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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?

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I edited your post to use TeX formatting. Please check that it reflects your original intent. –  David Mitra Apr 13 '12 at 15:14
    
yes, it's correct. –  paul Apr 13 '12 at 15:17
    
Where are the jumps of $F$? Hint: there are exactly two of them. –  Did Apr 13 '12 at 15:25
    
I know there are two jumps but I don't know how can I determine p1 and p2. –  paul Apr 13 '12 at 15:29
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1 Answer

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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 $x<0$ and for $x \ge 0.5$. Come to think of it, it is not particularly exciting for $0\le x<0.5$ 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.

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sorry, I still can't understand your answer precisely. There are two equations in F(x) and they are both continuous, right? Then should I multiply p2 both of them? –  paul Apr 13 '12 at 17:02
    
$p_1=\frac{3}{4}$. $F_c=0$ for $x<0$, $\frac{4}{3}x^2$ for $0\le x< 0.5$, $\frac{4}{3}(x-0.25)$ for $0.5\le x<1$, $1$ for $x\ge 1$. And $p_2=\frac{1}{4}$, $F_d$ is $0$ for $x<0$, $4/5$ for $0\le x<0.5$, $1$ for $x>0.5$. If added text explanation is needed, please leave message. (Sorry, was out, slow leak in tire needed fixing.) –  André Nicolas Apr 13 '12 at 18:00
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