I have a problem where I need to calculate the credible region for the odds ratio $\frac{\theta}{1-\theta}$. This is part of a bigger problem where I have already calculated 90% credible region for random variable $\theta$:
$$P(\theta\leq 0.30)=0.05$$ $$P(\theta\leq 0.52)=0.95$$ $$P(0.30\leq \theta\leq 0.52)=0.9.$$
Now I need to calculate the 90% credible region for $\frac{\theta}{1-\theta}.$ This is what I tried: first,
$$P\left(\frac{\theta}{1-\theta}\leq a\right)=P\left(\theta\leq \frac{1}{1+a}\right),$$
so now I took advantage of my earlier results: in order for
$$P\left(\frac{\theta}{1-\theta}\leq a\right)=P\left(\theta\leq \frac{1}{1+a}\right)=0.05,$$
I must have $\frac{1}{1+a}=0.30$ or $a=2.39$. Similarly, for the case
$$P\left(\frac{\theta}{1-\theta}\leq b\right)=P\left(\theta\leq \frac{1}{1+b}\right)=0.95,$$
I must have $\frac{1}{1+b}=0.52$ or $b=0.93$. So now I have:
$P\left(\frac{\theta}{1-\theta}\leq 2.39\right)=0.05$ and $P\left(\frac{\theta}{1-\theta}\leq 0.93\right)=0.95$. And now by looking at the numbers something seems to be wrong.
Question: Do you think I should suspect my earlier results for $\theta$ or did I do something illegal in my reasoning here?
Update: The distribution for $\theta$ is $Beta(21,31)$.
You should be careful in that your manipulation only works if $\theta \le 1$. You say that $\theta$ comes from a beta distribution though, so you're good.
You have a small algebra mistake: it should be
$$P\left(\frac\theta{1-\theta} \le a\right)=P\left(\theta \le \frac a{1+a}\right) = 0.05$$ which implies
$a=0.42$ and similarly
$$P\left(\frac\theta{1-\theta} \le b\right)=P\left(\theta \le \frac b{1+b}\right) = 0.95$$ which implies
$b=1.08$