Max(min()) random variable problem

Question

Let X a random variable with density

$f(x)= \begin{cases} \frac{1}{4}x,& \quad \text{if} \quad 0\le x \le 2\\ \\ \frac{1}{8},& \quad \text{if} \quad 2\lt x \le 6\\ \\ 0,& \quad \text{otherwise} \end{cases} $

Define $Y=\max\{\min(X,3),2\}$. Find $F_Y$ and decompose it in discrete, singular and continuous part.

Answer 1

There are three relevant possibilities: (a) $X\leq 2$ (b) $2\leq X<3$ (c) $X\geq 3$. With this

$$Y=\max(\min(X,3),2)=\begin{cases}2&\text{ if } &X\leq 2\\ X&\text{ if }&2<X\leq3\\ 3&\text{ if }&X>3.\end{cases}$$

Accordingly

$$P(Y<2)=0,$$ $$P(Y=2)=P(X\leq 2)=\int_0^2f(x)\ dx=\frac14\int_0^2 x\ dx=\frac12.$$ If $2\leq y<3$ then

$$P(Y<y)=P(Y=2)+\int_2^yf(x)\ dx=\frac12+\frac18\int_2^y\ dx=\frac12+\frac18(y-2).$$ $$P(Y=3)=P(X>3)=P(3\leq X\leq 6)=\int_3^6f(x)\ dx=\frac38.$$

The shape of the distribution function is

Let $Y'$ denote a random variable taking the values $2$ and $3$ with probabilities $\frac47$ and $\frac37$, respectively and let $Y''$ another random variable distributed uniformly over $[2,3]$. $Y$ is then a random variable equaling $Y'$ (the discrete part) with probability $\frac78$ and equaling $Y''$ (the continuous part) with probability $\frac18.$