Definition talk:Probability Density Function

Is this defined only when that complicated limit actually exists? Is it the case that it does always exist? --prime mover (talk) 13:55, 20 December 2012 (UTC)


 * Corrected to point out that it doesn't always exist. The definition is neater when the pdf is considered as the derivative of the cdf, but I was going to add that as a theorem, with a sufficient condition for the pdf existing is that the cdf is differentiable. If you have a more elegant approach by all means please suggest it. The motivation of this definition is to get around the problem that $Pr(X=c) = 0$, should I mention it on the page so that the limit isn't coming out of nowhere? --GFauxPas (talk) 14:37, 20 December 2012 (UTC)


 * Having thought about it, if $X$ is continuous, then it probably does exist througout. No worries, just thought I'd ask. If you're quoting what Bean says, then trust him not me ... --prime mover (talk) 17:37, 20 December 2012 (UTC)


 * It wasn't just what you said; I realize now I was confusing piecewise-continuous and continuous. I'll fix it later --GFauxPas (talk) 17:48, 20 December 2012 (UTC)


 * Okay, I have enough clarity to fix the page. Now, what's the best way to present the following definition? I'm not sure how to present it clearly and rigorously:

Let $X$ be piecewise continuous.


 * $\forall x \in \R: f_X \left({x}\right) = \begin{cases}

\displaystyle \lim_{\epsilon \to 0^+} \frac{\Pr \left({x-\frac \epsilon 2 \le X \le x + \frac \epsilon 2}\right)} \epsilon & : \text{all $x \in \Omega_X$ except for countably many $x$ at which point $f_X$ has a removable discontinuity} \\ \text {doesn't matter, as long as it's a real number between 0 and 1} &: \text{the countably many $x \in \Omega_X$ that I mentioned in the above line} \\ 0 & : x \notin \Omega_X \end{cases}$

--GFauxPas (talk) 14:18, 21 December 2012 (UTC)


 * So you are defining $f_X$, and in that definition, you refer to it having a discontinuity? That's not a very good idea. Not sure how to circumvent that, though. --Lord_Farin (talk) 15:39, 21 December 2012 (UTC)
 * Why is it a bad idea? It's just saying that $X$ need not be everywhere continuous to be a probability measure --GFauxPas (talk) 23:02, 22 December 2012 (UTC)