Discussion
Let X be a r.v. with cdf, \(F(x)\), where :
\[ F(x) = \begin{cases} 0 & \text{if } x \lt 0 \\ x/2 & \text{if } 0 \le x \lt 1 \\ 3/4 & \text{if } 1 \le x \lt 2 \\ 1 & \text{if } 2 \le x \end{cases} \]
Then E(X) is given by :
Answer: D
\(E(X) =\) 1.
The distribution is mixed. On [0, 1) the cdf rises continuously as \(x/2\), a density of 1/2 carrying total probability 1/2. At \(x = 1\) it jumps from 1/2 to 3/4, a mass of 1/4 at 1, and at \(x = 2\) it jumps from 3/4 to 1, a mass of 1/4 at 2. So
\[ \begin{aligned} E(X) &= \int_{0}^{1} \frac{x}{2}\,dx + 1 \cdot \tfrac{1}{4} + 2 \cdot \tfrac{1}{4} \\ &= \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} = 1 \end{aligned} \]
3/8 and 1/2 come from missing one or both point masses, and 3/2 from counting the mass at 2 twice.
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