# Transformation of probability distribution

I have a question about a snippet on page 526 in the PRML book of Bishop.

Can someone explain to me why the right-hand side of equation (11.6) equals $$z$$?

It's unclear to me where this derivation comes from. Thanks for your help!

Plugging $$p(z)=1$$, in equation (11.5), we get $$p(y)=|dz/dy|$$.
Taking the integral of a derivative of $$z$$ (right-hand side), should give back the original $$z$$, leading to equation (11.6). However, where do the bounds $$-\infty$$ and $$y$$ of the integral come from?
I also didn't take into account that there is an absolute value around $$dz/dy$$.
• This looks wrong to me. Consider the transformation $z=1-y$ for which $|dz/dy|=1,$ whence (by $(11.5)$) $p(y)=p(z)(1)=1$ when $0\le y\le 1$ (and is $0$ otherwise). For $0\le y\le 1$ and $y\ne 1/2$ equation $(11.6)$ gives $$1-y=z=\int_{-\infty}^y I(0\le \hat y\le 1) p(\hat y) d\hat y = \int_0^y d\hat y = y,$$ a clear contradiction. – whuber Oct 2 '18 at 14:15