Talk:Discrete probability distribution

Moved from the article:
 * ...its range consists entirely of point-masses. Let X be a random variable; then a "point-mass" in the range of X is a value x such that $$P(X>x)$$.  If the sum of all such point-masses is 1, then X is a discrete random variable.

Definition of point-mass is not clear. Is a point mass a probability or a value in the range of X? AxelBoldt 06:40 Jan 9, 2003 (UTC)

Problems with the math websites
These math pages are basically worthless. If you don't know the math already, you can't understand the pages. For example, in order to understand the page on probability mass distributions, you have to know the meaning of discrete number. The discrete variable uses the probability mass distribution in the definition. It's all very circular. The page discribing how to use the capital sigma notation is likewise confusing. Are these pages here to teach people who don't know about these concepts or just to fill up space? —Preceding unsigned comment added by 164.107.90.170 (talk) 18:01, 6 October 2007 (UTC)

Hear hear. This is rapidly becoming a real problem with Wikipedia. Increasingly the technical articles seem to be written by mathematicians for mathematicians. How about ordinary folk? This is supposed to be an encyclopedia, not a bloody maths textbook! Keep it simple, stoopid!

--84.9.73.211 (talk) 18:23, 1 February 2008 (UTC)

So true! I was hoping to get a quick definition for my stats exam tomorrow, and I get a page of big math words that mean bloody well nothing to me. Fine, keep the confusing math definition for all the scientists and mathematicians who understand what it means, and who want to make sure they have a full grasp on it. But why can't there be a definition in plain English??? I'd write it, except, thanks to the bloody uselessness of this page, I have no idea still what a discrete random variable is... —Preceding unsigned comment added by 75.70.219.214 (talk) 06:35, 18 December 2008 (UTC)

Incorrect definition?
What is the actual definition of DRV. Is it the discrete probability dist or is it the following:

''If a random variable is discrete then the set of all possible values that it can assume is finite or countably infinite, because the sum of uncountably many positive real numbers (which is the smallest upper bound of the set of all finite partial sums) always diverges to infinity. '' --Light current 04:56, 30 October 2005 (UTC)


 * Good points. I find this article unnecessarily complicated and partially incomprihensible.


 * I also believe that the definition is actually incorrect. It says that a random variable is discrete if


 * $$\sum_u \Pr(X=u) = 1$$


 * as u runs through the set of all possible values of the random variable X.


 * This does not preclude the random variable having a continuous part, for example, the random variable


 * f(x) = 8 for x in [-4, -3]
 * f(x) =x otherwise


 * seems to satisfy this "sum" definition, as in the sum above one has a 1 for u=8 and zeros ohterwise. Oleg Alexandrov (talk) 00:20, 27 November 2005 (UTC)

Oleg, I find your remarks incredibly confused. What random variable are you referring to when you write your piecewise definition above?? I can't figure it out, and I'm pretty good at understanding such things. I can't see anything in your piecewise definition that suggests something satisfying the "sum" definition, nor anything that defines or characterizes any particular random variable. "Discrete random variable" is conventionally understood to mean a random variable with a discrete probability distribution, and the definition you cut is correct. Michael Hardy 02:46, 27 November 2005 (UTC)

It looks as if when you wrote "1 for u=8 and zeros ohterwise", you probably meant the Lebesgue measure of the inverse-image under the piecewise-defined function f that you specified. But you haven't defined any random variable, nor any probability distribution, nor any discrete or continuous or other random variable. The probability distribution that assigns measure 1 to the set {8} and 0 to sets disjoint from that set, is a discrete distribution. Any random variable X having that distribution satisfies Pr(X = 8) = 1 and Pr(X = x) = 0 for x = anything except 8. So it's not a counterexample to the statement you called incorrect. Michael Hardy 02:53, 27 November 2005 (UTC)


 * I meant the random variable to be f. What I forgot about is a probability measure. Let us have the real line with the probability of a set being the Lebesgue measure of that set intersected with [-4, -3]. What I missed is that its complement is negligeable, so it does not matter what values the random variable has in there. So you are right. However, the defintion is still misleading, it should say that in addition to the countable number of values, the random variable can have some other values, but the set of all points on which it takes those values can have measure zero. No? Oleg Alexandrov (talk) 04:03, 27 November 2005 (UTC)

NO! You're still confused. Continuity, or its lack, of the random variable as a function on a probability space has absolutely nothing to do with it. In particular, the DISTRIBUTION OF the random variable you describe is continuous EVERYWHERE, NOT JUST almost everywhere.

Admittedly, the explanation on the page is skimpy. I'll get to it some time soon, I think. Michael Hardy 02:07, 29 November 2005 (UTC)


 * Who says continuity? Not me. Let us take it step by step.


 * 1. Consider the real numbers, with the probability measure:


 * $$P(A)=m(A\cap [-4, -3])$$


 * where m is the Lebesgue measure. This is a probability space, isn't it?


 * 2. Consider the function X:R->R, defined by X(t)=1 if t in [-4, -3], and X(t)=t otherwise. This is a random variable (a measurable function). Correct?


 * 3. This function satisfies


 * $$\sum_u \Pr(X=u) = 1$$


 * as u runs through the set of all possible values of the random variable X. Correct or not?


 * 4. The set of all possible values of this function is not countable. Do you agree with this?


