# Problem:

Assume that we want to estimate $$f(\theta)$$ with a pre-specified strictly increasing function $$f$$ and a parameter $$\theta$$.

Let $$\hat{\theta}_1$$ and $$\hat{\theta}_2$$ be unbiased estimators for $$\theta$$. My question is to compare which estimator is better for $$f(\theta)$$:

1. $$f\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2}\right)$$, or
2. $$\frac{f(\hat{\theta}_1) + f(\hat{\theta}_2)}{2}$$.

# My approach:

My approach is taking an expectation on both based on Taylor's expansion of $$f$$, provided that they are well-defined.

1. $$\mathbb{E}f\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2}\right) = f(\theta) + \frac{f''(\theta)}{2}\text{Var}\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2}\right) + \ldots$$

2. $$\mathbb{E} \left[\frac{f(\hat{\theta}_1) + f(\hat{\theta}_2)}{2} \right] = f(\theta) + \frac{f''(\theta)}{2}\left(\frac{\text{Var}(\hat{\theta}_1) + \text{Var}(\hat{\theta}_2)}{2}\right) + \ldots$$

Here, the first-order disappears in each expectation because $$\hat{\theta}_1$$ and $$\hat{\theta}_2$$ are unbiased.

Let's further assume $$f''(\theta)>0$$. Then, since $$\label{varineq} (0 \le) \text{Var}\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2}\right) \le \frac{\text{Var}(\hat{\theta}_1) + \text{Var}(\hat{\theta}_2)}{2},$$ the first estimator (i.e. a function of averages) seems less distant from $$f(\theta)$$, as long as the remaining terms (order $$\ge 3$$) are ignored. The last inequality comes from a simple observation: $$\frac{\text{Var}(\hat{\theta}_1) + \text{Var}(\hat{\theta}_2)}{2} - \text{Var}\left(\frac{\hat{\theta}_1 + \hat{\theta}_2}{2}\right) = \dfrac{1}{4}\left( \text{Var}(\hat{\theta}_1) + \text{Var}(\hat{\theta}_2) - 2\text{Cov}\left(\hat{\theta}_1, \hat{\theta}_2\right) \right) = \dfrac{\text{Var}(\hat{\theta}_1 - \hat{\theta}_2)}{4} \ge 0.$$

# Question:

I wonder if this is logical and there is another way to justify a better estimator.

# Edit:

Let's restrict the class of $$f$$ by the strictly increasing ones.

• An example would help, e.g.: Suppose someone gives you the mean $\theta_1$, median $\theta_2$, and size $n$ of a sample drawn from a normal parent $N(\mu,\sigma)$. What are your best estimates for $\mu$ and $\mu^2$? It should be possible to work this out using the joint distribution at math.stackexchange.com/questions/477115/… -- note the answers may well be weighted averages and the covariance may be significant. Mar 2, 2020 at 14:51
• Where do you obtain the last inequality about variances? It's not generally true. Counterexamples include the cases $\hat\theta_2 = -\hat\theta_1$ (for which the right hand side is zero) and when the $\hat\theta_i$ are independent (for which the right hand side is half the left hand side).
– whuber
Mar 2, 2020 at 14:54
• @whuber Oops, sorry for that. I used the opposite inequality. Now it is edited. Mar 2, 2020 at 19:43
• Bear in mind that if the function represents something other than a linear transformation of the inputs, the two will yield [different] answers to different questions. To get a sense of that, it may help to read my answer here: Difference between generalized linear models & generalized linear mixed models. Mar 2, 2020 at 19:57
• @gung-ReinstateMonica Thanks! Then, let's keep our focus on the strictly increasing $f$ (see my edit). Does this make sense to you? Mar 2, 2020 at 20:19

Suppose $$\theta_1$$ and $$\theta_2$$ are unbiased estimators of $$\mu$$ with a bivariate normal distribution.

Let $$\sigma_1 = k \sigma_2$$, with $$k<1$$, so $$\theta_1$$ is the more efficient estimator.

Let the correlation be $$r$$, with $$-1.

We seek the least-variance linear estimates for $$\mu$$ and $$\mu^2$$. The results are:

\begin{align} \min_{a,b}Var[\mu-(a\theta_1+b\theta_2)] \text{ has }&\mu \sim \frac{(1-kr)\theta_1+(k^2-kr)\theta_2}{1-2kr+k^2}\\ \min_{c,d,e}Var[\mu^2-(c\theta_1^2+d\theta_1\theta_2+e\theta_2^2)] \text { has }&\mu^2 \sim \left(\frac{(1-kr)\theta_1+(k^2-kr)\theta_2}{1-2kr+k^2}\right)^{\!2} \end{align}

So for functions $$f$$ that are quadratic or well approximated by quadratics, if $$\frac{\theta_1+\theta_2}{2}$$ is the best linear estimate for $$\mu$$, then $$f\left(\frac{\theta_1+\theta_2}{2}\right)$$ is the best linear estimate for $$f(\mu)$$. In general an appropriate weighted average inside $$f$$ works even better.

Example: $$\theta_1$$ is the mean of a sample from a normal parent with mean $$\mu$$, and $$\theta_2$$ is the median from the same sample. Then $$k=r=\sqrt{2/\pi}$$, and the above formula shows that the best estimate for $$\mu$$ is $$\theta_1$$, and the best estimate for $$\mu^2$$ is $$\theta_1^2$$, i.e. ignoring the median.

Example: $$\theta_1$$ is the mean of a sample from a normal parent with mean $$\mu$$, and $$\theta_2$$ is the median of a different sample of the same size. Then $$k=\sqrt{2/\pi}$$, $$r=0$$, so the best estimate for $$\mu$$ is $$(\pi\theta_1+2\theta_2)/(\pi+2)$$, or $$.611\theta_1 + .389\theta_2$$, and the best estimate for $$\mu^2$$ is the square of that.

Example: $$\theta_1$$ is the mean of a sample from a normal parent with mean $$\mu$$, and $$\theta_2$$ is the mean of a different sample of the same size. Then $$k=1$$, $$r=0$$, so the best estimate for $$\mu$$ is $$(\theta_1+\theta_2)/2$$, and the best estimate for $$\mu^2$$ is $$\left(\frac{\theta_1+\theta_2}2\right)^{\!2}$$.

• Thanks for your answer! I was more curious about general conditions of $f$ or the estimators, but this also helps. Mar 4, 2020 at 14:10
• What functions $f$ do you have in mind that aren't well approximated by quadratics, or what estimators do you have in mind that aren't distributed normally? Mar 4, 2020 at 14:59
• @inmybrain, if this is worthy of thanks, it is also worth an upvote, or a clearer specification of what situations you care about that it doesn't cover? Mar 5, 2020 at 17:28