Discussion
Suppose a distribution function \(F : \mathbf{R} \to [0, 1]\) 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 \(F(x) = l_{1}F_{1}(x) + l_{2}F_{2}(x)\) where :
Answer: B
The decomposition is \(l_{1} = l_{2} = 1/2\), \(F_{1}\) discrete on {0, 1} with probabilities 1/2, 1/2, and \(F_{2}\) uniform on (2, 3).
F jumps by 1/4 at 0 and by 1/4 at 1, so the discrete part has total mass 1/2, split equally between 0 and 1. Normalised, that is the distribution with probabilities 1/2, 1/2 on {0, 1}, with weight \(l_{1} = 1/2\).
From 2 to 3, F rises linearly from 1/2 to 1, so the continuous part has mass 1/2 spread uniformly over (2, 3): weight \(l_{2} = 1/2\) and \(F_{2}\) the U(2, 3) distribution function. Check: \(\frac{1}{2}F_{1} + \frac{1}{2}F_{2}\) gives 1/4 on [0, 1), 1/2 on [1, 2) and \(\frac{1}{2} + \frac{x - 2}{2}\) on [2, 3).
The options with \(l_{1} = 1/4\) count only one jump, and option D puts the jumps at 1 and 2 instead of 0 and 1.
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