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In the paper On the Bootstrap of the Sample Mean in the Infinite Variance Case by Keith Knight, on page 1170 at the bottom of the page before the theorem, the author mentions that the random variable X is well-defined with probability 1 since E(X) and E(X^2) exists.

I did not understand their fragment, but I concluded that I may say that a random variable X is said to be well-defined with probability 1 if E(X) and E(X^2) exists? If that was the case may anyone give me a reference that state this explicitly?

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    $\begingroup$ Sounds strange. How can there be an expectation of a random variable that is not "well-defined"? Maybe you should mention (in your question, not in a comment) what exactly is meant in this context with "well-defined". $\endgroup$
    – drhab
    Commented Oct 1, 2014 at 9:11
  • $\begingroup$ What @drhab said. Or indicate the source. $\endgroup$
    – Did
    Commented Oct 1, 2014 at 9:12
  • $\begingroup$ The paper I am currently reading is "On the Bootstrap of the Sample Mean in the Infinite Variance Case" that of Keith Knight and this is found on p. 1170 at the bottom of the page before the theorem $\endgroup$
    – Anonymous
    Commented Oct 1, 2014 at 9:14

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they mentioned that the random variable X is well-defined with probability 1 since E(X) and E(X^2) exists.

Not at all--instead, the authors mention that some random variable defined as $$\sum_kX_k$$ is "well-defined" because $E(X_k)=0$ for every $k$ and $\sum\limits_kE(X_k^2)$ converges. Additional independence assumptions are available, which make it direct that, indeed, under these two conditions, the series converges almost surely.

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  • $\begingroup$ but in this case wouldn't be the case that the random variable Y=sum(X) be well defined because E(Y)=sum(E(X))=0 and var(Y)< infinity since var(y)=sum(var(X))=sum(E(X^2)) which converge and hence exists ?!! or Am I interpreting in a wrong way? $\endgroup$
    – Anonymous
    Commented Oct 2, 2014 at 16:41
  • $\begingroup$ You seem to be repeating verbatim what is in the answer, no? $\endgroup$
    – Did
    Commented Oct 2, 2014 at 20:12
  • $\begingroup$ @ Did thank you very much $\endgroup$
    – Anonymous
    Commented Oct 4, 2014 at 10:19

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