Post-selection unbiased estimator

selective inference
Does a post-selection unbiased estimator for the mean of a normal distribution exist?

Kenneth Hung


December 8, 2023

Suppose we are interested in the mean \(\mu\) of a normal distribution \(N(\mu, 1)\), using only a sample of \(X\) drawn from this distribution. The answer is straightforward. We can just use \(X\), and it would be unbiased. But what if we only observe \(X\) when \(X\) exceeds some threshold \(c\), e.g. when \(X > c\)? In other words, we want to construct some estimator \(\delta(X)\) such that \(\mathbb{E}_\mu[\delta(X) \mid X > c] = \mu\). One way to see why we condition on \(\{X > c\}\) is that the sampling distribution of \(X\) is truncated at \(c\).

The setting here might seem highly stylized, but has practical implications. For example suppose we only look at experiments that were statistically significant and positive. We know that the estimate will be biased, but can we somehow debias it in a frequentist sense?

A more general form of this question was given to me as a class project in Stat 212 by Will. Specifically, we can also ask if there exists a \(\delta(X)\) such that \(\mathbb{E}_\mu[\delta(X) \mid X \in A] = \mu\) for some general measureable set \(A\). I did not have an answer to that general question, but I do have a negative answer to this particular form of selection (\(X > c\)).

Proof by contradiction

Suppose such \(\delta(X)\) exists. Note \[\mathbb{E}_\mu[\delta(X - c + c) - c \mid X - c > 0] = \mu - c\] and \(X - c \sim N(\mu - c, 1)\). So without loss of generality that \(c = 0\), otherwise by replacing \(\delta(X)\) with \(\delta(X + c) - c\), we can assume \(c = 0\).

Now consider the function \(\tilde\delta(X) = 1_{\{X > 0\}} \delta(X) - 1_{\{X < 0\}} \delta(-X)\), which is kind of like gluing together \(\delta\) and a reflected version of itself. The function \(\tilde\delta\) will have some interesting properties. For example, \[\mathbb{E}_\mu[\tilde\delta(X) \mid X > 0] = \mathbb{E}_\mu[\delta(X) \mid X > 0] = \mu,\] and \[\mathbb{E}_\mu[\tilde\delta(X) \mid X < 0] = \mathbb{E}_\mu[-\delta(-X) \mid X < 0] = -\mathbb{E}_{-\mu}[\delta(X) \mid X > 0] = \mu.\] So all around, we have \(\mathbb{E}_\mu[\tilde\delta(X)] = \mu\).

Now we pull out some results from mathematical statistics! Normal location family is an exponential family of full rank, with \(X\) being the sufficient statistic. So a normal location family is also complete. Since \(\mathbb{E}_\mu[\tilde\delta(X) - X] = \mu - \mu = 0\), by the definition of completeness, we must have \(\tilde\delta(X) = X\) and \(\delta(X) = X\).

On the other hand, we can always check on Wikipedia that \(\mathbb{E}_\mu[X \mid X > 0] = \mu + \frac{\phi(-\mu)}{1 - \Phi(-\mu)}\), where \(\phi\) and \(\Phi\) are the p.d.f. and c.d.f. of a normal distribution respectively. This is definitely not \(\mu\) as we assumed before, yielding a contradiction.

An interesting thing, that I did not notice at the time of taking Stat 212, is that this proof does not apply to randomized selection, i.e. finding a \(\delta(X)\) such that \(\mathbb{E}_\mu[\delta(X) \mid X + \epsilon > 0] = \mu\) where \(\epsilon\) is some independent Gaussian noise, which is very much possible (and you should try it yourself)!