mixed random variable cdf Let
$$F(x) =\cases{ 0, & $ x\lt0$ \cr
          x^2+0.2, &  $0\le x\lt 0.5 $\cr
          x,&  $0.5\le x\lt 1$ \cr
          1,&  $x\ge 1$. }$$
How do I rewrite $F(x)$ like $p_1F_c(x)+p_2F_d(x)$, where:  $p_1+p_2=1$,
$F_c$ is a continuous c.d.f,
and $F_d$ is a discrete c.d.f?    
 A: There is a mass of $0.2$ at $0$ and $0.05$ at $0.5$, total mass $0.25$. This $0.25$ will be your $p_2$. To make $p_2$ times the discrete part right, for the discrete part the $F_d$ is the cumulative distribution function which comes from putting $p_d(0)=4/5$, $p_d(0.5)=1/5$.  This is because the discrete mass of $0.20$ at $0$ bears the ratio $\frac{0.20}{0.25}$, that is, $4/5$, to the total combined discrete masses. 
Now $F_d$ is easy to write down. Do remember that $F_d$ is defined for all $x$, though admittedly it is pretty dull for $x&lt0$ and for $x \ge 0.5$. Come to think of it, it is not particularly exciting for $0\le x&lt0.5$ either.  
For the continuous part, the weight $p_1$ is $0.75$. Now subtract the discrete weights from the given cumulative distribution function, scale up by $4/3$ so that multiplication by $0.75$ makes things come out right.  
For the masses before scaling up, we will have mass $0$ up to $0$, then $x^2$ up to $0.5$. From $0.5$ to $1$ we have $x-0.2-0.5=x-0.25$, and from $1$ on we have $0.75$.  From this $F_c$ is easy to describe. It is $\frac{4}{3}x^2$ between 
$0$ and $0.5$, and $\frac{4}{3}(x-0.25)$ between $0.5$ and $1$, and the usual things elsewhere.
