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The beta density with parameters $\alpha$ and $\beta$ is given by $$ p(r) = \frac{\Gamma (\alpha + \beta)}{\Gamma (\alpha) \Gamma (\beta)} r^{\alpha - 1} (1-r)^{\beta - 1} $$ I want to find the parameters that results in 1) $p(r) = 2r$ and 2) $p(r) = 3r^2$. I can quickly figure out that the parameters needed for the first case is $\alpha = 2$ and $\beta = 1$ but how do I algebraically derive this result?

How to algebraically solve the two equations for $\alpha$ and $\beta$ and thus find the right parameters for the two cases?

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  • $\begingroup$ Basically just look at the powers of $r$ and $(1-r)$ that you want. $\endgroup$ – Minus One-Twelfth Dec 16 '19 at 21:35
  • $\begingroup$ @MinusOne-Twelfth But can't you solve the left hand side for α and β instead of just verbally stating the solutions without any math`? $\endgroup$ – noflow Dec 16 '19 at 22:28
  • $\begingroup$ It is effectively done by inspection. This is fine. $\endgroup$ – Minus One-Twelfth Dec 16 '19 at 22:30
  • $\begingroup$ Hmm okay. But is it possible to do without just inspection - like by solving the equations $å(r) = 2r$ and $p(r) = 3r^2$? If so, do you know how? $\endgroup$ – noflow Dec 16 '19 at 22:38
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You have targets of the form $k_1 r^{2-1} (1-r)^{1-1}$ and $k_2 r^{3-1} (1-r)^{1-1}$

so (assuming the targets are proper pdfs) the solutions must be $\alpha=2, \beta=1$ and $\alpha=3, \beta=1$

If your target was of the form $k_3 r^{x} (1-r)^{y}$ then you would want $\alpha=x+1, \beta=y+1$

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  • $\begingroup$ But can't you solve the left hand side for $\alpha$ and $\beta$ instead of just verbally stating the solutions without any math`? $\endgroup$ – noflow Dec 16 '19 at 22:07
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    $\begingroup$ I never told you the targets since they would give the parameters directly (and those are the ones I am trying to find by mathematically solving p(r)=something) $\endgroup$ – noflow Dec 16 '19 at 22:24
  • $\begingroup$ @noflow You have said the targets are $2r$ and $3r^2$ in your examples. Given the powers of $r$ and $(1-r)$ in the targets, you can find $\alpha$ and $\beta$ $\endgroup$ – Henry Dec 16 '19 at 23:15

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