# When a probability density function is defined to be finite?

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?

• This statement from Bishop is incorrect in that the Ga$(a,1)$ density is finite on its support $(0,\infty)$ albeit unbounded when $0<a<1$. – Xi'an Apr 27 at 7:48
• @Xi'an What do you mean by "finite in an (infinite) support"? – rtrtrt Apr 27 at 9:49
• I did not write "finite in an (infinite) support"... – Xi'an Apr 27 at 10:08
• @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 $b \neq \pm \infty$? – rtrtrt Apr 27 at 10:31
• 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] – Xi'an Apr 27 at 12:02

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.

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 $$\alpha:$$

• 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? – rtrtrt Apr 27 at 8:23
• 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. – BruceET Apr 27 at 17:59