The risk of an estimator $\delta$ is defined as $$E_\theta[L(\theta,\delta(X))],$$ where, say, $L(\theta,\delta(X)) = (\theta-\delta(X))^2$, and $E_\theta(X)$ is defined as $\int XdP_\theta$, namely the expectation of random variable $X$ when the parameter is $\theta$.

I wonder why people never considered $$Var_\theta[L(\theta,\delta(X))]$$ in addition to the risk as a way to evaluate an estimator given a predefined loss function?

Say $\delta_1$ and $\delta_2$ have similar risk, but different variance of loss. Intuitively I would choose the one with a smaller variance of loss. However, I've never seen people ever talked about it. Have they?

  • $\begingroup$ That would make a very precise but inaccurate predictor look very good. For example, if I consistently predict tomorrow's temperature to be exactly 20 degrees too high, the variance of the difference would be 0. A precise but inaccurate predictor needs to be calibrated. $\endgroup$ Nov 4, 2020 at 17:16
  • $\begingroup$ @RobbytheBelgian I understand. My question is "in addition" to the expectation of loss. Suppose two estimators give the same expectation of loss for some $\theta$, then why people never consider using their variances of loss to further pick the best of these two? $\endgroup$
    – Tan
    Nov 4, 2020 at 17:19
  • $\begingroup$ It would essentially mean using a different loss function. By changing the loss function you can penalize occasional high errors more. Choosing the correct loss function for a given problem is not trivial. Any crucial risk assessment needs to be taken into account. $\endgroup$ Nov 4, 2020 at 17:27
  • $\begingroup$ Yes, I hear you. However, we always predefined a loss before we talk about the quality of an estimator, e.g. squared loss for the risk of an estimator. We basically assess an estimator under a specific loss. $\endgroup$
    – Tan
    Nov 4, 2020 at 17:35
  • $\begingroup$ It's an interesting question. You would, effectively, optimize an estimator that balances the error across samples (not one that achieves minimal error overall). Perhaps you can frame it along the likes of a loglikelihood too $\endgroup$
    – Firebug
    Nov 4, 2020 at 18:12

1 Answer 1


I saw something similar in two papers:

The second one is more on point (and easier to read to me).

They consider the following problem statement: minimize the empirical risk given by $$R(\theta):=\mathbb E[\ell(\theta,x)]$$

The following provides a lower-bound, given a sample $X$ from the population $x$ and a parameter space $\Theta$:

$$R(\theta)\leq\mathbb E[\ell(\theta,X)] + C_1 \sqrt\frac{\operatorname{Var}(\ell(\theta,X))}{n}+\frac{C_2}{n} \quad \forall \quad \theta \text{ in } \Theta$$

According to authors, the bound can be interpreted in regards to the bias-variance tradeoff. They then use this bound to define a robustly regularized risk.

  • $\begingroup$ It looks that the paper is not assessing an estimator but rather to estimate $\theta$ given $X$ using risk. $\endgroup$
    – Tan
    Nov 4, 2020 at 19:28
  • $\begingroup$ @Tan true, but the lower bound is still true (given the constraints), even though it's not the main point of the paper. $\endgroup$
    – Firebug
    Nov 4, 2020 at 19:30

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