Discussion
Suppose a distribution function \(F : \mathbf{R} \to [0, 1]\) of a random variable X is as follows :
\[ F(x) = \begin{cases} 0, & \text{if } x \lt 0, \\ 1/4, & \text{if } 0 \le x \lt 1, \\ 1/2, & \text{if } 1 \le x \lt 2, \\ 1/2 + (x - 2)/2, & \text{if } 2 \le x \lt 3, \\ 1, & \text{if } x \ge 3. \end{cases} \]
Then E(X) is :
Answer: C
\(E(X) =\) 3/2.
Read the distribution off F. It jumps by 1/4 at 0 and by 1/4 at 1, is flat on [1, 2), and rises linearly with slope 1/2 on [2, 3), from 1/2 to 1. So X is 0 with probability 1/4, 1 with probability 1/4, and uniform on [2, 3] with total probability 1/2.
\[ \begin{aligned} E(X) &= 0 \cdot \tfrac{1}{4} + 1 \cdot \tfrac{1}{4} + \tfrac{1}{2} \cdot 2.5 \\ &= \tfrac{1}{4} + \tfrac{5}{4} = \tfrac{3}{2} \end{aligned} \]
Check with \(E(X) = \int_{0}^{\infty}(1 - F(x))dx\), which is \(\tfrac{3}{4} + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{2}\). The other values come from treating F as purely continuous or dropping one of the jumps.
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