1

Given $$F(x)= \begin{cases} 0 &x<0\\ x &0\leq x\leq1\\ 1 &x>1\end{cases}$$

Why this distribution function is discontinuous at $x=0$ and $x=1$

But according to me $F(1)=1$ and $$\lim(1-h)=1 F(1+)=1$$ and similarly for $0$

Then how can we say that it is discontinuous at $x=0$ and $x= 1$

lioness99a
  • 4,943
Akanksha Singh
  • 11
  • $F$ is continuous at all points. – Kavi Rama Murthy Mar 10 '20 at 09:39
  • The $F$ you give is a cdf, not a pdf. It is continuous everywhere. Its associated pdf (its derivative) $f$ is not even defined in $x=0$ and $x=1$ (characteristic fonction of interval $[0,1]$). – Jean Marie Mar 10 '20 at 09:41
  • This is a CDF, a cumulative distribution function. This function is continuous everywhere as you have seen. However, the CDF has an associated density which cannot be defined at $0$ and $1$ but can be done everywhere else. If you like, I can write an answer for this question. – Sarvesh Ravichandran Iyer Mar 10 '20 at 10:45

0 Answers0