Given $A_n =\{\omega\in\Omega: n \leq |X(\omega)|< n+1\} $. I need to show $\sum_ {n=1}^{\infty}n p(A_n)$ $\leq$ $E|X| \leq \sum_ {n=0}^{\infty} (n+1)P(A_n)$

I try by begin with ${n \leq |X(\omega)|< n+1} $ with taking integral to all side

$$\int_\Omega n \,dP \leq \int_\Omega|X|\,dP\leq \int_\Omega( n+1 )dP$$

$$\int_{A_n} n dP + \int_{A^c_n}n dP \leq\int_{A_n}(n+1)dP+\int_{A^c_n}(n+1)dP$$

what about this integral its equal zero $\int_{A^c_n}(n+1)dP$ \, $\int_{A^c_n}ndP$.

now how I can transform integration to summation ??

  • $\begingroup$ What are $A_{n}$ and x($\omega$) and $E|x|$? $\endgroup$ – Mostafa Ayaz Jan 4 '18 at 13:56
  • $\begingroup$ $A_n$ is aset that defined above and x is arandom variable and E|x| is the expectation of absolute value of x @MostafaAyaz $\endgroup$ – nikola Jan 4 '18 at 14:02
  • $\begingroup$ Do you mean $A_n = \{\omega : |X(\omega)| \in [n, n+1) \}$? $\endgroup$ – dEmigOd Jan 4 '18 at 14:03
  • $\begingroup$ Now, it at least make sense to me $\endgroup$ – dEmigOd Jan 4 '18 at 14:04
  • $\begingroup$ @dEmigOd yes that i mean $\endgroup$ – nikola Jan 4 '18 at 14:08

Using the fact that $\Omega = \bigcup A_n$ and $\{A_n\}$ are pairwise disjoint, $$ \int_\Omega |X| dP = \sum_{n=0}^\infty \int_{A_n} |X| dp \le \sum_{n=0}^\infty (n+1) P(A_n)$$

  • $\begingroup$ exchanging limits and integrals need to be justified $\endgroup$ – dEmigOd Jan 4 '18 at 14:26
  • $\begingroup$ this is easy answer ..... the idea is the union equal summation as disjoint thnx alot $\endgroup$ – nikola Jan 4 '18 at 14:31


Regarding your approach - i don't agree with it. Specifically 3rd row is wrong. As $|X|$ could be much larger, than some specific $n$ on almost all $\Omega$


$|X|$ is a non-negative variable, so at least integrable.

Define a sequence of $Y_n = \sum\limits_{i=0}^n (n+1) \cdot 1_{[n, n+1)}$ - simple functions. All are non-negative, and the sequence is non-decreasing.

Let $0 \leq \varphi \leq |X|$ be some bounded function with finite support, then $\exists N, \forall n \geq N$ $\varphi \leq Y_n$.

Which leads us to state $\int \varphi~\mathrm{d} \mathbb{P} \leq \int Y_n~\mathrm{d} \mathbb{P} = \sum\limits_{i=0}^n (n+1) \mathbb{P}(A_n)$

Since the above is right for all $n \geq N$, then $\int \varphi~\mathrm{d} \mathbb{P} \leq \lim\limits_{n \to \infty}\int Y_n~\mathrm{d} \mathbb{P} = \sum\limits_{i=0}^{\infty} (n+1) \mathbb{P}(A_n)$

As usual, such claim is right for any $\varphi$ which is $0 \leq \varphi \leq |X|$ bounded function with finite support, hence by definition $\int |X|~\mathrm{d} \mathbb{P} = \sup\limits_{\text{such } \varphi}\int \varphi~\mathrm{d} \mathbb{P}\leq \sum\limits_{i=0}^{\infty} (n+1) \mathbb{P}(A_n)$

  • $\begingroup$ what about my attempt $\endgroup$ – nikola Jan 4 '18 at 14:22

Let us start by defining $Y=|X|$. What you can do is to notice that $\lceil Y \rceil = \sum_{n=0}^\infty (n+1)\mathbf{1}_{\{n+1> Y\geq n\}}$ (verify this!). Moreover we have $Y \leq \lceil Y \rceil$. Hence by monotonicity we have: \begin{align} E[Y]\leq E[\, \lceil Y \rceil\, ] \end{align} Moreover by Fatou's Lemma we have: \begin{align} E[\, \lceil Y \rceil\, ] \leq \sum_{n=0}^\infty (n+1)P(n+1>Y\geq n) \end{align} We are there since $P(n+1>Y\geq n)=P(A_n)$ we can conclude: \begin{align} E|X|\leq \sum_{n=0}^\infty (n+1)P(A_n) \end{align}

  • $\begingroup$ In fact one has equality by using MCT, but Fatou is enough for this setting.. $\endgroup$ – Shashi Jan 4 '18 at 14:47

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