Converting a probability density function into its Probability Distribution

Question

I have the following probability density function

$$f(x) =\begin{cases} 4x & \mbox{for }0< x < 1/2 \\ 4-4x & \mbox{for }1/2 \leq x < 1 & \\ 0 & \mbox{otherwise}\end{cases}$$

I am tasked to now find its probability distribution function in the same piece wise format. I have the solution to this problem however I do not really understand the solution.

The solution is the integral from 0 to x of 4x(for the first interval of 0 < x<= 1/2)+ integral of 0 to 1/2 of 4x + integral of 1/2 to x of 4-4x(for the second interval).

I understand the first interval but I am stumped as to why we add the integral of 0 to 1/2 of 4x to the integral of 1/2 to x for the second interval.

Any explanation would be appreciated.

Answer 1

I guess it is asking for the CDF. It's hard to tell what you are writing since you didn't format.

Essentially, it should be

$$F_X(x) = P(X\leq x)=\begin{cases} 0& x<0\\ \int_0^x 4t\,dt& 0\leq x< \frac{1}{2}\\ \int_0^{1/2} 4t\,dt+\int_{1/2}^x4-4t\,dt&\frac{1}{2}\leq x <1\\ 1& x\geq 1\end{cases}$$

Loosely speaking, this is the case because the CDF is the accumulation of probability (area) under the curve up to $x$.

Answer 2

You have $$f(x) =\begin{cases} 4x & \mbox{for }0< x < 1/2 \\ 4-4x & \mbox{for }1/2 \leq x < 1 & \\ 0 & \mbox{otherwise}\end{cases}$$

Then you want to find

$$F(x) =\begin{cases} 0 & \mbox{for } x\leq 0 \\[1ex] \int_0^x 4s\operatorname d s & \mbox{for }0< x < 1/2 \\[1ex] \int_0^{1/2} s\operatorname d s + \int_{1/2}^x 4-4s\operatorname d s & \mbox{for }1/2 \leq x < 1 & \\[1ex] 1 & \mbox{for } 1\leq x\end{cases}$$