Finding the expectancy of a distribution function

Question

I have the function below and I'm trying to calculate the expectancy of X. The answer is E(X) = $\frac{59}{40}$ but I don't know what I'm doing wrong because I cannot achieve it.

F(x) = \begin{cases} 0, & \text{if $x$ < 0} \\ \frac{1}{4}, & \text{if 0 $\le$ $x$ < 1} \\ \frac{2}{5}, & \text{if 1 $\le$ $x$ < 2} \\ \frac{2x-3}{2}, & \text{if 2 $\le$ $x$ < 2.5} \\ 1, & \text{if $x$ $\ge$ 2.5} \\ \end{cases}

I did like that

$$E(X) = \frac{1}{4}*1 + \frac{2}{5}*1 + \int_2^{2.5} \frac{2x-3}{2} dx$$

which gave me thw wrong answer. I'm sure if the problem is in the computation of the integral (I'm a little bit rusty in calculus) or of the problem is the concept of my calculations as a whole

Answer 1

Its the Cumulative distribution function you have there not the probability density function. You have to convert the CDF first if you want to calculate the expected value of $X$. Or use

$$E(X) = \int_0^\infty (1-F_X (x)) \, dx = \frac{3}{4} + \frac{3}{5} + \frac{1}{8} = \frac{59}{40}$$

Answer 2

This is a mixed distribution to begin with. It has atoms at $0$, $1$ and $2$, meaning that the cumulative distribution function jumps at these points. To calculate the expected value of such a distribution, you would like to look at $$ E[X] = \sum_{x_i \mbox{ is a mass point}}^{}p_i x_i + \int_{x \mbox{ is not a mass point}} x f(x) dx $$ So here, the mass points are $x\in \{0,1,2\}$ such that $p(0)=1/4$, $p(1)=2/5$ and $p(2)=1/2-2/5=1/10$. When $x\in [2,2.5)$, $f(x)=F'(x)=1$. Therefore, your answer is $$ E[X] = 0\times 1/4+1\times 2/5+2\times 1/10+\int_2^{2.5}x \frac{2x-3}{2}dx = \frac{59}{40} $$