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Let x have a piecewise right hand continuous pdf, $f_X(x)$, which is defined as follows.

$$f_X(x) = \begin{cases} \frac{1}{2}(x+2)^2 & -2 \leq x < -1 \\ -\frac{1}{2}x +\frac{1}{12}\delta (x+1) & -1 \leq x < 0 \\ \frac{1}{2}x & 0 \leq x < 1 \\ \frac{1}{2}(x-2)^2 + \frac{1}{12}\delta(x-1) & 1 \leq x \leq 2 \\ \end{cases} $$

Let Y = $F_X(x)$. Compute $f_Y(y)$.

My approach: I have tried writing this:

$$ F_Y(y) = Pr\{Y \leq y\} = Pr \{F_X(x) \leq y\} $$

But I honestly have no idea how to handle this piecewise function in this situation. So far, I've tried to compute the derivative of $F_Y(y)$ using: $$ f_Y(y) = lim_{dy->0^+}\frac{Pr\{Y \leq y + dy \} - Pr\{Y \leq y \}}{dy} $$

to no avail.

Shooting Stars
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Farhood ET
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1 Answers1

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To explain the solution step by step, a couple of drawings are needed but with this sketch I am sure you will be able to conclude by yourself.

First observe, that IF your X rv was continuous, $Y\sim U(0;1)$ by integral transform, being

$$F_Y(y)=F_X\left[F_X^{-1}(y) \right]=y$$

Your case is very similar but you have 2 discontinuity points in $F_X$ or, equivalently, you have 2 dirac impulses of $1/12$ each.

Immediately the result is that

$$f_Y(y)=\frac{1}{12}\delta\left( y-\frac{3}{12} \right)+\frac{1}{12}\delta\left( y-\frac{10}{12} \right)+\mathbb{I}_{\left(0;\frac{2}{12} \right)\cup \left(\frac{3}{12};\frac{9}{12} \right)\cup\left(\frac{10}{12};1 \right)}$$

or alternatively, using special function $\delta$ and $\text{Rect}$,

$$f_Y(y)= \text{Rect}\left( \frac{12y-1}{2} \right)+ \frac{1}{12}\delta\left( y-\frac{3}{12} \right)+ \text{Rect}\left( \frac{12y-6}{6} \right) +\frac{1}{12}\delta\left( y-\frac{10}{12} \right)+\text{Rect}\left( \frac{12y-11}{2} \right) $$

Observe also that $f_Y(y)$ is not a pdf, being it not absolutely continuous. In some branches as Signal Theory they define these function as "mixed densities"

To derive analytically this solution I suggest you

  • First derive $F_X(x)$

$$F_X(x) = \begin{cases} 0, & \text{if $x<-2$} \\ \frac{(x+2)^3}{6}, & \text{if $-2\le x<-1$} \\ \frac{1}{4}+\frac{1-x^2}{4}, & \text{if $-1\le x<0$} \\ \frac{1}{2}+\frac{x^2}{4}, & \text{if $0\le x<1$} \\ 1+\frac{(x-2)^3}{6}, & \text{if $1\le x<2$} \\ 1, & \text{if $x\ge 2$ } \end{cases}$$

  • Do a drawing of $F_X$

  • Do a drawing of its inverse, $F_Y$

  • Derive $f_Y$, considering that it is discrete in 2 points

tommik
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