Proving upper bound for Bias of truncated sample mean

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.

• Is $\tau$ a real number or strictly positive number? Apr 4, 2023 at 11:27
• Yes I should of added this $\tau >0$ Apr 4, 2023 at 11:31
• Do you know anything else on the distribution of $X$ ? Like the mean or variance? Apr 4, 2023 at 11:43
• stats.stackexchange.com/search?q=markov+inequality. Chebyshev's Inequality is also helpful.
– whuber
Apr 4, 2023 at 12:49
• @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 "$E$". Apr 4, 2023 at 16:01

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:

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)

• I wonder if you can now show that the Bias is strictly bounded away from zero? Apr 26, 2023 at 10:35
• I have made a new question here now Apr 26, 2023 at 12:13