How to combine two integrals containing the PDFs of a variable and its linear transform?

Question

Original Post:

Suppose we have two random variables $X$ and $Y$ with cumulative distribution functions $F(x)$ and $G(y)$. We know that $Y = aX + b$.

I want to compute $Z(x) = F(x) - G(y)$.

What I have so far is the following: $$F(x) = \int_{-\infty}^{x}f(x)dx $$

$$G(y) = \int_{-\infty}^{y}g(y)dy = \int_{-\infty}^{y}\frac{1}{|a|}f\left(\frac{y-b}{a}\right)dy $$

$$Z(x) = \int_{-\infty}^{x}f(x)dx - \int_{-\infty}^{y}\frac{1}{|a|}f(x)dy $$

How should I go on doing the subtraction of the integrals? Is this even possible as the integrales belong to different variables (dx and dy)?

Solution thanks to jbowman:

Suppose $a>0$ and $y=ax+b$

$$G(y) = P(aX+b <= y)$$ $$G(y) = P(aX+b<=ax+b) = P(X<=x) = F(x)$$ And thus: $$Z(x,y) = F(x) - G(y) = 0$$

Corrected Post

Suppose we have two random variables $X$ and $Y$ with cumulative distribution functions $F(k)$ and $G(k)$. We know that $Y = aX + b$.

I want to compute $Z(k) = F(k) - G(k)$.

Proposed solution

Assuming $a>0$:

$$F(k) = \int_{-\infty}^{k}f(x)dx $$

$$G(k) = \int_{-\infty}^{k}g(y)dy = \int_{-\infty}^{\frac{k-b}{a}}f(x)dx $$

$$Z(k) = \int_{-\infty}^{k}f(x)dx - \int_{-\infty}^{\frac{k-b}{a}}f(x)dx $$

And thus

$$Z(k) = \int_{\frac{k-b}{a}}^{k}f(x)dx $$

Answer 1

Let's consider the case $a > 0$. In this case, $Z(x) = 0$ for all $x$ as $y$ is a deterministic monotonically increasing function of $x$. The probability that $y \leq 1$, for example, is equal to the probability that $x \leq (1-b)/a$, i.e., $G(1) = F((1-b)/a)$, and when $y = 1$, $x$ does $ = (1-b)/a$, so the two cumulative probabilities are equal.

If this isn't clear, consider the case $a = 1, b = 0$.

You should be able to work out the case $a < 0$ from here!

Addition to take into account @yahiro's comment: If $a = 0$, then things work differently, as $g$ is a point mass at $b$, so $G = 0$ to the left of $b$ and $1$ at $b$ and above. This affects $Z(x)$ in an obvious way.