Redistributing probability mass Consider some canonical probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a random variable $X:\Omega\rightarrow[0,\infty)$ and let's assume that $X$ is unbounded, but has finite expectation.
That is, $X\in\mathcal{L}^{1}$, and for all $M\in[0,\infty)$ we have $\mathbb{P}(X \geq M)>0$.
I am wondering, if there is a way to construct a bounded random variable $\hat{X}$ with


*

*$\mathbb{E}[X] = \mathbb{E}[\hat{X}]$

*$\hat{X}(\Omega)\subset X(\Omega)$.


Intuitively, one should redistribute probability mass from the tail of the random variable into some bounded domain. But I expect some assumption on the underlying probability space for this to be possible.
Have you seen sothing similar or can you point me in the right direction? Any comments, ideas and suggestions are highly appreciated. Thank you!
 A: This is not really an answer to your question, but might lead to the conclusion that such $\hat X$ cannot always be found. In the sequel $\mathbb N$ denotes the set of positive integers.
Let  $(\Omega,\mathcal F,\mathbb P)=(\mathbb N,\wp(\mathbb N),\mathbb P)$ where $\mathbb P(\{n\})=p_n$. 
Let $X:\mathbb N\to[0,\infty)$ be prescribed by $n\mapsto n$. 
For $n=\mathbb N=\{1,2,\dots\}$ let $p_n>0$ with (of course) $\sum_{n=0}^{\infty}p_n=1$ and also $\sum_{n=0}^{\infty}np_n<\infty$.
Question: does there exist a finite partition $\{P_1,\dots, P_m\}$ of $\mathbb N$ such that  $$\sum_{k=1}^m\sum_{n\in P_k}kp_n=\sum_{n=0}^{\infty}np_n\tag1$$
The RHS of $(1)$ equals $\mathbb EX$.
The LHS of $(1)$ equals the expectation of $\hat{X}$ determined by $\hat X^{-1}(\{k\})=P_k$.
If you can find a case of non-existence of such partition then you have found an example where no such $\hat X$ can be found.
A: Provided that your $X$ is a continuous variable, I think that the simplest strategy would be to cut not only the tail but also the head of your distribution so that the expectation value is unchanged. Then you can concentrate the probability mass you have taken away exactly on the expectation value. This should satisfy your requests: $\hat X$ goes to some interval $[\epsilon,k]$ and is constructed so that $\mathbb{E}[\hat{X}] = \mathbb{E}[X]$.
