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I am trying to derive the CDF of a Beta distribution defined on the interval $[a,b]$. I managed to get the PDF for this more general version of Beta, with the following result: $$ f(\alpha, \beta, x) = \frac{(x-a)^{\alpha-1} (b-x)^{\beta-1}}{B(\alpha, \beta) (b-a)^{\alpha+\beta+1}} $$ Now, I was trying to get the CDF from that, by calculating the integral, but whatever I was getting, was likely incorrect as it wouldn't have the properties of the CDF such that $F(a) = 0$, $F(b) = 1$. Is there an easier way to approach that, i.e. perhaps doing a transformation of the CDF of the standard Beta distribution?

mrp
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Paul
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  • If $X$ is Beta on $[a,b]$ then $X=a+(b-a)X_0$ where $X_0$ is Beta on $[0,1]$. In terms of CDF, this reads $$F(x)=P(X\leqslant x)=P(X_0\leqslant(x-a)/(b-a))=F_0((x-a)/(b-a))$$ – Did Jul 16 '17 at 19:09
  • If $\alpha$ and $\beta$ are integers $\geq 1$, $F$ is clearly a polynomial on $[a,b]$, otherwise $F$ is, at best, expressible with transcendental functions (elliptic functions, etc...). – Jean Marie Jul 16 '17 at 19:29
  • Incomplete Beta function, in particular. @JeanMarie is right, it is transcendental. – kimchi lover Jul 16 '17 at 20:31
  • @kimchi lover You are right : I was forgetting that this cdf has this name (http://mathworld.wolfram.com/IncompleteBetaFunction.html). – Jean Marie Jul 16 '17 at 20:34
  • @Did, thank you, this is what I was looking for but was not sure I can do that. – Paul Jul 19 '17 at 19:12

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