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

And I saw the inequality:

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

But I am not sure how this was derived.

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  • $\begingroup$ Is $\tau$ a real number or strictly positive number? $\endgroup$
    – lulufofo
    Apr 4, 2023 at 11:27
  • $\begingroup$ Yes I should of added this $\tau >0$ $\endgroup$
    – Dylan Dijk
    Apr 4, 2023 at 11:31
  • $\begingroup$ Do you know anything else on the distribution of $X$ ? Like the mean or variance? $\endgroup$
    – lulufofo
    Apr 4, 2023 at 11:43
  • $\begingroup$ stats.stackexchange.com/search?q=markov+inequality. Chebyshev's Inequality is also helpful. $\endgroup$
    – whuber
    Apr 4, 2023 at 12:49
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    $\begingroup$ @DylanDijk Thanks for your clarification. I added an answer below. Only rank ordering of the two functions is needed -- it is simpler than Markov or Chebyshev as the inequality of interest actually does not contain "$P$", but only "$E$". $\endgroup$
    – Zhanxiong
    Apr 4, 2023 at 16:01

1 Answer 1

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First note that \begin{align*} & |E[(X - \operatorname{sign}(X)\tau)I(|X| > \tau)]| \\ =& |E[(X - \tau)I(X > \tau) + (X + \tau)I(X < -\tau)]| \\ \leq & E[|(X - \tau)I(X > \tau) + (X + \tau)I(X < -\tau)|]. \end{align*}

Now note the function $f(x) = |(x - \tau)I_{(\tau, \infty)}(x) + (x + \tau)I_{(-\infty, -\tau)}(x)|, x \in \mathbb{R}$ is dominated by the function $g(x) = x^2/\tau, x \in \mathbb{R}$ (draw a picture). The inequality then follows by taking "$E$" on both sides of the inequality $f(X) \leq g(X)$.


When $\tau = 2$, the graphs of $f(x)$ and $g(x)$ are shown as follows:

enter image description here

f <- function(x, tau) {
  abs((x - tau) * (x > tau) + (x + tau) * (x < -tau))
}

g <- function(x, tau) {
  x^2 / tau
}

x <- seq(-5, 5, len = 1000)
y1 <- f(x, 2)
y2 <- g(x, 2)

plot(x, y1, type = 'n', xlab = '', ylab = '', ylim = c(0, 3))
lines(x, y1, lty = 1)
lines(x, y2, lty = 2)
legend('bottomright', c("f(x)", "g(x)"), lty = 1:2)
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  • $\begingroup$ I wonder if you can now show that the Bias is strictly bounded away from zero? $\endgroup$
    – Dylan Dijk
    Apr 26, 2023 at 10:35
  • $\begingroup$ I have made a new question here now $\endgroup$
    – Dylan Dijk
    Apr 26, 2023 at 12:13

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