# Finding Distribution function from probability density function

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....

## 2 Answers

Discern the following cases:

• $x\leq 0$ $$F(x)=\int_{-\infty}^{x}0 dt=0$$

• $0<x\leq1$ $$F(x)=F(0)+\int_{0}^{x}tdt$$

• $1<x\leq2$ $$F(x)=F(1)+\int_{1}^{x}(2-t)dt$$

• $x>2$ $$F(x)=F(2)=1$$

Given $$f(x) = \begin{cases} x & 0 < x <1\\ 2-x & 1 \leqslant x < 2\\ 0 & \text{ everywhere else} \end{cases}$$ Then $$F(x) = \begin{cases} 0 & x \leqslant 0\\ \int_0^x x\operatorname d x & \; 0 < x <1\\ F(1)+\int_1^x 2-x \operatorname d x & 1 \leqslant x < 2\\ 1 & 2\leqslant x \end{cases}$$

• $\int_{0}^{x}xdx$ must be changed into something like $\int_{0}^{x}tdt$ . Sortlike for $\int_{1}^{x}2-xdx$ – drhab Sep 24 '14 at 6:39