Given $$f(x) = \begin{cases}
x, \; 0 < x <1\\
2-x, \; 1 \leq x < 2\\
0 \text{ everywhere else}
\end{cases}$$
as our P.D.F, I must find the corresponding distribution function.
I know that $F(x) = P(X \leq x) = \int_{-\infty}^{x}f(t) dt$ is the distribution function, but I don't know how to apply it to this particular probability density function. The P.D.F has three different cases. How do I handle that? A sum of integrals with the appropriate bounds for each case of $f(x)$?
I'm kind of confused on how to create those bounds. How do we get rid the of the $-\infty$?
I'm thinking:
For $x > 0$:
$$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)dt = \int_{0}^{1} x dx + \int_{1}^{2} 2-x dx$$
Eh….
Best Answer
Discern the following cases:
$x\leq 0$ $$F(x)=\int_{-\infty}^{x}0 dt=0$$
$0<x\leq1$ $$F(x)=F(0)+\int_{0}^{x}tdt$$
$1<x\leq2$ $$F(x)=F(1)+\int_{1}^{x}(2-t)dt$$
$x>2$ $$F(x)=F(2)=1$$