Finding CDF of $Y = \min({X, a/2})$

Question

Let $X$ be a uniformly distributed random variable on $[0, a]$. Let $Y$ be a random variable such that $Y = \min(X, a/2)$.

I quickly identified that it was not a discrete random variable. However, I'm not sure if it is a continuous random variable. It is able to take infinitely many values in the range $[0, a/2]$, but the CDF that I found was not differentiable everywhere, so I don't think it has a PDF. Unless we can break down the function piecewise and treat the integral of the function accordingly?

The CDF that I obtained was $$\begin{cases} 0 & \text{for } y < 0 \\ 2y/a & \text{for } 0 \leq y < a/2 \\ 1 & \text{for } y \geq a/2 \end{cases}$$

Is this the correct CDF? It is not differentiable everywhere due to the sharp point at $a/2$.

Answer 1

The probability that $P(Y < c)$ for $c < \frac{a}{2}$ is $$P(Y < c) = P(X < c) = \int_0^c \frac{1}{a}\;dx = \frac{c}{a}$$ Note that $Y$ can only take on values in $[0,\frac{a}{2}]$ so it is always true that $Y\leq \frac{a}{2}$. So its CDF is: \begin{cases} \frac{y}{a} & \text{for }0\leq y<\frac{a}{2} \\ 1 & \text{for } y = \frac{a}{2} \end{cases}