations are measured from $\bar{X}$, not the true $\mu$, and $\bar{X}$ sits inside the data, so the squared deviations are systematically too small. You spent one degree of freedom estimating the mean; divide by $n-1$ to correct.

The number $1.5>1$ is correct but the reading is backwards: relative efficiency of $A$ to $B$ is $\text{Var}(B)/\text{Var}(A)$; here $A$ has the larger variance, so $B$ is the more efficient one.

A tiny fixed bias that does not shrink with $n$ blocks consistency — the estimator converges to $\theta +0.02$, not $\theta$. Consistency needs the bias to $\to 0$, not merely to be small.

Consistency is convergence in probability, a statement about the sampling distribution as $n\to \infty$, not about a handful of runs. A few lucky samples prove nothing about the limit.

The sign is wrong: $\text{MSE}=\text{Var}+{\text{Bias}}^2$. Both terms are non-negative and add; bias never reduces mean squared error.

Equal weighting is optimal only when the observations have equal variance. If some $X_i$ are noisier, a variance-weighted average beats plain $\bar{X}$.

Because $\bar{X}$ is fitted to the data and hugs it, the deviations $(X_i-\bar{X})$ are on average smaller than $(X_i-\mu )$; the algebra gives $\mathbb{E}[\sum (X_i-\bar{X}{)}^2]=(n-1){\sigma }^2$, so dividing by $n-1$ restores unbiasedness.

Each observation carries only finite information about $\theta$; the Fisher information $I(\theta )$ measures how sharply the likelihood peaks, and a flatter peak means $\theta$ can't be pinned down tightly — so variance can't drop below $1/I(\theta )$.

An unbiased estimator can still be wildly variable (like $X_1$), scattering far from $\theta$ on any given sample. Being centred correctly is worthless if the wobble never shrinks — you also want low variance and consistency.

Variance-only comparison rewards biased estimators unfairly (a constant has zero variance). MSE = Var + Bias² charges the estimator for both its wobble and its off-centredness, so it compares them on equal footing.

Shrinking toward a value trades a small bias for a larger drop in variance; when the variance saving exceeds the squared bias added, total MSE falls — this is the Bias–Variance Tradeoff in action.

The Law of Large Numbers says $\bar{X}\stackrel{P}{\to }\mu$; that is precisely the definition of $\bar{X}$ being consistent for $\mu$, driven by $\text{Var}(\bar{X})={\sigma }^2/n\to 0$.

Independent observations contribute additively to the log-likelihood, and the information is the expected squared score, so information adds: $I_n(\theta )=n{I}_1(\theta )$ — more data, proportionally more information, hence the CRLB floor drops like $1/n$.

Only in the single degenerate case $\theta =7$; for every other $\theta$ its expectation is $7\ne \theta$, so it is biased. It does have zero variance, showing why "smallest variance" alone is a bad criterion.

Every observed value is $\le \theta$, so their maximum can never exceed $\theta$ and almost surely falls short; hence $\mathbb{E}[\max ]=\frac{n}{n+1}\theta <\theta$, a strictly-below bias that shrinks as $n$ grows.

The MSE argument via ${\sigma }^2/n$ collapses since ${\sigma }^2=\infty$. $\bar{X}$ may still be consistent under the Law of Large Numbers (which only needs a finite mean), but the neat variance-based reasoning fails — you need a stronger tool.

No — attaining the CRLB requires being unbiased with variance $1/(n{I}_1(\theta ))\to 0$, so its MSE vanishes and it is automatically consistent. Efficiency in this classical sense is the strongest of the three properties.

It is $\frac{0}{0}$ — undefined. With a single point $\bar{X}=X_1$, so the deviation is zero and there are zero degrees of freedom; you cannot estimate spread from one observation.

No. Unbiasedness needs only linearity of expectation, and consistency needs only $\text{Var}\to 0$. The CLT describes the shape (approximate normality) of $\bar{X}$'s sampling distribution, a separate and finer question.

Neither — they are equally efficient by definition, since efficiency compares variances among unbiased estimators and here they tie. Relative efficiency equals exactly $1$.