So we have 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:
$\begin{align} \psi_{\tau}(x) = (|x| \wedge \tau) \; \text{sign}(x), \quad x \in \mathbb{R} \end{align}$
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.
The assumptions on the data are that they are i.i.d and have finite variance.
Here's my attemp.
$$ \begin{equation} \begin{split} |\text{Bias}| &\le \frac{1}{n} \; \mathbb{E}\sum_{i=1}^{n}\bigg|\tau \; \text{sign}(X_i) - |X_i| \bigg| \; \mathbb{I}_{\{ |X_i| > \tau \}} \\ &\le \frac{1}{n} \; \mathbb{E}\sum_{i=1}^{n} \bigg(\tau \; \mathbb{I}_{\{ |X_i| > \tau \}} + |X_i| \mathbb{I}_{\{ |X_i| > \tau \}} \bigg) \\ &= \tau \; \mathbb{P}(|X| > \tau) + \frac{\text{cost}}{n} \qquad \text{for n large enough}\\ &\le \tau \; \mathbb{P}(|X| > \tau) \\ &\le \frac{\mathbb{E}[X^2]}{\tau} \end{split} \end{equation} $$
where in the last inequality I used the Generalized Markov's inequality with the second moment