How are the ranges of these cdf determined?

Question

In the solution, it says $a > 0$ is $P(aX+b\leq y)$ and $a < 0$ is $P(aX+b\geq y)$. Can I get an explanation for why that is, please?

Answer 1

It is because, when you divide by a, the inequality will change sign, so you get for a<0 : $$P(aX+b\leq y)=P\left(X\geq\frac{y-b}{a}\right)=1-F_X\left(\frac{y-b}{a}\right) $$

Answer 2

Let $Y=aX+b$, where $a,b\in\mathbb{R}$ with $a\ne 0$.

By definition, $$ F_Y(y)=P(Y\le y)=P(aX+b\le y) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\; $$ hence if $a > 0$ we get \begin{align*} F_Y(y) &=\, P(aX+b\le y) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\; \\[4pt] &=\, P(aX\le y-b) \\[4pt] &=\, P\Bigl(X\le \frac{y-b}{a}\Bigr) \\[4pt] &=\, F_X\Bigl(\frac{y-b}{a}\Bigr) \\[4pt] \end{align*} and if $a < 0$, correcting a typo in the referenced solution, we get \begin{align*} F_Y(y) &=\, P(aX+b\le y) \\[4pt] &=\, 1-P(aX+b\ge y)&&\text{[assuming $P(aX+b=y)=0$]} \\[4pt] &=\, 1-P(aX\ge y-b) \\[4pt] &=\, 1-P\Bigl(X\le \frac{y-b}{a}\Bigr)&&\text{[since $a < 0$, dividing by $a$}\\[-4pt] &&&\;\text{reverses the direction of the inequality]} \\[4pt] &=\, 1-F_X\Bigl(\frac{y-b}{a}\Bigr) \\[4pt] \end{align*}