# Example when globally unbiased estimator does not exist while locally unbiased estimator exists?

The locally unbiased(l.u.) estimator $$\hat{\theta}\left( x \right)$$, with $$x$$ stands for the experiment result, refers to the estimator that satisfies(see Eq(5) of this paper for multiparameter case) $$\sum_x{p\left( x|\theta =\varphi \right) \hat{\theta}\left( x \right)}=\varphi ,\sum_x{\partial _{\theta}p\left( x|\theta \right) |_{\theta =\varphi}\hat{\theta}\left( x \right)}=1$$ where I have used $$\varphi$$ to stands for true value. It is a weak version of globally unbiased, i.e. unbiased at any possible true value of $$\theta$$. The aim of the l.u. estimator is that there might be cases when a globally unbiased estimator does not exist. So I want to know if there is some specific example that we cannot find a globally unbiased estimator while we can find a l.u. estimator?

• A trivial example might be (I am not so sure about the second condition) $\hat\theta(x)=7$ when $\varphi=7$, i.e., the trivial estimator that does not consider the data and happens to be "lucky" and one true parameter value. Mar 28 at 9:33
• @ChristophHanck Thanks for the comment. In this case, the estimator does not satisfy the second condition since $\sum_x{\partial _{\theta}p\left( x|\theta \right) |_{\theta =\varphi}7}=\partial _{\theta}\sum_x{p\left( x|\theta \right) |_{\theta =\varphi}7}=0$. The partial derivative is used to require the estimator not to act so weirdly at least around the true value. I'm not sure if this l.u. normally used in statistical inference, but in the field named quantum metrology, this condition is used a lot. Mar 28 at 10:04

You can always construct a locally unbiased estimator at a point $$\varphi$$ from an arbitrary (non constant) estimator by shifting and scaling.
suppose that $$\tau(x)$$ is some estimator, then define $$\hat \theta(x)= \varphi + \alpha( \tau(x) - E_\varphi[\tau(x)])$$.
Clearly $$E_\varphi[\hat\theta(x)] = \varphi$$, and
$$\partial_\theta E_\theta[\hat \theta(x)]_{\theta=\varphi} =\alpha \partial_\theta E_\theta[\tau(x)]_{\theta=\varphi}$$.
So by choosing $$\alpha=1/\partial_\theta E_\theta[\tau(x)]_{\theta=\varphi}$$ the second condition is satisfied (assuming the derivative of the expectation at $$\varphi$$ is not zero).