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I'm trying to understand conditional density functions.

Let $X$ and $Y$ be jointly continuous random variables with joint density function $f_{X,Y}$. We condition on the event $x\leq X\leq x+\delta x$ and take the limit as $\delta x\downarrow 0$. Thus,

$\begin{aligned}\mathbb P(Y\leq y\mid x\leq X\leq x+\delta x)=&\frac{\mathbb P(Y\leq y,x\leq X\leq x+\delta x)}{\mathbb P(x\leq X\leq x+\delta x)}\\ =&\frac{\int_{u=x}^{x+\delta x}\int_{v=-\infty}^y f_{X,Y}(u,v)\,\mathrm d u\,\mathrm d v}{\int_x^{x+\delta x}f_X(u)\,\mathrm d u}. \end{aligned}$

Up until here I can follow it. However, then they say they divide both the numerator and the denominator by $\delta x$ and take the limit as $\delta x\downarrow 0$ to obtain

$\begin{aligned}\mathbb P(Y\leq y\mid x\leq X\leq x+\delta x)\rightarrow \int_{-\infty}^y\frac{f_{X,Y}(x,v)}{f_X(x)}\,\mathrm d v.\end{aligned}$

This is not a rigorous proof, and I'm having difficulty following it. I can see that the $x$ integral has disappeared, but what exactly happens when you divide by $\delta x$?

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    You should actually avoid integrals from $x$ to $x+\delta x$ and simply replace them by $f_X(x)\delta x$ (denominator) and $\int_{-\infty}^yf_{X,Y}(x,v)\mathrm dv\;\delta x$ (numerator) and the $\delta x$s will cancel out. This gives the same result as you wrote. – 2017-01-26
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    Yeaa, thanks, got it – 2017-01-26

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