How to compute expected value

Question

How do I solve the expected value of this problem, if I have already calculated the pmf?

Let $X$ be a random variable with cumulative distribution function given below: $$F_X(x) = \begin{cases} 0 & : x < 0 \\ 1/2 & : 0\leqslant x < 1 \\ 3/5 & : 1 \leqslant x < 2 \\ 4/5 & : 2 \leqslant x < 3 \\ 9/10 & : 3 \leqslant x < 3.5 \\ 1 & : x \geqslant 3.5 \end{cases}$$ Compute $\mathsf E[X]$

Answer 1

You say you have calculated the PMF $p_{X}(x)$ so you can use it to find the expected value:

$$E[X] = \sum_{x}xp_{X}(x)$$

Answer 2

these data are well-suited to using the formula $$ E[X] = \int_0^{\infty}(1- F_X(x))dx $$ which here gives: $$ E[X] = (1-0).\frac12+(2-1).\frac25+(3-2).\frac15 +(3.5-3).\frac1{10} = \frac{23}{20} $$

Answer 3

You probability mass function should look like: $f_X(x) = \begin{cases} \Box & : x=0 \\ \Box & : x=1 \\ \Box & : x=2 \\ \Box & : x=3 \\ \Box & : x= 3.5 \\ 0 & : \text{elsewhere} \end{cases}$

Since you have filled in the boxes for this, then you can just evaluate:

$$\begin{align}\mathsf E[X] &= \sum_{x\in\{0,\,1,\,2,\,3,\,3.5\}} x\, f_X(x) \\[1ex] & = 0\, f_X(0) + 1\, f_X(1) + 2\, f_X(2) + 3\,f_X(3)+3.5\,f_X(3.5) \end{align}$$