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 $\mathbb E(X)$


  • $\begingroup$ Hint: Notice the massive point. Then apply the usual definitions. $\endgroup$ – Graham Kemp Nov 6 '17 at 20:40

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 $F(x)$.


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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.