Function of Uniform distribution

Question

If I have a random variable $Z \sim U(0,1)$ how can I find the distribution of $Y=2-Z$? Is correct to say $P(2-Z \leq y) = P(Z \leq 2-y)$ and then integrate $2-Y$?

Answer 1

For computing the distribution of $Y = 2-Z$, at the first step we compute $F_{Y}(y)$.

$$F_{Y}(y) = P(Y \leq y) = P(2-Z\leq y) = P(2-y \leq Z) = \int_{2-y}^{1}f_{Z}(z)dz = \int_{2-y}^{1}1dz = 1- (2-y) = y-1$$

We know $PDF = \frac{d}{dy}CDF$, Thus $$f_{Y}(y) = \frac{d}{dy}(y-1) = 1$$

Answer 2

Intuitively, you are just shifting a uniform r.v.. Formally, let $Y=2-X$, then \begin{align} F_Y(y) &= P(Y\le y )\\ &=P(2-X \le y)\\ &= P(X\ge 2-y)\\ &= 1- P(X < 2-y)\\ &= 1 - (2-y)\\ &=y-1. \end{align} Namely, $f_Y(y) = 1$ for $1\le y\le 2$, because $$ 0\le X \le 1 \to -1 \le -X \le 0 \to 2-1\le 2 - X \le 2-0. $$ That is, due to one-to-one relation between the CDF which is $$ F_Y(y) = \frac{y-1}{2-1}, $$ and the r.v., we deduce that $Y \sim U[1,2]$.

Answer 3

Hint:

Let $f$ denote a function that serves as PDF of random variable $X$ and let $a,b\in\mathbb R$ with $a\neq0$.

Then the function prescribed by: $$x\mapsto\frac1{|a|}f\left(\frac{x-b}{a}\right)$$

serves as PDF of random variable $b+aX$.