# Find expected value from given PDF (CDF)

The probability distribution function (or Cumulative Distributions Function) of a discrete random variable $X$ is given by $$$$F_X(x) = \begin{cases} 0, & \text{for x<-2.5}.\\ 0.3, & \text{for -2.5\le x< 0}.\\ 0.6, & \text{for 0\le x< 1}.\\ 1, & \text{for } x\ge 1\\ \end{cases}$$$$ a) Find the probability mass function for this random variable, i.e $P(X = x_j ) = p (x_j)$
b) $𝑃(𝑋<1 ) =$ ?
c) Find the Expected value of $X$. ($𝐸(𝑋)=$?)
d) Find the variance and standard deviation of X. ( $\operatorname {𝑣𝑎𝑟}(𝑋)=$? and $\sigma_𝑋=$?)

here what I tried:
a) $$$$p(x) = \begin{cases} 0, & \text{for x<-2.5}.\\ 0.3, & \text{for -2.5\le x< 0}.\\ 0.3, & \text{for 0\le x< 1}.\\ 0.4, & \text{for } x\ge 1\\ \end{cases}$$$$

b) $P(X<1 )= P(X<1 )=0.3+0.3=0.6$
c) I know this : $E(X) = p*f(x_i) +...$ but i dont think it is suitable neither this: $\int_{a}^{b}xf(x)dx$
so to find expected value what method i can use ? and for question a) and b) are correct right ? Thank you

• Part (a) isn't right. The pmf for a discrete random variable should be defined by point masses, not over intervals. Once you fix that, it should help you with (c) and (d). Part (b) is ok. – Mick A Dec 22 '15 at 20:37

$X$ is a discrete random variable. So it assumes discrete integral values only.
Hence your $p(x)$ can be correctly written as: $$$$p(x) = \begin{cases} 0.3, & \text{for x=-2}.\\ 0.3, & \text{for x=-1}.\\ 0.3, & \text{for x= 0}.\\ 0.4, & \text{for x= 1}.\\ \end{cases}$$$$
As a result, $E(X)=-0.6-0.3+0.4=-0.5$