0
$\begingroup$

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 ??

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

Update:

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 $\Omega$

Original:

$|X|$ 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 $\int |X|~\mathrm{d} \mathbb{P} = \sup\limits_{\text{such } \varphi}\int \varphi~\mathrm{d} \mathbb{P}\leq \sum\limits_{i=0}^{\infty} (n+1) \mathbb{P}(A_n)$

$\endgroup$
  • $\begingroup$ what about my attempt $\endgroup$ – nikola Jan 4 '18 at 14:22
0
$\begingroup$

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}

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

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.