I have the following expectation
$E[\bar{X}^2 \cdot \mathbb{1}\{|\sqrt{n}\bar{X}|\geq 1.96\} ]$
where $\bar{X}$ is the sample mean of $x_1, \cdots, x_n$ and $x_i\sim iid$
My doubt is wether this is equivalent to
$E[\bar{X}^2]\cdot Pr(|\sqrt{n}\bar{X}|\geq 1.96)$
I'm not sure how to account for the fact that $\bar{X}$ and $\mathbb{1}\{|\sqrt{n}\bar{X}|\geq 1.96\}$ are correlated.
Thanks in advance for your help!
In general, $\mathbb E\left[Y^2\mathbf 1\left\{Y\geqslant R\right\}\right]$ is not the same as $\mathbb E\left[Y^2\right]\mathbf P\left\{Y\geqslant R\right\}$ for a non-negative square integrable random variable. For example, when $R$ is large, we are sure that $\mathbb E\left[Y^2\right]\mathbf P\left\{Y\geqslant R\right\}$ decays at least as $R^2$ where $R$ goes to infinity, but the decay of $\mathbb E\left[Y^2\mathbf 1\left\{Y\geqslant R\right\}\right]$ can be arbitrarily slow. For example, if we choose $Y$ such that $\mathbb E\left[Y^2\log\left(1+|Y|\right)\right]=+\infty$, then the decay cannot better than $ 1/\log R$.