statistics - Find cumulative distribution function of random variable

Let the random variable $Y$ be defined as $Y = aX$ where $a$ is some non-zero constant (may be a positive or negative real number) and $X$ is some known random variable. How would we find the cumulative distribution function of $Y$?

We can say...
$$F_Y(y) = \mathbb P(Y \le y)$$

$$= \mathbb P(aX \le y)$$

$$\text{if } a > 0\text{, then}$$
$$= \mathbb P(X \le y / a)$$

$$\text{if } a < 0 \text{, then}$$
$$= \mathbb P(X \ge y / a) \approx 1 - \mathbb P(X \le y / a)$$

$$\text{So thus...}$$
$$\text{if } a > 0\text{, then}$$

$$F_Y(y) = F_X(y / a)$$

$$\text{if } a < 0 \text{, then}$$
$$F_Y(y) = 1 - F_X(y / a)$$

Is my formulation correct, and if so how can we make this disjointed function into a single line?

EDIT:
Could we say something like...
$$F_Y(y) = (0.5 - 0.5 * sgn(a)) + sgn(a) * F_X(y / a)$$

Answer

Yes, that is correct if $X$ is a continuous random variable. In this case your $\approx$ sign is actually an $=$ sign. If $X$ is discrete, though, then the $\approx$ is not an $=$, and you should instead have $\mathbb{P}(X \geq y/a) = 1-\mathbb{P}(X \leq y/a) + \mathbb{P}(X=y/a)= 1-\lim_{x \to (y/a)^-}F_X(y/a)$.

I would not recommend trying to "make this disjointed function into a single line" the way you did. Your one-line formula really obscures what's going on in the two cases. If anything, I would write a "piece-wise" function:
$$F_Y(y) = \begin{cases}F_X(y/a) & a>0 \\ 1-F_X(y/a) & a<0 \end{cases}$$
(in the continuous case).