# Random variable with zero variance

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

• 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) – user83346 Aug 7 '16 at 7:17
• 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. – Mico Aug 7 '16 at 20:15
• 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 $\frac12$. – Henry Aug 7 '16 at 20:47
• Ignoring the math, it doesn't even make sense. "x with property y" is still something of type x – Batman Aug 7 '16 at 21:54

$E[(X-E[X])^2] =0 \implies X = E[X]$
Thus $X$ is almost surely constant. A better description for such random variables is that it follows a degenerate distribution.
• I am slightly confused: $X=E[X]$ is constant, not "almost surely constant", isn't it? – amoeba Aug 8 '16 at 10:05