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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Zhanxiong
    Commented Apr 26, 2023 at 12:47
  • $\begingroup$ What does the AND operator $x\land y$ mean for real numbers $x$ and $y$? Bitwise AND? How is this related to "truncation"? $\endgroup$
    – cdalitz
    Commented Apr 26, 2023 at 13:29
  • $\begingroup$ @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 $\max(x, y)$. $\endgroup$
    – Zhanxiong
    Commented Apr 26, 2023 at 14:26

1 Answer 1


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.

  • 2
    $\begingroup$ Thank you, I guess that if a distribution is symmetric with mean zero then the truncated mean will always be unbiased. $\endgroup$
    – Dylan Dijk
    Commented Apr 26, 2023 at 14:32
  • $\begingroup$ @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. $\endgroup$
    – Zhanxiong
    Commented Apr 26, 2023 at 14:37
  • 1
    $\begingroup$ If the truncation is instead defined in terms of symmetric order statistics, as in trimmed means, the property stands for a distribution symmetric around its unknown expectation. $\endgroup$
    – Xi'an
    Commented Apr 26, 2023 at 17:01

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