Suppose $\pi=\frac{\theta}{1-\theta}$ where theta is between $[0,1]$.

If we set a uniform prior for $\pi$ ($p(\pi) \propto 1$), what is the induced prior on $\theta= \frac{\pi}{1+\pi}$? Is this prior proper?

I'm stuck on this problem. Can someone help me out?


If $\pi$ is uniformly distributed over $(0,a)$ then $\theta\in\left(0,\frac{a}{1+a}\right)$. Then for $0<x<\frac{a}{1+a}$ $$ F_\theta(x)=\mathbb P(\theta\leq x)=\mathbb P\left(\frac{\pi}{1+\pi}\leq x\right) = \mathbb P\left(\pi\leq \frac{x}{1-x}\right)=F_\pi\left(\frac{x}{1-x}\right)=\frac{x}{a(1-x)}. $$ And the pdf of $\theta$ should be $f_\theta(x)=\frac{1}{a(1-x)^2}\mathbb 1_{\left(0,\frac{a}{1+a}\right)}$.

Note that $\left(0,\frac{a}{1+a}\right)\subset (0,1)$. Say, for $a=1$, $\left(0,\frac{a}{1+a}\right)=\left(0,\frac12\right)$.

You cannot obtain $\theta$ with positive pdf over whole $(0,1)$ if $\pi$ is uniformly distributed since there cannot be uniform distribution over positive halfline.


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