How do I get the expected value of this random value knowing its c.d.f.

Question

Can anyone answer with steps how to get the expected value of this random variable?

Let $X$ be a random variable with following c.d.f,

$$F(x) = \begin{cases}0 &x < -1\\ \dfrac{1-x^2}4 & -1 \le x < \dfrac{-1}{\sqrt 2}\\ \dfrac12 - x^4 & \dfrac{-1}{\sqrt 2} \le x < 0 \\ \dfrac34 + x & 0 \le x \lt\dfrac14 \\ 1 &x \ge \dfrac14 \end{cases}$$

Find $\mathbb E(X)$

Source

Answer 1

You can compute the expectation directly from the cdf using the formula: \begin{align*} \operatorname E[X] &= \int_0^\infty (1-F(x)) \, dx - \int_{-\infty}^0 F(x)\, dx \end{align*} In order to solve these integrals, you have to break the region of integration into the regions that were used to define $F(x)$.

Answer 2

In the interiors of these intervals, you can differentiate to get the p.d.f. But in addition there may be point masses at the boundaries between the intervals. Those give you a discrete part of the probability distribution. Thus the expected value is $$ \operatorname E(X) = \int_{-\infty}^\infty x F'(x)\, dx + \sum \Big\{ x f(x) : x \text{ is a boundary point where there is a point mass} \Big\}. $$ Notice that the integral is only over a bounded interval since $F'$ is $0$ on the two unbounded components.