expectation value of a specific distribution function

Question

for a random variable X with this distribution function : F(x) = \begin{cases} 0, & x < 0 \\ \frac{(x+1)}{4}\, & 0 \leq x < 1 \\ 1, & x \geq 1 \end{cases} i want to find the expectation value and i know it is 5/8 but i don't know how to solve it!

Answer 1

I think you are possibly thrown off by the fact that $X$ is neither a discrete nor a continuous random variable. Note that $F(x)$ has jumps at $0$ and $1$, i.e., $X=0$ with probability $\dfrac{1}{4},$ and $X=1$ with probability $\dfrac{1}{2}.$ $X$ is continuous elsewhere. You can think of $X$ as

$$X=\begin{cases} 0 \hspace{2.5cm}\text{ with prob. }\dfrac{1}{4} \\\text{Unif}(0,1) \hspace{1cm}\text{ with prob. }\dfrac{1}{4} \\1 \hspace{2.5cm}\text{ with prob. }\dfrac{1}{2}. \end{cases}$$

By linearity of expectation, $\mathbb{E}(X)=0\times \dfrac{1}{4}+\mathbb{E}(\text{Unif}(0,1))\times \dfrac{1}{4}+1\times \dfrac{1}{2}=0+\dfrac{1}{2}\times\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{5}{8}.$

In any case, as pointed out in the comments, you can use $$\mathbb{E}(X)=\int_{0}^{\infty}(1-F(x))dx-\int_{-\infty}^{0}F(x)dx=\int_{0}^1\dfrac{3-x}{4}dx=\left[\dfrac{3x-x^2/2}{4}\right]_0^1= \dfrac{5}{8}.$$