0
$\begingroup$

Consider a random variable $X$ defined on $[0,1]$, which is $ X(\omega)= \begin{cases} 0.5, \omega \in [0,0.5]\\ \omega, \omega \in (0.5,1] \end{cases}$. Intuitively, this definition makes sense to me. However, $X$ is $0.5$ for a fixed fraction of the domain. Thus, $\Pr[X = 0.5] = 0.5$, which cannot be true, right? Also, if I'm not mistaken the CDF would not be continuous.

Does this mean there cannot be a continuous random variable which has one value for a fraction of its domain? Or what am I confusing here?

Edit: Sorry I forgot. The pdf of $X$ is the uniform distribution on $[0,1]$.

$\endgroup$
  • $\begingroup$ Is $X(\omega)$ your PDF, i.e. $f_X(x)$? $\endgroup$ – gunes Mar 4 at 8:32
  • $\begingroup$ $X$ is not a continuous random variable; it is a mixed random variable that takes on values in $[\frac 12,1]$ with a probability mass or atom of $0.5$ at $\frac 12$. The CDF has a jump discontinuity at $\frac 12$ and is right-continuous on $[\frac 12, \infty)$. $\endgroup$ – Dilip Sarwate Mar 4 at 13:57

Your Answer

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

Browse other questions tagged or ask your own question.