What is the difference between a consistent estimator and an unbiased estimator?

The precise technical definitions of these terms are fairly complicated, and it's difficult to get an intuitive feel for what they mean. I can imagine a good estimator, and a bad estimator, but I'm having trouble seeing how any estimator could satisfy one condition and not the other.

To define the two terms without using too much technical language:

An estimator is consistent if, as the sample size increases, the estimates (produced by the estimator) "converge" to the true value of the parameter being estimated. To be slightly more precise - consistency means that, as the sample size increases, the sampling distribution of the estimator becomes increasingly concentrated at the true parameter value.

An estimator is unbiased if, on average, it hits the true parameter value. That is, the mean of the sampling distribution of the estimator is equal to the true parameter value.

The two are not equivalent: Unbiasedness is a statement about the expected value of the sampling distribution of the estimator. Consistency is a statement about "where the sampling distribution of the estimator is going" as the sample size increases.


It certainly is possible for one condition to be satisfied but not the other - I will give two examples. For both examples consider a sample $X_1, ..., X_n$ from a $N(\mu, \sigma^2)$ population.

Unbiased but not consistent: Suppose you're estimating $\mu$. Then $X_1$ is an unbiased estimator of $\mu$ since $E(X_1) = \mu$. But, $X_1$ is not consistent since its distribution does not become more concentrated around $\mu$ as the sample size increases - it's always $N(\mu, \sigma^2)$!

Consistent but not unbiased: Suppose you're estimating $\sigma^2$. The maximum likelihood estimator is $$ \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \overline{X})^2 $$ where $\overline{X}$ is the sample mean. It is a fact that $$ E(\hat{\sigma}^2) = \frac{n-1}{n} \sigma^2 $$ which can be derived using the information here. Therefore $\hat{\sigma}^2$ is biased for any finite sample size. We can also easily derive that $${\rm var}(\hat{\sigma}^2) = \frac{ 2\sigma^4(n-1)}{n^2}$$ From these facts we can informally see that the distribution of $\hat{\sigma}^2$ is becoming more and more concentrated at $\sigma^2$ as the sample size increases since the mean is converging to $\sigma^2$ and the variance is converging to $0$. (Note: This does constitute a proof of consistency, using the same argument as the one used in the answer here)


