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In a class hand-out of "review" material (which, in theory, should simply be concepts that we learned last semester in class but in practice contains a lot of stuff that we never covered due to what must be some sort of miscommunication as a result of the previous professor retiring), we were given the following definition of a conditional density:

Suppose that $(X,Y)$ is a random pair where $X$ has a continuous distribution and the distribution of $Y$ is discrete. The conditional density of $X$ given $Y=y$ is:

$f[x|y]=$lim$_{h\rightarrow{0}}\frac{1}{h}P(x\leq X\leq x+h|Y=y)$

My main issue is that I don't understand where the $\frac{1}{h}$ part is coming from.

I understand the concept in general that, for a continuous random variable, $P(X=x)=0$, and thus we use lim$_{h\rightarrow{0}}[F(x+h)-F(x)]$ to "approximate" $P(X=x)$. But I don't understand why in the conditional case we have this $\frac{1}{h}$ term in the limit? To be more precise, I don't understand how to DERIVE this given equation.

I have searched around this site and the Internet but don't know if I am falling victim to terminology or what. I have searched phrases like "mixed conditional density" and "continuous conditioned on discrete" but can't seem to find anything talking about this. The closest I've been able to find are cases where one continuous distribution is conditioned on another continuous distribution, which use limits and l'Hopital's rule, but in none of those problems can I find the source of the $\frac{1}{h}$ term.

I am sure the answer is pretty simple, but I don't know why I can't find sources on it. Is there some specific term for this sort of problem I am not aware of, or does anybody know a source that gives the proof? I have fiddled around with Bayes' rule to rewrite the conditional distribution, but this isn't terribly helpful when speaking in generalities like this.

EDIT:

The solution was noted by Kim Jong Un in the comments. But for posterity's sake, it has to do with remembering the formal definition of a derivative. That is,

$f'(x)=$lim$_{h\rightarrow{0}}\frac{f(a+h)-f(a)}{h}$

From there, it is straightforward substitution.

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  • $\begingroup$ Without the conditioning, $f(x)$ is obtained from $\lim_{h\to 0}\frac{1}{h}[F(x+h)-F(x)]$ (in other words, $F'(x)=f(x)$) so you missed the $\frac{1}{h}$. With conditioning, the logic is still the step. You just divide everything by $Pr(Y=y)$ in this case. $\endgroup$ Commented Aug 27, 2014 at 0:17
  • $\begingroup$ I'm still confused as to where the 1/h comes from? Shouldn't it just be h? $\endgroup$ Commented Aug 27, 2014 at 0:18
  • $\begingroup$ Think about how you define a derivative. $\endgroup$ Commented Aug 27, 2014 at 0:19
  • $\begingroup$ Ah! I see. I had completely forgotten about the formal definition of the derivative (calculus was a long time ago). Thanks! If you make your comment an answer I will upvote it. Otherwise, editing my main post to include the answer. $\endgroup$ Commented Aug 27, 2014 at 0:29
  • $\begingroup$ Fine. I want an upvote please. :) (Doing it now.) $\endgroup$ Commented Aug 27, 2014 at 0:31

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Without the conditioning, $f(x)$ is obtained from $\lim_{h\to 0}\frac{1}{h}[F(x+h)-F(x)]$ (in other words, $F'(x)=f(x)$) so you missed the $\frac{1}{h}$. With conditioning, the logic is still the same. You just divide everything by $\Pr(Y=y)$ in this case.

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  • $\begingroup$ Voted. Always realized I had forgotten to log in to my account when I posted the question in the first place... $\endgroup$ Commented Aug 27, 2014 at 0:34
  • $\begingroup$ Appreciated. Good luck with your work. $\endgroup$ Commented Aug 27, 2014 at 0:37

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