 * 5. This function X nevertheless coicides almost everywhere with the indicator function of the interval [-4, -3], which is a discrete variable. Correct or not correct? Oleg Alexandrov (talk) 02:17, 29 November 2005 (UTC)


 * It seems to me that statement 3 above is incorrect. In your example you defined a continuous stochastic variable. Statement 3 is a sum, therefor of a countable number of terms. As you have taken a continuous stoch. var., all $$\Pr(X=u)$$ are zero. For instance $$\Pr(X=-3.1)=0$$ despite the fact that the PDF is not equal to zero for $$X=-3.1$$. Bob.v.R 23:15, 29 November 2005 (UTC)


 * No, "sum" does not mean a countable number of terms, and that fact is explained in the article. But how should one construe "possible values"?  From one simple and plausible point of view there is only one possible value of this random variable. Michael Hardy 23:56, 29 November 2005 (UTC)


 * Right Michael, the issue of what is a "value" for a random variable is a big question. You are right that this random variable has only the value 1 on [-4, -3]. It does not matter what values it takes outside this interval, as the outside has measure zero. However, how to explain that well in the article. And I hope you do agree with me that the function I provided has as range the set of all real numbers except for some interval. Oleg Alexandrov (talk) 00:00, 30 November 2005 (UTC)


 * The ´value´ that a random variable assumes is a number for which there is a certain probability of occurrence. The value is not the probability itself!! For the stoch. var. in Oleg´s example the important values of ´u´ are between -4 and -3; so these values are relevant when ´u´ is running through the set of possible values. And this case, because Oleg has chosen a continuous stochastic variable, we have that $$\Pr(X=u)=0$$ for every individual value ´u´. Bob.v.R 11:54, 30 November 2005 (UTC)

Oleg, I agree with the obvious parts of what you say and disagree with your main point. The points you raise have no place in this article, as will be seen as soon as you realize that the article title is incorrect; it needs to get moved. I'll do that right away. Michael Hardy 20:08, 30 November 2005 (UTC)

... and now I've made Oleg's objection irrelevant by moving the page. Michael Hardy 20:16, 30 November 2005 (UTC)


 * Now I am happy. Two random variables which coincide almost everywhere have the same probability distribution, and the thing I defined above clearly has a discrete probability distribution. Oleg Alexandrov (talk) 20:33, 30 November 2005 (UTC)

Alternative
Would it be a valid alternative definition to say that "a discrete random variable is a random variable whose cumulative distribution function increases only by jump discontinuities"? If not, would it be sensible to include it as a slightly more intuitive but less rigorous description, marked as such? The current page may be strictly mathematically correct, but it is hard to decipher if you are just someone who has, say, a working knowledge of calculus but no in-depth knowledge of analysis, indicator functions, all this other stuff that is referenced. 81.159.124.90 00:05, 6 December 2005 (UTC)
 * What you are say is correct, from what I can tell. But please notice that those bumps can be a dense set (the set of rational numbers for example). I think a function which is piecewise constant, increasing, and having bumps all over the place is enough to give everybody a headache. Wonder what others think. Oleg Alexandrov (talk) 02:14, 6 December 2005 (UTC)

Since no-one has made any objections to this addition, I will implement it. -Hammerite

What happened to the discrete uniform distribution?
The list of common discrete distributions fails to mention the Uniform distribution (discrete) distribution. Is there a reason for this, or just an oversight. Dale Gerdemann 15:57, 28 March 2007 (UTC)

Revert
Hi Oleg, thanks for respectfully reverting and being careful enough to preserve some of my changes :) I read your earlier comment "I find this article unnecessarily complicated and partially incomprehensible." and was wondering whether you still feel that way?  Does this article offer anything that is not offered at probability distribution? MisterSheik 20:09, 7 May 2007 (UTC)
 * I think this article has been edited for better since my earlier ranting. It is still not satisfactory enough, but I feel that this article has the potential to grow, and the concept of discrete probability distribution is important enough to have its own article.


 * To answer your question, the bigger probability distribution article is missing the section titled "Representation in terms of indicator functions" from this article, for example. Comments? Oleg Alexandrov (talk) 02:19, 8 May 2007 (UTC)


 * Honestly, I think that the indicator function definition is much less accessible than the "is characterized by a pmf" definition. :( Would you honestly read it if you knew nothing about the topic, or if you knew something about the topic, or if you were an expert on the topic?  MisterSheik 14:01, 9 May 2007 (UTC)
 * That's why it is at the bottom, because it is harder than the others. :) But it makes a good point that a random variable is discrete if and only if it is a combination of indicator functions. Oleg Alexandrov (talk) 01:52, 10 May 2007 (UTC)

Making it more definable to us younger people using this as a reference, which is still better than MathWorld's "A discrete probability distribuion is a probability distribution that is discrete."
I think that making a simple kind of example to help define it would help. I, as an eighth grade mathematician, would like to offer one myself. "Basically, to be a discrete probability distribution, it uses values like 1, 2, 3, 4, 5... rather than real numbers. In a geometric distribution, for example, you wouldn't have p(1-p)^(3/2); you would have p(1-p)^n where n is an integer. However, this example should not imply that all discrete distributions use 1, 2, 3, 4, 5...

Heero Kirashami 02:25, 2 November 2007 (UTC)

Riemann–Stieltjes integral
Is the following is true?


 * $$\operatorname{P}(X \in A) = \int_A^\ \, \mathrm{d} \operatorname{P}(X \le x)$$

Then how about A=[a,a]? Is the Riemann–Stieltjes integral well defined in this case?Nick C. (talk) 06:08, 30 July 2009 (UTC)