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This question already has an answer here:

NOTE: I see it suggested that this is a duplicate question. Certainly there is substantial overlap between the two questions. This one explicitly mentions the Beta function and explicitly invites arguments for doing it the other way. One of the more able regular posters here stated today in a comment that it ought to be done the other way; I would be curious to see that case made.
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One sometimes sees it opined that instead of $$ \Gamma(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x} \,dx $$ one should follow a convention that puts simply $\alpha$ where $\alpha-1$ appears. Similarly with both $\alpha$ and $\beta$ in $$ \operatorname{B}(\alpha,\beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1} \, dx. $$

What facts can be adduced on which to base arguments for one or the other convention?

(I will post a brief answer of my own, but possibly some very different answers can be readily written.)

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marked as duplicate by Simply Beautiful Art, Did, Shailesh, Leucippus, Lord Shark the Unknown Aug 21 '17 at 3:40

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Of course it's just a convention, so open to change in principle.

One reason for the current convention is that when working with such functions one often makes changes of variables of the form $x = cy$, with attendant $dx = c\,dy$ substitutions. When working this way it is often handy to know that the form $dx/x$ turns into $dy/y$. So think of the obnoxious $x^{\alpha-1}\dots dx$ as the user-friendly $x^\alpha\dots dx/x$.

Another reason is that such a change would confuse everyone, a lot.

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On the "pro" side I would mention two facts:

  • For $\alpha>0,$ let $$f_\alpha(x) = \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x}.$$ Then for $\alpha,\beta>0$ we have the convolution $$f_\alpha * f_\beta = f_{\alpha+\beta}.$$ (And recall that the convolution of two probability density functions is the density of the sum of two independent random variables.)

  • For $A\subseteq[0,1],$ suppose $$\Pr(X\in A) = \frac 1 {\operatorname{B}(\alpha,\beta)} \int_A x^{\alpha-1} (1-x)^{\beta-1} \, dx.$$ Then $$\operatorname{E}(X) = \frac \alpha {\alpha+\beta}.$$

(I wonder if the usual convention can be made less painful to those who don't like it by writing it like this: $$ \Gamma(\alpha) = \int_0^\infty x^\alpha e^{-x} \, \frac{dx} x. $$

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    $\begingroup$ I generally write the gamma function it like your last line (though the beta function can't be fully saved by this). I would add that in both cases this puts the region of convergence $\alpha>0$ rather than the less friendly $\alpha > -1.$ The big con is of course having to remember which way it goes when translating to a factorial. $\endgroup$ – spaceisdarkgreen Aug 20 '17 at 21:37
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    $\begingroup$ @spaceisdarkgreen : $$ \int_0^1 x^\alpha (1-x)^\beta\, \frac{dx}{x(1-x)} $$ The measure $dx\!\left/\Big(x(1-x)\Big)\right.$ can be said some interesting things about. $\endgroup$ – Michael Hardy Aug 20 '17 at 23:25
  • $\begingroup$ All right, I'm in. $\endgroup$ – spaceisdarkgreen Aug 20 '17 at 23:28
  • $\begingroup$ @spaceisdarkgreen : $$ \begin{align} \Pr( \text{hypothesis } H \mid \text{new data } D) & = \frac{\Pr(H) \Pr(D\mid H)}{ \Pr(H) \Pr(D\mid H) + \Pr(\text{not } H)\Pr(D \mid \text{not }H) } \\ \\ & = \frac{pa}{pa + (1-p)b} \end{align} $$ The group of all of these mappings of the form $p \mapsto \dfrac{pa}{pa + (1-p)b}$ with composition of functions is an abelian Lie group whose Haar measure is $\dfrac{dp}{p(1-p)}. \qquad$ $\endgroup$ – Michael Hardy Aug 20 '17 at 23:47
  • $\begingroup$ @spaceisdarkgreen : Note that the mapping depends on $a$ and $b$ only through $a/b,$ so it's a one-dimensional Lie group. $\qquad$ $\endgroup$ – Michael Hardy Aug 20 '17 at 23:48

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