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Probability function question

  • Thread starter jk_zhengli
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  • #1
Hi guys, I have a question.

E(|X|) < infinity iff E|X|I(|X| > n) -> 0 as n goes to infinity, where I is the indicator function.


=> this direction is easy and I have it solved.

I wonder if anyone has any idea of how to deal with <=. Thanks.
 
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Answers and Replies

  • #2
Ray Vickson
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Hi guys, is the following statement true?

E(|X|) < infinity iff I(|X| > n) -> 0 as n goes to infinity.


=> this direction is easy and I have it solved. I wonder if <= is true? I think the additional assumption of X is bounded would make it true, but this is condition necessary? Thanks.
What do you mean by I(|X| > n) ? Is it P{|X| > n}, or something else? If that is what you mean, I think the result is false: P{|X| > n}→ 0 is true for ANY random variable on ℝ, because we have P{-n ≤ X ≤ n}→ 1 as n → ∞ for any real-valued X. In order to have E |X| < ∞ you need P{|X| > n} to go to zero quickly enough. For example, the discrete random variable X with probability mass function p(n) = c/n2, n = 1,2,3, ... has E X = ∞. Note, however, that it is not necessary to have X bounded, because many familiar random variables are unbounded but have finite expectations (or even finite moments of all orders).

RGV
 
  • #3
Mute
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What is I(|X| > n)? The probability that the absolute value of X is greater than n? If that's the case, have you heard of the Cauchy distribution?

If not, what is I(|X| > n)?
 

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