Talk:Singular distribution

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"A singular distribution is not a discrete probability distribution because each discrete point has a zero probability" Is this correct?I think each discrete point has nonzero probability. —The preceding unsigned comment was added by Srgvie (talk • contribs).

Well, taking the Cantor distribution as an example, each point indeed has a zero probability. Indeed the definition "a singular distribution is a probability distribution concentrated on a measure zero set where the probability of each point in that set is zero" makes this clear: every point where the distribution is concentrated has zero probability. --Henrygb 21:02, 4 February 2007 (UTC)Reply

Actually, there is a little confusion here, at least for me. So let's make the things clear: Discrete probability distribution: each point has a nonzero probability. Singular distribution:each point has a zero probability. Ok? —Preceding unsigned comment added by Srgvie (talk • contribs) 21:00, 6 April 2008 (UTC)Reply

I am not convinced that the requirement that the distribution be non-vanishing only on a set of measure zero is germane/important/correct. One example is the CDF built from the fat Cantor set, which I think qualifies as a singular distribution w.r.t. the Lebesgue measure, but clearly is singular on a set of measure greater than zero. I give two more examples on the talk page for singular function, which appears to have the same 'measure-zero' problem ... linas (talk) 05:40, 11 February 2010 (UTC)Reply

In two dimensions, the lower bound looks to have the upper right quadrant of the unit square as a subset of its set of concentration. Similarly, it looks like the whole unit square is a subset of the upper bound's set of concentration. Since these have nonzero Lebesgue measure, I recommend removing the Frechet Hoeffding bounds as examples. Can someone show me my error? Analytical Steven (talk) 11:31, 11 June 2025 (UTC)Reply