1
$\begingroup$

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?

  • 0
    I know there are two jumps but I don't know how can I determine p1 and p2.2012-04-13

1 Answers 1

0

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.

  • 0
    $p_1=\frac{3}{4}$. $F_c=0$ 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.)2012-04-13