# If odd is uniformly distributed, what is the distribution of proportion?

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 $$0 $$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.