calculate CDF from given PDF

Question

I am trying to understand the calculate the CDF from the given PDF

$f(x) = \begin{cases} 0.5& 0\le x<1\\ 1& 1\le x<1.5\\ 0& \text{otherwise}\end{cases}$

The CDF is

$F(x) = \begin{cases}0 & x<0\\ 0.5x & 0\le x<1\\ x-0.5 & 1\le x<1.5\\ 1 & x\ge 1.5\end{cases}$

Can anyone please describe how

$0.5x$ for $0\le x<1$

and

$x-0.5$ for $1\le x<1.5$

are calculated?

Answer 1

If you know the pdf $f(x)$ then the cdf is obtained via integration

$$ F(x) = \int_{-\infty}^x {f(t)}~{\rm d}t \tag{1} $$

For your case

$$ f(x) = \begin{cases} 1/2 & 0 \leq x <1\\ 1 & 1 \leq x < 3/2\\ 0 & \mbox{otherwise} \end{cases} \tag{2} $$

It is just a matter on calculating the integral

$x < 0$

If $x < 0$ then, $f(x) = 0$ and

$$ F(x) = \int_{-\infty}^{x}f(t)~{\rm d}t = 0 \tag{3} $$

$ 0 \leq x < 1$ in this range the $f(x) = 1/2$ so that

$$ F(x) = \int_{-\infty}^{x}f(t)~{\rm d}t = \int_{-\infty}^{0}f(t)~{\rm d}t + \int_{0}^{x}f(t)~{\rm d}t = 0 + \int_{0}^{x}\frac{1}{2}~{\rm d}t = \frac{x}{2}\tag{4} $$

$1 \leq x < 3/2 $

As before

\begin{eqnarray} \int_{-\infty}^{x}f(t)~{\rm d}t &=& \int_{-\infty}^{0}f(t)~{\rm d}t + \int_{0}^{1}f(t)~{\rm d}t + \int_{1}^{x}f(t)~{\rm d}t \\ &=& 0 + \int_{0}^{1}\frac{1}{2}~{\rm d}t + \int_{1}^{x}1~{\rm d}t \\ &=& \frac{1}{2} + (x - 1) = x - \frac{1}{2} \tag{5} \end{eqnarray}

$x > 3/2$

Finally

\begin{eqnarray} \int_{-\infty}^{x}f(t)~{\rm d}t &=& \int_{-\infty}^{0}f(t)~{\rm d}t + \int_{0}^{1}f(t)~{\rm d}t + \int_{1}^{3/2}f(t)~{\rm d}t + \int_{3/2}^{x}f(t)~{\rm d}t\\ &=& 0 + \int_{0}^{1}\frac{1}{2}~{\rm d}t + \int_{1}^{3/2}1~{\rm d}t + 0\\ &=& \frac{1}{2} + \frac{1}{2} = 1 \tag{6} \end{eqnarray}

If you put everything together you'd get

$$ F(x) = \begin{cases} 0 & x < 0 \\ x/2 & 0 \leq x < 1 \\ x - 1/2 & 1 \leq x < 3/2 \\ 1 & x > 3/2 \end{cases} $$