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I found lots of references that say, "the probability that a continuous random variable equals some single value is always zero". Why is that?

Here is a counterexample I thought of: supposing $X\sim N(0,1)$, define $Y=min(X,0)$. Then Y is a continuous random variable but the probability of $Y$ at a single point $0$ should be $0.5$, not zero.

Also, I think any CDF would be left continuous if "the probability that a continuous random variable equals some single value is always zero".

What is wrong with my thoughts?

P.S. Examples of the references are:

  1. http://www.henry.k12.ga.us/ugh/apstat/chapternotes/7supplement.html
  2. http://mathinsight.org/probability_distribution_idea
gung - Reinstate Monica
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user112758
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  • thanks for the editing @gung, and what is your opinion? – user112758 Jan 16 '16 at 05:19
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    see definition 1.32 of the link http://math.arizona.edu/~jwatkins/probnotes.pdf for the definition of a continuous random variable –  Jan 16 '16 at 09:12
  • @fcop, which part of that definition is helpful? I have looked at it, and I do not immediately see. – Richard Hardy Jan 16 '16 at 10:39
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    By the way, the distribution of your $Y$ is not continuous, it is mixed continuous-discrete :) – JohnK Jan 16 '16 at 10:56
  • @Richard Hardy: part 2, the definition of continuous random variable. I think JohnK makes a similar remark –  Jan 16 '16 at 12:59
  • Asking about the $Y$ counterexample may be sufficiently unique to merit this Q remaining open. – gung - Reinstate Monica Jan 16 '16 at 14:02
  • According to the definition of a continuos variable given in the link in my comment supra, $Y$ is not a continuous variable. –  Jan 16 '16 at 16:04
  • @fcop, thanks for the link, I agree with you now. It seems I missed the rigorous definition of continuous random variable. – user112758 Jan 16 '16 at 22:23

2 Answers2

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The thing is $Y$ is not that continuous to begin with. To be continuous, the distribution function of $Y$ must be absolutely continuous (see definition 1.32, page 10 of link http://math.arizona.edu/~jwatkins/probnotes.pdf by @fcop). You see the distribution of Y has a half impulse (Dirac delta) function at zero. When you approach zero on the negative side there is a jump in distribution function value. So the distribution function of $Y$ is not continuous.

If $f(x)$ is continuous, $g(f(x))$ is not necessarily continuous.

Jariani
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  • Hi, I believe Y is a continuous random variable. And I think you mean the density function of Y has a half impulse. Am I correct? – user112758 Jan 16 '16 at 05:17
  • (+1) $Y$ does not have a density function and therefore is not continuous. – whuber Jan 16 '16 at 15:01
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Let's just go with a simple intuitive explanation. But it can get really mathy really fast (if you prefer).

A continuous distribution is a line between two points A and B. On this line there are infinitely many points, no matter if the distance between the points (A, B) is extremely small. If all of those infinite points had a probability larger than 0, then the sum of probabilities would be infinity.

But if that where the case, then the said probability distribution would violate the (Kolomogorov) axioms of probability. So it would not be measure of probability in the modern understanding.

Edit, the derivative of the function $\text{min}(x,y)$ is not defined for $x=y$. So your example is not continuous.

Richard Hardy
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Repmat
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  • OP does not ask for *all* the points to have non-zero probability. The question is about *at least one* point ("some single value") with positive probability. – Richard Hardy Jan 16 '16 at 10:42
  • @Richardhardy then it is not a contiounes distribution – Repmat Jan 16 '16 at 12:18
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    OK, let me rephrase: you say that if *all* points had positive probabilities then the sum would be infinity. However, if only a fixed number of points had positive probabilities, the sum would be finite. So your argument does not go through for the case of the OP. Also, mind the spelling of *continuous*. – Richard Hardy Jan 16 '16 at 12:29