0
$\begingroup$

What are the ways of characterising that a continuous random variable $X$ is constant?

For example, if $Y$ is a discrete random variable with pdf $p_Y$ then we can say $Y$ is constant almost surely if $p_Y(c) = 1$ for some $c$.

For a continuous random variable with cdf $F_X$, we could say that $F_X$ is constant if there exists $c$ such that $F_X(c) = 1$ and for all $a < c$ we have $F_X(a) = 0$. However, this characterisation seems clumsy. I'm wondering which characterisations are more straight-forward to verify, or could make a clearer definition.

For instance, here's a characterisation that I think is correct and is arguably more elegant than the previous one:

$X$ is almost surely constant if $F_X(\mathbb R) = \{0,1\}$.

I'd intuitively like to say that $P(X = c) = 1 \implies X$ is constant, but I'm not sure I can safely consider the event $\{X = c\}$ for a continuous r.v. $X$.

$\endgroup$
2
$\begingroup$

A continuous random variable can't be almost surely constant. If $X$ is continuous then $P(X=c)=0$ for all $c\in\mathbb{R}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What about $X:[0;1]\longrightarrow \mathbb R, x \longmapsto 1$ with probability space $([0;1],\mathfrak B \cap [0;1],\lambda)$ and measure space $(\mathbb R, \mathfrak B)$? Then $X$ is a continuous random variable that is constant and thus also almost surely constant. $\endgroup$ – C. Brendel Jun 6 at 16:12
  • $\begingroup$ This random variable is not continuous. It doesn't have a continuous CDF, there is a discontinuity at the point $t=1$. (because $P(X\leq t)=0$ for all $t<1$ but $P(X\leq 1)=1$) $\endgroup$ – Mark Jun 6 at 17:05
  • $\begingroup$ Oh ok, I was thinking of topological continuity. Good to know there is a difference! $\endgroup$ – C. Brendel Jun 6 at 18:04
  • 1
    $\begingroup$ Yes, these are different terms. Actually, in most probability spaces we don't even have a topology. $\endgroup$ – Mark Jun 6 at 18:07
  • $\begingroup$ ah interesting, so via the cdf you are basicially pulling back the continuous property $\endgroup$ – C. Brendel Jun 6 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.