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Moved from the article:
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)
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)
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)
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)
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)
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)
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)
Since no-one has made any objections to this addition, I will implement it. - Hammerite
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)
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 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)
Is the following is true?
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)
![]() | This redirect does not require a rating on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||
|
Moved from the article:
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)
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)
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)
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)
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)
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)
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)
Since no-one has made any objections to this addition, I will implement it. - Hammerite
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)
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 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)
Is the following is true?
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)