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According to my understanding regarding discrete and continuous random variables is that discrete random variables are variables that can take finite values that have resulted from a finite number of outcomes or if the values are infinite but still countable ( whole numbers) then the variable is also classified as a discrete random variable. On the other hand , if the random variable can take any value from an infinite outcomes set or if it can be assigned any value from a finite interval ( whole and decimal numbers so that the possible values are infinite ) then it is called continuous random variable. Now for a certain random experiment , let the set of possible values of a random variable to be {1 , 1.2 , 1.2222 ,1.2554 , 2 , 2.54 ,2.66666 ,3} , then the random variable is classified as a discrete or continuous random variable ? I am confused here , since this sample space is finite and hence it is discrete random variable , however , the possible values include whole and decimal numbers so can it be classified as a continuous random variable ?

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    $\begingroup$ Discrete distributions do not have to consist of values that are only integers, whole numbers, or even regularly spaced. For example, a discrete distribution might return the value $\pi = 3.14159 \dots$ half the time, 1, a third of the time, and 0 the rest of the time. There are only three possible values—hence discreet because countable, but these are not regularly spaced, and are not even all rational numbers. $\endgroup$
    – Alexis
    Commented Feb 16, 2022 at 20:47
  • $\begingroup$ Possible duplicate The difference between discrete and continuous variables $\endgroup$
    – Glen_b
    Commented Feb 17, 2022 at 9:50

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It depends whether the set of possible outcomes $\Omega$ is countable (which includes some infinite sets) or uncountable. In the former case, you have a discrete random variable (RV), in the latter case a continuous one.

For the geometric distribution with $P(X=m)=p(1-p)^{m-1}$, e.g., the set of possible values $m=1,2,\ldots$ is infinite, but only countably infinite and it is thus a discrete RV with $\sum_k P(X=k)=1$. This is an example of countably infinte set, which means that there is a bijective mapping between the set and the natural numbers.

A finite interval of real values, in conmtrast, is an uncountable set because there are uncountable possible values in between. It can be proven that such a set is uncountably infinite, i.e., there is no bijective mapping onto the natural numbers.

In case of interest, please have look at Cantor's set theory (the link is to an informal introduction).

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  • $\begingroup$ You can have uncountable possible outcomes without having a continuous random variable. For example consider $Y=\max(X,0)$ where $X$ has a normal distribution $\endgroup$
    – Henry
    Commented Feb 16, 2022 at 12:00
  • $\begingroup$ @Henry AFAIK, the extreme value distribution is continuous. But you are right in pointing out that my classification is overly simplistic (and excludes mixed cases like energy distributions in solid states physics). OTOH I think it is better to give a simple explanation that is wrong in unusual cases than a complicated explanation that the addresse cannot (possibly) understand. This is a question of didactical philosophy, though, and I am aware that almost all mathemticians would disagree ;-) $\endgroup$
    – cdalitz
    Commented Feb 16, 2022 at 12:09
  • $\begingroup$ @cdalitz Thanks alot , Just to make sure that I have understand the concept well. This set of possible values {1 , 1.2 , 1.2222 ,1.2554 , 2 , 2.54 ,2.66666 ,3} belongs to a discrete random variable , since they are countable . Is that right ? $\endgroup$
    – AAA
    Commented Feb 16, 2022 at 12:21
  • $\begingroup$ My example was not an extreme value distribution but a rectified distribution, a mixture of a discrete distribution and a continuous distribution $\endgroup$
    – Henry
    Commented Feb 16, 2022 at 12:24
  • $\begingroup$ @AAA Yes this is a discrete distribution, provided these are indeed the only possible outcomes (which looks supicious, though). $\endgroup$
    – cdalitz
    Commented Feb 16, 2022 at 12:31

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