1
$\begingroup$

I was researching why it is that the cumulative distribution function of a discrete random variable could not exist at a median of $0.5$ in some cases. I came across this answer by the user V.Vancak, which answered my initial question. But at the end of his post he says

... hence you have to chose $0$ or $1$ as the median. Any value in $(0,1)$ will not change a thing as $X$ cannot receive these values.

If the cumulative distribution function does not exist at a median of $0.5$, then why, according to V.Vancak, must we select $0$ or $1$ as the new median? How does that solve the issue of there being no value of the CDF that will satisfy a median of $0.5$? Why will no other value in $(0, 1)$ suffice?

I would greatly appreciate it if people could please take the time to clarify this.

$\endgroup$
  • $\begingroup$ The example he mentioned is a Binomial random variable with support $\{0, 1, 2, 3\}$. For a discrete random variable like this we know that there is a jump in the CDF at those support points, with the jump size equals to the pmf, while at the other point the CDF is just a horizontal line. So it is easy to see that in general, there may not be any point satisfying $F_X(x) = 0.5$ because the CDF may directly jump over the $0.5$ level. That's why you have the inequality definition to cater these cases. $\endgroup$ – BGM Feb 21 '18 at 6:23
  • 1
    $\begingroup$ CDF always exists. The linked answer is not about CDF (it exists) but about definition of the median of discrete distribution. $\endgroup$ – kludg Feb 21 '18 at 6:28
  • 1
    $\begingroup$ And for any $m \in (0, 1)$, as mentioned in above, the probability $\Pr\{X \leq m\}$ and $\Pr\{X \geq m\}$ will remain a constant. Using his example of $n = 3, p = 1/4$, you just need to check the support points without need to search for other points. You can easily check that $\Pr\{X \leq 1\} \approx 0.84 > 0.5$ and $\Pr\{X \geq 1\} \approx 0.58 > 0.5$ so $1$ is the median. For any other points in the small neighborhood of $1$, you will lost the mass of $\Pr\{X = 1\} \approx 0.42$ so you do not need to check that. For the other support points you can easily verify them. $\endgroup$ – BGM Feb 21 '18 at 6:35
  • $\begingroup$ @kludg Ahh, you’re right. I must have confused myself. $\endgroup$ – The Pointer Feb 21 '18 at 12:36
  • $\begingroup$ @BGM Thank you. I understand now. $\endgroup$ – The Pointer Feb 21 '18 at 14:33

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.