The question stems from section '30.6.2 Continuous random variables' of Riley Hobson and Bence (Mathematical Methods for Physics and Engineering). They are deriving the following:
If $X$ is a continuous RV, then so too is the new random variable $Y = Y (X)$. The probability that $Y$ lies in the range $y$ to $y + \mathrm dy$ is given by
$$g(y) \mathrm dy = \int_{\mathrm dS} f(x) \mathrm dx$$
where $\mathrm{dS}$ corresponds to all values of $x$ for which $Y$ lies in the range $y$ to $y + \mathrm dy$.
I get this part. In the next part I'm having trouble with following their steps from the integrals and I'm not sure if I'm just being silly as I'm quite rusty.
Once again the simplest case occurs when $Y (X)$ possesses a single-valued inverse $X(Y )$. In this case, we may write
$$g(y) \mathrm dy = \left| \int_{x(y)}^{x(y+\mathrm dy)}f(x')\mathrm dx' \right|= \int_{x(y)}^{x(y) + |\frac{\mathrm dx}{\mathrm dy}| \mathrm dy)}f(x')\mathrm dx' $$
from which we obtain
$$g(y) = f(x(y)) \left| \frac{\mathrm dx}{\mathrm dy} \right| $$
So I get the expansion of $x(y + \mathrm dy) \approx x(y) + \left| \frac{\mathrm dx}{\mathrm dy} \right| \mathrm dy$ but then that coming out of the integral I cant seem to follow! Any pointers would be greatly appreciated.
EDIT: It seems like its quite similar to just a change of integration variables where as $x(y)$ t.f. $\mathrm dx = \mathrm dy \frac{\mathrm dx}{\mathrm dy}$ and then $x(y)$ in $f$ goes to $f(x(y))$ but then the integral out front still bothers me!
EDIT 2:
I think I got it, would be great if someone could verify.
From the definition $$f'(x)=\frac{\mathrm df(x)}{\mathrm dx} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} $$
therefore if we consider above the indefinite integral of $f(x')$ as a function $g$, when we put in the limits we get $g \big(x(y) + |\frac{dx}{dy}|dy \big) - g(x(y)$ which can be written as
$$\begin{align} \frac{g \big(x(y) + |\frac{dx}{dy}|dy \big) - g(x(y) }{|\frac{dx}{dy}|dy} \left|\frac{dx}{dy}\right|dy &= \frac{dg}{dx} \left|\frac{dx}{dy}\right|dy \\ &= f(x(y)) \left|\frac{dx}{dy}\right|dy \end{align}$$
This doesn't sit truly well with me, feels a bit like physics maths not real maths!
Edit 3:
$$ \int_x^{x+\Delta x} f(u)\mathrm du \approx f(x)\Delta x$$
if $\Delta x$ is small
Solved cheers was being very rusty!
