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!

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My first thoughts about this:

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?

It is mentioned that it is an indefinite integral, why do we have these bounds?

I also didn't take into account that there is an absolute value around $dz/dy$.

Can we just discard this as if it isn't there?

  • $\begingroup$ 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. $\endgroup$ – whuber Oct 2 '18 at 14:15

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