# Solving the Beta density function?

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?

• Basically just look at the powers of $r$ and $(1-r)$ that you want. – Minus One-Twelfth Dec 16 '19 at 21:35
• @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? – noflow Dec 16 '19 at 22:28
• It is effectively done by inspection. This is fine. – Minus One-Twelfth Dec 16 '19 at 22:30
• 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? – noflow Dec 16 '19 at 22:38

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 $$\alpha=x+1, \beta=y+1$$
• But can't you solve the left hand side for $\alpha$ and $\beta$ instead of just verbally stating the solutions without any math? – noflow Dec 16 '19 at 22:07
• @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 $\beta$ – Henry Dec 16 '19 at 23:15