# Is $p(x > a)$ for a discrete PDF necessarily linear?

We got these solutions to a class problem, and I'm having trouble seeing why the (b) is true: problem solutions (sorry it's not inline, not enough rep)

For example, let $\alpha_1 = 0.25, \alpha_2 = 0.5, \alpha_3 = 0.75$. This satisfies the constraint that $\alpha_1 < \alpha_2 < \alpha_3$. Now, let $p(x = \alpha_1) = 0.25, p(x = \alpha_2) = 0.5, p(x = \alpha_3) = 0.25$.

Then,

\begin{equation*} p(x \geq \alpha) = \begin{cases} 1 & 0 \leq \alpha \leq \alpha_1\\ 0.75 & \alpha_1 < \alpha \leq \alpha_2\\ 0.25 & \alpha_2 < \alpha \leq \alpha_3\\ 0 & \alpha_3 < \alpha \leq 1 \end{cases} \end{equation*}

Plotting this out, it's pretty clearly not linear...so I don't understand the solution's claim that this is a linear function of p. It definitely doesn't look convex. Am I missing something?

$\mathbf{p}$ is a vector. I think linear in $\mathbf{p}$ just means that it can be expressed as a product of a matrix and $\mathbf{p}$, i.e. like $\mathbf{A}\mathbf{p}$. And $P(x\ge \alpha)$ indeed can be.