Is truncated mean a biased estimator

We have data $$X_1, \dots, X_n$$ which are i.i.d copies of $$X$$. Where we denote $$\mathbb{E}[X] = \mu$$, and $$X$$ has finite variance.

We define the truncated sample mean:

\begin{align} \hat{\mu}^{\tau} := \frac{1}{n} \sum_{i =1}^n \psi_{\tau}(X_i) \end{align}

Where the truncation operator is defined as:

\begin{align} \psi_{\tau}(x) = (|x| \wedge \tau) \; \text{sign}(x), \quad x \in \mathbb{R}, \quad \tau > 0 \end{align}

The bias for this truncated estimator is then defined as:

Bias $$:= \mathbb{E}(\hat{\mu}^{\tau}) - \mu$$

In previous question it is shown that we can upper bound the truncation

\begin{align} |\text{Bias}| = |\mathbb{E}[(X - \text{sign}(X)\tau) \mathbb{I}_{\{|X| > \tau\}}]| \leq \frac{\mathbb{E}[X^2]}{\tau} \end{align}

I was now wondering if it can be shown that:

\begin{align} 0 < |\text{Bias}| \end{align}

That is, is the truncated mean estimate biased?

• It depends on the distribution of $X$ and the choice of $\tau$. I believe in most cases it is biased, which can be verified by some simple simulations. Apr 26, 2023 at 12:47
• What does the AND operator $x\land y$ mean for real numbers $x$ and $y$? Bitwise AND? How is this related to "truncation"? Apr 26, 2023 at 13:29
• @cdalitz In probability theory, for $x, y \in \mathbb{R}$, "$x \wedge y$" is a commonly used shorthand for $\min(x, y)$. Similarly, $x \vee y$ means $\max(x, y)$. Apr 26, 2023 at 14:26

To elaborate my comment, consider first that $$X \sim U(-2, 2)$$ and $$\tau = 1$$. In this case $$E[X] = 0$$ and $$E[\psi_\tau(X)] = 0$$, hence the bias is $$0$$.
On the other hand, if $$X \sim U(-1, 2)$$ and $$\tau = 1$$, then $$E[X] = \frac{1}{2}$$ but $$E[\psi_\tau(X)] = \frac{1}{6}$$, which gives the bias of $$-\frac{1}{3}$$.
Therefore, if you are faced with a non-trivial statistical inference problem, I tend to believe $$\hat{\mu}^\tau$$ is biased (as unbiasedness requires $$E_\mu(\hat{\mu}^\tau) = \mu$$ for every distribution in the parametric family). To see this more clearly, I suggest you tabulating a table of biases for the family of normal distributions $$\{N(\mu, 1): \mu > 0\}$$. Of course, the question may have a more definitive answer if you can restrict the underlying distribution family to a specific one.
• @DylanDijk That's true. But as I stated, then it is not a "non-trivial" estimation problem. If a distribution is symmetric with mean $0$, then its mean is certainly $0$, there is nothing to infer about. But as a probabilistic statement, that's fine. Apr 26, 2023 at 14:37