Probability Density Function from: $F(x)=x , \text{ for } 0\leq x\leq \frac12$ Probability Density Function from:
F(x)=begin{cases}0 & \text{if }x<0\
x & \text{if }0\leq x\leq\frac{1}{2}\
1 & \text{if }x>\frac{1}{2}
\end{cases}.
Do somebody know how to determine the p.d.f from that $F(x)$?
actually, I have tried to solve it using simple derivative of $F'(x)$
with p.d.f appear following below
f(x)=begin{cases}1 & \text{if }0\leq x\leq \frac{1}{2}\0 & \text{else}\end{cases}.
but when i tried to show that those p.d.f is satisfied by integration,
then unfortunately i found that the integration of $f(x)$ not equals to $1$.
please show me the way if you have the others ideas. thanks.
 A: As several people have pointed out, this is not valid as you have it because one requirement of a CDF is that it is right continuous.  This means that if there is a jump in the graph of the CDF, the open hole is on the left and the point is on the right.  Your function is the opposite since $F(1/2)$ is defined based on the left side of $\frac{1}{2}$.  So, let me start by redefining your $F(x)$ to make it right continuous.  Perhaps there was a typo somewhere along the way.  This problem is still interesting after we change that.
Your $F(x)$ is equal to $x$ for $0 \leq x < \frac{1}{2}$, but then $F(x) = 1$ for $x \geq \frac{1}{2}$.  This is perfectly valid, including being right continuous.  The problem is, it is not a discrete distribution and it is not a continuous distribution.  Since it is not a continuous distribution, it does not have a PDF technically.  But, you could say that the PDF is 1 for $0 \leq x < \frac{1}{2}$.  The problem is, this distribution also has a point mass.  That is,
$$P(X = \frac{1}{2}) = P(\frac{1}{2} \leq X \leq \frac{1}{2}) = F(\frac{1}{2}) - \lim_{x \to \frac{1}{2}^- }F(x) = 1 - \frac{1}{2} = \frac{1}{2}.$$
This point mass portion is more like a discrete distribution.  With continuous distributions, you integrate the PDF.  With discrete distributions, you sum the probability function or probability mass function, whatever you call it.  So, here we would do a combination
$$\int_0^{1/2} 1 \,dx + \frac{1}{2} = 1.$$
It still works out.
Many textbooks don't even bother with this type of distribution.  They talk about continuous ones and discrete ones but don't bother with the mixed distributions that are partially continuous and partially discrete.
