Questions: Asymptotic Normality of the MLE

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A researcher uses two estimators for the same parameter θ. For large samples, the MLE achieves asymptotic variance I(θ)^{-1}/n. Estimator B achieves variance 2·I(θ)^{-1}/n. What does 'asymptotic efficiency' of the MLE mean in this context?

AThe MLE is faster to compute than estimator B
BThe MLE converges to θ at rate √n while estimator B converges at rate n
CNo unbiased estimator can achieve smaller asymptotic variance than the MLE, so estimator B is suboptimal
DThe MLE's asymptotic variance does not depend on the sample size
Question 2 Multiple Choice

Under regularity conditions, √n(θ̂_n − θ) → N(0, I(θ)^{-1}). A statistician has n = 400, θ̂_n = 3.0, and I(θ̂_n) = 16. Which expression gives the approximate 95% confidence interval?

A3.0 ± 1.96 × 16
B3.0 ± 1.96 / √(400 × 16)
C3.0 ± 1.96 × 1/√16
D3.0 ± 1.96 / √400
Question 3 True / False

The asymptotic normality theorem implies that θ̂_n is exactly normally distributed for any sample size n, provided the model is correctly specified.

TTrue
FFalse
Question 4 True / False

Higher Fisher information at the true parameter θ implies the MLE will be more precisely estimated asymptotically, since the asymptotic variance is I(θ)^{-1}/n.

TTrue
FFalse
Question 5 Short Answer

Why does the Fisher information appear in the asymptotic variance of the MLE? What would it mean for inference if a parameter had very low Fisher information?

Think about your answer, then reveal below.