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From the definition, the probability density function of truncated distribution is

$$ f_{a,b}(x) = \frac{f(x)}{F(b) - F(a)} $$

Now, the truncation interval is referred $(a, b]$, e.g. in Wikipedia, but sometimes as $(a,b)$, or even as $[a,b]$.

Why does the ambiguity arise? Is this just a lousy use of notation, since in each case they seem to be using the same definition? What is the proper interval? This seems to be too trivial to lead to any ambiguity...

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Tim
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1 Answers1

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If the probability density function (pdf) $f(x)$ includes atoms (point masses) as well as regions where $f(x)$ is continuous, then how we refer to the truncation interval matters. If the random variable is absolutely continuous and there are no point masses, then the details of the truncation interval do not matter as much.

Dilip Sarwate
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  • (+1) Agree, but the problems arise in discrete distributions. – Tim Dec 14 '16 at 15:37
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    @Tim if thats the case we must appeal to the fact that cdfs are CADLAG. So the interval would *include* b and *exclude* a from the support unless you changed the denominator by adding or subtracting the point mass probabilities. – AdamO Dec 14 '16 at 15:43
  • @AdamO thanks, again, this is clear for me, but the question is about the source of ambiguity and it's consequences. – Tim Dec 14 '16 at 15:45
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    @Tim Yes, and in your own answer that you cited in the first line of your question, you very carefully specify $(a,b]$ as the interval of truncation, and your work there matches what Wikipedia says. So, do you have any citations for where the interval is _not_ a semiopen interval? The formula _can_ be fudged to take into account the cases $(a,b), [a,b)$ and $[a,b]$ for discrete or mixed random variables, as AdamO also tells you. – Dilip Sarwate Dec 14 '16 at 15:46
  • @DilipSarwate you're right, maybe it's splitting hair, it seems that all the "ambiguous" intervals were mentioned when continuous distributions were described, or when it was not stated directly if it is a general case or continuous distribution. – Tim Dec 14 '16 at 16:07
  • @Tim nobody has said you're splitting hair. We're all in agreement $(a,b)$ and $[a, b)$ are *wrong*. – AdamO Dec 14 '16 at 16:27