# Existence of certain event

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 that I have received previously and I am trying to understand every part of it.

• If you consider events as above depending on $X$ then you have to assume that the distribution function of $X$ is continuous. Otherwise there need not exist any $F$ with $P(F) = q$. – Dieter Kadelka Jan 14 at 12:47
• @DieterKadelka Thank you for the comment. I revised my question to make the notation a little bit less ambiguous. I do not think that we necessarily need to assume that the distribution function of $X$ is continuous. Only the distribution of $X$ matters, so the probability space can be chosen freely. The event $F$ cannot be of the form $\{|X|>a\}$. As a starting point, I consider an event of this form and wonder if it is possible to modify such an event so that both of the conditions are satisfied. – Cm7F7Bb Jan 14 at 13:39

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 $$$$P(|X|>b)=Q(b)\le p\le Q(b-)=P(|X|\ge b)\le1/b^q, \tag{1}$$$$ by Markov's inequality, so that $$b\le p^{-1/q}.$$ Also, (1) implies that $$$$P(|X|=b)=P(|X|\ge b)-P(|X|>b)\ge p-P(|X|>b)\ge0. \tag{2}$$$$ Since the probability space can be chosen freely, assume that it is non-atomic. Then, by (2), there is an event $$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.