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What is the correct definition of bounded probability density function:

  1. $\sup_{x} f(x)<\infty$. If this is the correct definition of bounded probability density function, can you give the example of a "well-known" probability density function which is not bounded?

  2. An unbounded probability distribution has support on all real numbers. In this case, he Normal distribution would be unbounded, for instance. Instead, the uniform distribution would be bounded.

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    $\begingroup$ 'When I use a word,’ Humpty Dumpty said in rather a scornful tone, ‘it means just what I choose it to mean — neither more nor less.’ So it depends on the author's definition. I might naturally read the words as meaning (1) since they are talking about a density function and that is how I interpret a bounded function, while I might describe a random variable with bounded support in (2) as a "bounded random variable", but others are free to define their meanings as they wish so long as they are clear. $\endgroup$
    – Henry
    Commented Dec 20, 2022 at 16:49
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    $\begingroup$ A bounded function in mathematics is always (1), but we don't know in what context you found this. As an example for an unbounded probability density, take the beta with either $\alpha<1$ or $\beta<1$ or both. $\endgroup$ Commented Dec 20, 2022 at 16:59
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    $\begingroup$ A site search for unbounded density picks up some good examples (among many other hits concerning unbounded support). By distinguishing a random variable or a distribution function from a density function -- all three are different kinds of things, albeit related -- you will see what the answer to the question must be. What makes this interesting is the fact that there exist probability density functions defined and nonzero everywhere that are unbounded in every nonempty open set, no matter how small! $\endgroup$
    – whuber
    Commented Dec 20, 2022 at 17:29

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Here is an example of an unbounded density function from the Gamma distribution:

$$f(x) = \text{Ga}(x|\tfrac{1}{2},1) = \frac{1}{\sqrt{\pi x}} \cdot \exp(-x) \quad \quad \quad \quad \quad \text{for } x > 0.$$

For this density function we have $\lim_{x \downarrow 0} f(x) \rightarrow \infty$ so $\sup_x f(x) = \infty$.

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