A biased estimator can be consistent.

A biased estimator can be consistent.

• A. True
• B. False
• C. Sometimes
• D. Never

Answer: (A)

Consistency means the estimator converges in probability to the true parameter as the sample size grows. An estimator can be biased at finite samples but still be consistent if its bias shrinks to zero as n increases (and its variance also shrinks). In other words, the overall error can vanish in the limit even though the expectation is not exactly the parameter for finite n. A concrete example helps: the usual sample variance computed with denominator n is biased for finite samples, since its expected value is ((n−1)/n) times the true variance σ². The bias is −σ²/n, which goes to zero as n → ∞. At the same time, the variance of this estimator also goes to zero as n grows. Therefore, it converges in probability to σ², making it consistent despite being biased for finite n. That’s why a biased estimator can be consistent. The other options would ignore the possibility that bias vanishes in the limit, which is precisely what consistency requires.

Consistency means the estimator converges in probability to the true parameter as the sample size grows. An estimator can be biased at finite samples but still be consistent if its bias shrinks to zero as n increases (and its variance also shrinks). In other words, the overall error can vanish in the limit even though the expectation is not exactly the parameter for finite n.

A concrete example helps: the usual sample variance computed with denominator n is biased for finite samples, since its expected value is ((n−1)/n) times the true variance σ². The bias is −σ²/n, which goes to zero as n → ∞. At the same time, the variance of this estimator also goes to zero as n grows. Therefore, it converges in probability to σ², making it consistent despite being biased for finite n.

That’s why a biased estimator can be consistent. The other options would ignore the possibility that bias vanishes in the limit, which is precisely what consistency requires.