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In "Pattern recognition and machine learning" by Cristopher Bishop, Chapter 2.3.6 (pag. 100) says that

The gamma distribution has a finite integral if $a>0$, and the distribution itself is finite if $a \ge 1$

When a probability density function is defined to be finite? Can you make me an example of a finite pdf, and one of an infinite pdf?

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  • $\begingroup$ This statement from Bishop is incorrect in that the Ga$(a,1)$ density is finite on its support $(0,\infty)$ albeit unbounded when $0<a<1$. $\endgroup$ – Xi'an Apr 27 at 7:48
  • $\begingroup$ @Xi'an What do you mean by "finite in an (infinite) support"? $\endgroup$ – rtrtrt Apr 27 at 9:49
  • $\begingroup$ I did not write "finite in an (infinite) support"... $\endgroup$ – Xi'an Apr 27 at 10:08
  • $\begingroup$ @Xi'an well, what does it mean that "the Ga(a,1) density is finite"? That its image lies in a bounded vectorial space, that is, that $Ga(\cdot): \mathcal{R} \to [a,b]$ where $a \neq \pm \infty$ and $b \neq \pm \infty$? $\endgroup$ – rtrtrt Apr 27 at 10:31
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    $\begingroup$ As stated in an answer, to another question, $f$ is finite if $$\forall x\in\text{supp}(f)\quad f(x)<\infty$$[Elias Stein's Real Analysis, at the beginning of Chapter 4.1] $\endgroup$ – Xi'an Apr 27 at 12:02
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I suppose you mean the function $f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}e^{-\beta x},$ for $x > 0,$ and $0$ otherwise, as stated in Wikipedia. There it is stated that parameters $\alpha >0$ and $\beta > 0,$ assuring that $f(x)$ is a density function.

If that is true, then the integral $J = \int_0^\infty f(x)\,dx$ converges and the constant factor $\frac{\beta^\alpha}{\Gamma(\alpha)}$ is chosen so that $J = 1.$ That together with the fact that $f(x) \ge 0,$ makes $f(x)$ the density function of a probability distribution.

Furthermore, if $\alpha \ge 1,$ then $f(x) \le M$ for some finite $M$ and for all $x\in (0,\infty).$ By contrast, for $0 < \alpha < 1,$ the function $f(x)$ has a vertical asymptote at $0.$

By way of illustration we show density functions of the distributions $\mathsf{Beta}(\alpha = 1/2, \beta = 1)$ on the left, $\mathsf{Beta}(\alpha = 1, \beta = 1)$ in the center, and $\mathsf{Beta}(\alpha = 2, \beta = 1)$ at the right.

enter image description here

Note: The parameter $\alpha$ is called a shape parameter because it determines the fundamental shape of the density curve. As you can see from the right-hand plot above, the density function has one inflection point for $\alpha = 2.$ Can you find a value of $\alpha$ above which the density function has two inflection points (and thus a 'left tail')?

Addendum: Plots for three additional values of $\alpha:$

enter image description here

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  • $\begingroup$ ah ok, so the distribution is finite if the image of the pdf lives in a finite vectorial space. For your question, I guess for $\alpha=3$ has two inflection points? $\endgroup$ – rtrtrt Apr 27 at 8:23
  • $\begingroup$ Many commonly used density functions are are unbounded (often near $x=0).$ In computation that is sometimes an inconvenience, but does not interfere with the practical use of the probability model. // Yes, on $\alpha = 3.$ There are 5 distinct behaviors of gamma distn's near 0, taking into account asymptotic values & slopes. // I would not want my books judged by single sentences taken out of context, but the sentence you quote is far too 'chatty' and colloquial and could be misinterpreted. Unless a function integrates to unity it can't be the density function of a probability dist'n. $\endgroup$ – BruceET Apr 27 at 17:59

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