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If any random variable has zero variance, then is it right to say that:

A random variable with zero variance is not a random variable

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    $\begingroup$ According to kolmogorov's definition a random variable can have 1 outcome $\Omega=\{o\}$, then $\sigma$-algebra is the set of subsets of $\Omega$ and the measure of $\{o\}$ is 1. So a random variable with zero variance is a random variable (any map that maps the above o to a single real number is an example) $\endgroup$ – user83346 Aug 7 '16 at 7:17
  • $\begingroup$ It's important to keep in mind that there's nothing random (in the usual senses of this word) about a "random variable". Non-zero variation is not a requirement for a random variable. $\endgroup$ – Mico Aug 7 '16 at 20:15
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    $\begingroup$ Suppose $X \sim U(0,1)$ and $Y=X$ if $X \in \mathbb{Q}$ but $Y=\frac12$ otherwise. Then $Y$ is a random variable with mean $\frac12$ and variance $0$ but can take values which are not $\frac12$. $\endgroup$ – Henry Aug 7 '16 at 20:47
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    $\begingroup$ Ignoring the math, it doesn't even make sense. "x with property y" is still something of type x $\endgroup$ – Batman Aug 7 '16 at 21:54
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$E[(X-E[X])^2] =0 \implies X = E[X]$

Thus $X$ is almost surely constant. A better description for such random variables is that it follows a degenerate distribution.

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  • $\begingroup$ I am slightly confused: $X=E[X]$ is constant, not "almost surely constant", isn't it? $\endgroup$ – amoeba Aug 8 '16 at 10:05
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    $\begingroup$ 'almost surely' is a technical term meaning that any deviation from the stated assertion has zero probability. An example would be, let's say you have an indicator variable indicating whether a number is an integer or not, and then you take the expectation of this over the real numbers from 0 to 10. Of course there are 9 or so locations where this indicator is 1, but the probability of choosing one of these locations is infinitely unlikely (not a technical term, more hand-waving), so the expectation of this indicator will be 0. The indicator takes values that are almost surely equal to 0. $\endgroup$ – Hugh Perkins Sep 2 '17 at 15:10

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