$X$ is a random variable and $E|X|=1$. Do we know any relation between $P(|X|\leq M)$ and $E[|X| \cdot 1\{|X| \leq M \}]$? Do we only have one direction, i.e., either $P(|X|\leq M) \geq E[|X| \cdot 1\{|X| \leq M \}]$ or $P(|X|\leq M) \leq E[|X| \cdot 1\{|X| \leq M \}$? Or in fact, both cases can happen? EDIT: Especially when $M \to \infty$, what would be the asymptotic comparison?
2026-03-29 03:23:06.1774754586
Compare $E[|X| \cdot 1\{|X| < M\}]$ and $P(|X| < M)$ with $E|X|=1$
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(Chebyshev's Association Inequality) $f$ is nonincreasing and $g$ is nondecreasing. $X$ is a real-valued random variable and $Y$ is a nonnegative random variable. Then $$ E[Y] E[Yf(X)g(X)] \leq E[Yf(X)] E[Yg(X)] .$$ Take $Y \equiv 1$, $f(\cdot) = 1\{|X|\leq \cdot\}$ and $g(\cdot) = |\cdot|$. The result would be similar to Did's comment.