I'm reading a book about Bayesian statistics and in this book is being explained why the Uniform distribution does not make for a good prior. The following example is given:
Example: Suppose that $X|\Theta = \theta \sim \operatorname{Bin}(n,\theta)$ and take the uniform prior over the parameter set: $\Theta\sim \operatorname{Unif}(0,1)$. The prior is supposed to express ignorance about $\theta$. However, if we are ignorant about $\theta$, we are ignorant about $\theta^2$ as well (for example). If we define $\Psi= \Theta^2$, then $$f_\Psi(\psi) = f_\Theta(\sqrt{\psi})\dfrac{1}{2\sqrt{\psi}} = \dfrac{1}{2\sqrt{\psi}}\mathbb{1}_{(0,1)}(\psi)$$
this implies that small realisations of $\Psi$ are more probable than large realisations and we are no longer ignorant about $\Psi$.
Question: Why is $f_\Psi(\psi)$ the correct density function of $\psi$, the way it's given in the example? Isn't the density function of the uniform distribution supposed to be a constant? I understand why the uniform distribution would not work if this is in fact the correct density, I don't know why it is though.
Thanks in advance!
Indeed, but it is $\Theta$ that is supposedly uniformly distributed, not $\Theta^2$, and we have definied $\Psi$ as being $\Theta^2$. So by the chain rule we have: $$\begin{split}f_\Psi(\psi) &=\left\lvert\dfrac{\partial~\mathsf P(\Theta^2\leqslant \psi)}{\partial~\psi\hspace{9.5ex}}\right\rvert\\[1ex] &= \left\lvert\dfrac{\partial~F_\Theta(\surd \psi)}{\partial~\psi\hspace{7.5ex}}\right\rvert \\[1ex] &=\left\lvert\dfrac{\partial~\surd \psi}{\partial~\psi~~~~}\right\rvert~ f_\Theta(\surd\psi)\\[1ex] & = \dfrac 1{2\surd\psi}~\mathbf 1_{(0;1)}(\psi)\end{split}$$