Let $X$ be a random variable with the following cumulative distribution function:

$$F(x)= \begin{cases} 0 & \quad x<0\\ x^2 & \quad 0\leq x<\frac{1}{2}\\ \frac{3}{4} & \quad \frac{1}{2}\leq x<1\\ 1 &\quad x\geq1. \end{cases}$$

Then what is the value of $P(\frac{1}{4}<X<1).$

I am trying to solve this problem as:

$P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)$, since for continuous random variable probability at a point is always zero. Thus $P(\frac{1}{4}<X<1)=P(\frac{1}{4}<X\leq1)=F(1)-F(\frac{1}{4})=\frac{15}{16}=0.9375$. But the answer didn't match. Answer is 0.68. So where did i wrong. Any suggestion or solution regarding this should be highly appreciated.

  • 1
    $\begingroup$ Who told you the CDF is continuous at $x=1$? More care, please... $\endgroup$ – Did Jan 27 '18 at 8:57
  • $\begingroup$ @Did I guess it as the range of the random variable $X$ is in piece-wise continuous form. $\endgroup$ – SAHEB PAL Jan 27 '18 at 17:12
  • $\begingroup$ Stop guessing then, and start using solid definitions. Just my two cents. $\endgroup$ – Did Jan 27 '18 at 18:25


| cite | improve this answer | |
  • $\begingroup$ Using \left( and \right) delimiters might make formatting better. $\endgroup$ – TheSimpliFire Jan 27 '18 at 9:28
  • $\begingroup$ Tnx............ $\endgroup$ – Mostafa Ayaz Jan 27 '18 at 9:34
  • $\begingroup$ $F(x)=F(x^+)$ for every $x$ and every CDF $F$. $\endgroup$ – Did Jan 30 '18 at 7:34
  • $\begingroup$ Nope this is completely true. This also gets important specially if you have Dirac delta function in PDF. $\endgroup$ – Mostafa Ayaz Jan 30 '18 at 17:16
  • $\begingroup$ That's why the inequalities in the question have been defined so. Also $F(x)$ $F(x^{-})$ and $F(x^+)$ are three distinct number generally and they are equal only if $F(x)$ is continuous. $\endgroup$ – Mostafa Ayaz Jan 30 '18 at 17:17

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.