I have 2 questions:

  1. Suppose I have a finite mean but an infinite variance for a discrete distribution w/support $\{1,2,\dots\}$. Is there any probability inequality tighter than Markov in this case? The motivating example is a Yule-Simon with parameter $\lambda \in (1,2)$.

  2. Suppose now the mean is infinite but a fractional moment may exist, $\mathbb{E}(X^\alpha) <\infty$, for $0<\alpha<1$. The distribution is still discrete with same support as question 1 above. In this case Markov will not apply but something else might. What would one typically use for an ineuqality in this scenario?

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    $\begingroup$ You switch between "mean ... is infinite" and "mean is finite but variance is infinite...", you should probably clarify and pick one of those two, as, for example, Markov applies in the latter case to nonnegative random variates but not in the former case. $\endgroup$ – jbowman Sep 27 '18 at 3:15
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    $\begingroup$ @jbowman has a good point - if the variance is infinite and expectation is finite you have some results to draw from, but if the random variable is not integrable you are not able to even state many concentration results, which tend to be phrased in terms of bounds on $P \left( | X - \mathbb{E}(X)| > t \right)$. $\endgroup$ – Titus Sep 27 '18 at 3:39
  • $\begingroup$ @jbowman, good point. I was hoping for some general inequalities that hold with infinite variances, the Markov obviously still applies here but not Chebyshev. I'd rather learn something from any answer to the post so I'll phrase as mean finite, variance infinite, and not any of the 3: Chebyshev, Chernoff, or Markov. $\endgroup$ – Lucas Roberts Sep 27 '18 at 12:55
  • $\begingroup$ @LucasRoberts I don't follow your response to jbowman. If you're choosing mean finite, Markov would apply, would it not? $\endgroup$ – Glen_b Sep 27 '18 at 16:17
  • $\begingroup$ @Glen_b, yes the Markov inequality provides a bound if the first moment is finite that is why I rephrased the question to read "not Markov,..." is it still not clear? or maybe is your question, why wouldn't you use Markov in this case? $\endgroup$ – Lucas Roberts Sep 27 '18 at 17:04

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