We know that CDF, $F_X(x)=P_X((-\infty,x])$ (cumulative distribution function) of a continuous random variable, ie. a function $X:\Omega \to \mathbb{R}$ that follows all the properties that a random variable should, is right continuous, and moreover, we have the equation:$$F_X(x_0)-F_X(x_0-)=P_X[\{x_0\}]$$
where $$F_X(x_0-) \equiv \lim_{x\to x_0^{-}} F_X(x)$$
Proof that CDF is continuous for continuous random variables. This question is similar, but the answer is not satisfactory, as in, sure, if the $P[\{x\}]$is $0$, then from the above equation its obvious that the cdf is continuous (lhl=rhl). On the other hand, the answerer assumes that the pdf (probability density function), being the derivative of the cdf, exists, and is riemann integrable. This claim is not obvious. I think the author of that answer conflated the definition of continuous random variables with the existence of PDF, instead of the (correct) definition that says its range(X) is continuum.
I would like to know two things:
- Is there another pathway to somehow show that $P[\{x\}]$ is $0$?
- Is there more to this "existence of pdf", like, is there a theorem of analysis that allows this?
Best Answer
We usually define a random variable as continuous if its cumulative distribution function (CDF) is continuous, so this is simply the definition.
It is true that there are random variables of uncountable range that do not have continuous CDF; we call them mixed random variables.