
> 
> Suppose that $X$ is an unbounded random variable such that $\operatorname EX=0$ and $\operatorname E|X|^q=1$ with some $q>2$. Only the distribution of $X$ matters, so the probability space can be chosen freely. Choose any $p\in(0,1)$. I want to prove that there exists an event $F$ such that
> 
> 
> 1. $P(F)=p$;
> 2. $|X'|\le p^{-1/q}$, where $X'=X\chi\_{F^c}+p^{-1}\operatorname E[X\chi\_F]\chi\_F$.
> 
> 
> How can I prove that such an event exists?
> 
> 
> 


Since $|X'|$ has to be $\le p^{-1/q}$, this means that $|X(\omega)|\le p^{-1/q}$ when $\omega\in F^c$. Using Jensen's inequality and the assumption that $\operatorname E|X|^q=1$, $|p^{-1}\operatorname E[X\chi\_F]|\le p^{-1/q}$.


As a starting point, consider $G=\{|X|>p^{-1/q}\}$. Clearly, $|X(\omega)|\le p^{-1/q}$ when $\omega\in G^c$. However, using Markov's inequality and the fact that $\operatorname E|X|^q=1$, we only have that $P(G)\le p$. So $G$ only satisfies one of the required conditions.


Consider $\tilde G$ which is obtained by taking some $\omega$'s from $G^c$ and putting them into $G$ so that $G\subset\tilde G$. Since $\tilde G^c\subset G^c$, we have that $|X(\omega)|\le p^{-1/q}$ when $\omega\in\tilde G^c$. Since $G\subset\tilde G$, $P(\tilde G)\ge P(G)$. So perhaps if we take, loosely speaking, the right amount of $\omega$'s from $G^c$ and put them into $G$, we might have that $P(\tilde G)=p$ and $|X(\omega)|\le p^{-1/q}$ when $\omega\in\tilde G^c$. Is it possible to make this argument rigorous?


Maybe there is a better and simpler way to prove the existence of such an event. Any help is much appreciated.


P.S. This is a part of a very nice [answer](https://mathoverflow.net/a/310232/46211) that I have received previously and I am trying to understand every part of it.



Existence of certain event
For real $a$, let $Q(a):=P(|X|>a)$. Then the function $Q$ is right-continuous. So, 
\begin{equation\*}
 b:=a\_p:=\inf\{a\in\mathbb R\colon Q(a)\le p\}=\min\{a\in\mathbb R\colon Q(a)\le p\}\in\mathbb R 
\end{equation\*}
and hence 
\begin{equation}
 P(|X|>b)=Q(b)\le p\le Q(b-)=P(|X|\ge b)\le1/b^q, \tag{1}
\end{equation}
by Markov's inequality, so that 
$$b\le p^{-1/q}.$$ 
Also, (1) implies that 
\begin{equation}
 P(|X|=b)=P(|X|\ge b)-P(|X|>b)\ge p-P(|X|>b)\ge0. \tag{2}
\end{equation}
Since the probability space can be chosen freely, assume that it is non-atomic. Then, by (2), there is an event $H\subseteq\{|X|=b\}$ such that $P(H)=p-P(|X|>b)$. Let $F:=H\cup\{|X|>b\}$. Then 
$$P(F)=p$$ 
and $|X|\le b\le p^{-1/q}$ on $F^c$. Also, by Hölder's inequality, 
$\frac1p\,|EX\chi\_F|\le \frac1p\,P(F)^{1-1/q}=p^{-1/q}$. 
Thus, your conditions 1 and 2 both hold. 


