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!?? $:)$
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$\begingroup$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, $$F(y) = \begin{cases} 0 & y <-1\\ \frac12 + y + \frac{y^2}{2} & y \in[-1,0]\\ \frac12 + y - \frac{y^2}{2} & y \in[0,1]\\ 1 & y > 1\end{cases}$$
$\endgroup$ $\begingroup$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 rightof $x_0$ 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.
$\endgroup$ 2 $\begingroup$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 $y\ge+1$...
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