How to compute the probabilities of discrete distribution

Question

$$F_X(x)=\begin{cases} 0, \quad \quad \quad \quad x<-1 , \\ \\ \frac{x}{4} +1/4, \ \ −1 ≤ x < 0 \\ \\ \frac12 ,\ \ \ \ \ \quad \quad 0 ≤ x < 1 \\ \\ \frac{x}{12}+\frac{7}{12}, \ \quad 1 ≤ x < 2 \\ \\1, \ \ \ \ \ \quad \quad 2≤ x \end{cases}$$

Here's the solution to it which i don't understand how to derive:

I'm not sure how to compute these discrete probabilities, since I'm confused how to calculate the value of $x$. Could you please hint me?

Answer 1

Firstly you have to check the conditions that the cdf is right-continuous. For instance, $P(X< 1)=\frac12$. This can be read off directly from the cdf. And $P(X\leq 1)=\frac{1}{12}+\frac{7}{12}=\frac{8}{12}$ Thus there is a gap of $\frac{8}{12}-\frac12=\frac16$. And this probability corresponds to $P(X=1)$.

For continuous, discrete and mixed distributions we have the following relations:

• $P(a\leq X\leq b)=P(X\leq b)-P(X<a)$
• $P(X>c)=1-P(X\leq c)$

I hope these hints help you to comprehend the solution.