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Consider a random variable $X$ with pdf \begin{equation} f(x)= \begin{cases} 3/2 &\text{ if } x\in[0,1/3]\cup[2/3,1] \\ 0 & \text{ otherwise} \end{cases} \end{equation} Here, $P(X=1/6)=P(X=1/2)=0$, but $1/6$ and $1/2$ are somehow different, because the event $X=1/6$ can actually happen, but $X=1/2$ can't. How do you differentiate between values like $1/2$ that can't happen, and $1/6$ that can, and what do I need to know about a random variable in order to know whether a value is zero probability of one type or the other?

Jimmy Jam
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    If only the pdf or cdf are given, it is impossible to differentiate between your two cases. You'd have to look at the function X itself to tell. From my experience, not much attention is paid to this in probability or statistics, but who knows something has been found. – Bib-lost Feb 15 '16 at 21:57
  • The following links might provide some insight: wiki:Null Set, wiki:Almost Everywhere. Perhaps you could use the phrase "almost impossible" and "impossible" to describe the two outcomes. – JMoravitz Feb 15 '16 at 22:00
  • This question seems to go into some details related to your question and the answers there might be useful to you. – JMoravitz Feb 15 '16 at 22:06
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    I think the way to tell them apart is with the PDF. If $f(x) \gt 0$ then $X = x$ can actually happen, even though $P(X = x) = 0$. – Dan Brumleve Feb 15 '16 at 22:08
  • Do you say $X=\frac{1}{2}$ cannot happen because its pdf has value 0 when $x=\frac{1}{2}$? Why does it follow that it can happen if the pdf has some non-zero value? If you look at the cdf it takes a constant value between $\frac{1}{3}$ and $\frac{1}{2}$ rather than increasing. Does an increasing cdf mean a value can happen and a constant cdf mean it can't? – Paul Feb 15 '16 at 22:14
  • Why do you think that $X=1/6$ can happen and $X=1/2$ not? – user251257 Feb 16 '16 at 00:36

1 Answers1

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One way to differentiate these two cases would be to compute the probability of observing an outcome in some $\epsilon$ neighborhood around each point. For instance, with $\epsilon=\frac{1}{100}$, we see that the probability of observing an outcome between $\frac{1}{6}-\frac{1}{100}$ and $\frac{1}{6}+\frac{1}{100}$ is $\frac{3}{100}$. Likewise, for the same $\epsilon$ neighborhood around $\frac{1}{2}$, the probability is zero. Thus, while the likelihood of observing an outcome of exactly $\frac{1}{6}$ is zero, you have a non-zero likelihood of observing an outcome in some small neighborhood around $\frac{1}{6}$, which is not the case with $\frac{1}{2}$ (provided you choose $\epsilon$ sufficiently small).

rywit
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