Split an estimator's error into bias and variance

Question

A desk wants to estimate the true fair value $\theta$ of an illiquid bond. Estimator A is unbiased (right on average) but jumps around, with variance 16. Estimator B is biased 3 units too high on average but is much steadier, with variance only 4.

The mean squared error of any estimator splits cleanly into two pieces: $\text{MSE} = \text{Variance} + \text{Bias}^2$ .

Which estimator has the smaller MSE, and what is that value?

Accepted Answer

The idea. MSE measures the average squared distance between your estimate and the truth. That distance comes from two separate problems: your estimate can be scattered (variance) and it can be systematically off-center (bias). Squaring the bias puts both on the same footing, and they simply add up. Estimator A. It is unbiased, so its bias is 0. All its error is scatter: 	ext{MSE}A = 16 + 0^2 = 16. Estimator B. It carries a bias of 3 but very little scatter: 	ext{MSE}B = 4 + 3^2 = 4 + 9 = 13. The winner. Estimator B wins, 13 versus 16, even though it is biased. Being right on average is not …

Split an estimator's error into bias and variance

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