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.

  • 1
    $\begingroup$ 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 $P(F) = q$. $\endgroup$ – Dieter Kadelka Jan 14 at 12:47
  • $\begingroup$ @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. $\endgroup$ – 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 \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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.