Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The PDF for $Y$ is $$f_Y(y) = \begin{cases} 0 & |y|> 1 \\ 1-|y| & |y|\leq 1 \end{cases}$$

How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get $$F_Y(y)=\begin{cases} 1/2 -y/2-y^2/2 & [-1,0] \\ 1/2-y/2+y^2/2 & [0,1] \end{cases} $$ by using the fact that $F_Y(y)=\int _{-\infty}^{y}{f_Y(y)}\,dy$, however my text claims the answer is $$F_Y(y)=\begin{cases} 1/2 +y+y^2/2 & [-1,0] \\ 1/2+y-y^2/2 & [0,1] \end{cases} $$ I am struggling with pdf and cdfs, so I asssume I did something wrong other than the simple integration. Who's correct? Me or the Text!?? $:)$

share|improve this question
1  
HINT:Does you solution have the property $F_Y(-1) = 0$ and $F_Y(1) = 1$ ? –  Sasha Mar 11 '12 at 1:47
    
It does not, and I know those are properties my answer should have. But why should my answer have to be fiddled with if I used the definition of the cdf? –  cap Mar 11 '12 at 1:50
1  
You probably did the integration wrong, and were too distracted by your struggle with PDFs and CDFs to notice :) By the way, I edited your answer to use \begin{cases}...\end{cases}, which is a better way to typeset piecewise functions. Feel free to click 'edit' to see what I did and use it in the future. –  Rahul Mar 11 '12 at 1:53
    
@Sasha I found my mistake. I had forgot about that fact and I could've used it to find my constants of integration, but for some reason I integrated by actually breaking up the graph to find the areas. –  cap Mar 11 '12 at 1:58
    
@RahulNarain, is there a software I can download to type in TeX for practice and HW, etc? Thanx –  cap Mar 11 '12 at 1:59
show 2 more comments

3 Answers

up vote 3 down vote accepted

We have that $F(y) = \displaystyle \int_{-\infty}^y f(x) dx$. In your case, we are given that $$f(x) = \begin{cases} 0 & x <-1\\ 1 + x & x \in[-1,0]\\ 1-x & x \in [0,1]\\ 0 & x > 1\end{cases}$$

  • If $y < -1$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^y 0 dx =0 $. We have the integrand $f(x) = 0$ since $x \leq y < -1$.
  • If $y \in [-1,0]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-1}^{y} f(x) dx = \displaystyle \int_{-1}^{y} \left( 1+x \right) dx = \left( x + \frac{x^2}{2} \right)_{-1}^{y} $$ $$F(y) = \left(y + \frac{y^2}{2} \right( - \left( -1 + \frac12 \right) = \frac12 + y + \frac{y^2}{2}.$$
  • If $y \in [0,1]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx = \displaystyle \int_{-1}^{0} \left( 1+x \right) dx + \displaystyle \int_{0}^{y} (1-x) dx$$ Hence, $$F(y) = \frac12 + \left( x - \frac{x^2}{2}\right)_0^{y} = \frac12 + y - \frac{y^2}{2}$$
  • For $y > 1$, since $f(x) = 0$ for all $x>1$, we have that $F(y) = F(1)$ for all $y > 1$. Hence, $F(y) = F(1) = 1$.

Hence, $$F(y) = \begin{cases} 0 & y <-1\\ \frac12 + y + \frac{y^2}{2} & y \in[-1,0]\\ \frac12 + y - \frac{y^2}{2} & y \in[0,1]\\ 1 & y > 1\end{cases}$$

share|improve this answer
add comment

This is the kind of problem that gives integration a bad name among students.

  • Draw a graph of the density function. It looks like an isoceles right triangle with hypotenuse $2$ and apex at $(0,1)$ and very obviously has area $1$ (useful as a check on one's work.)

  • For any $x_0$, $F(x_0)$ is the area under the density function to the left of $x_0$. It should be obvious that $F(x_0) = 0$ if $x_0 \leq -1$ and $F(x_0) = 1$ if $x_0 > 1$.

  • Pick an $x_0$ in $[-1,0]$. The area to the left of $x_0$ is a right triangle with altitude $1+x_0$ and base $1+x_0$ (or $x_0 - (-1)$ if you like, and so $F(x_0) = \frac{1}{2}(1+x_0)^2$. Quick check: value is $\frac{1}{2}$ at $x_0 = 0$ and $0$ at $-1$.

  • Pick an $x_0$ in $[0,1]$. By symmetry, the area to the right of $x_0$ is $\frac{1}{2}(1-x_0)^2$. Quick check: value is $\frac{1}{2}$ at $x_0 = 0$ and $0$ at $1$. Hence, $F(x_0) = 1 - \frac{1}{2}(1-x_0)^2$.

Putting it all together, we get the same answer as Sivaram Ambikasaran. It takes longer to write out the instructions than to just work off the diagram.

share|improve this answer
    
Nice answer. I like the fact that you try and avoid integration as much as you can. +1. :-) –  user21436 Mar 11 '12 at 5:17
1  
@KannappanSampath Thanks. Since I taught probability over many years to recalcitrant engineering students who invariably fail to distinguish between antiderivatives and integrals, I like to keep things as simple and visual as possible. Understanding and remembering $F(x)$ as the area under $f(x)$ to the left of $x$ is easier than wondering what is this strange $t$ that suddenly showed up in $$F(x) = \int_{-\infty}^x f(t)\mathrm dt$$ and how $f(x) and $f(t)$ are related. Trivial to a mathematician; difficult for an engineering student to whom calculus is a dimly remembered haze of formulas. –  Dilip Sarwate Mar 11 '12 at 11:46
add comment

First work with $y\le 0$ to obtain

$$F_Y(y)=\int_{-1}^y 1-|u|du=\int_{-1}^y 1+u du=y-(-1)+\frac{y^2-(-1)^2}{2}=\frac{1}{2}+y+\frac{1}{2}y^2 $$

Now work with $1\ge y\ge0$ by splitting (using the fundamental theorem of calculus)

$$F_Y(y)=\int_{-1}^y f_Y(u)du=F_Y(0)+\int_0^y 1-u du $$

Now figure out what $F_Y$ must be for $y\le-1$ and $y\ge+1$...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.