# Clarifying the elementary calculus used in this statistics problem

Let $X \sim N(\mu, \sigma^{2})$ and $Y = \alpha X + \beta$ for $\alpha > 0$. I'm looking at a demonstration that $Y = \alpha X + \beta \sim N(\alpha\mu + \beta, (\alpha\sigma)^{2})$, and find myself not following all the mechanical steps. I have a very weak calculus background, so I think this might clarify things for me.

\begin{align*} F_{Y}(c) &= P(Y \leq c) \\ &= P(\alpha{X} + \beta \leq c) \\ &= P \left( X \leq \frac{c - \beta}{\alpha} \right) \\ &= F_{X} \left( \frac{c - \beta}{\alpha} \right) \\ &= \int_{-\infty}^{\frac{c - \beta}{\alpha}} \frac{1}{\sqrt{2\pi}\sigma} \exp \left\{ \frac{-(x-\mu)^{2}}{2\sigma^{2}} \right\}dx \end{align*}

Using the change of variables $y = \alpha{x} + \beta$, we reduce this expression to $$\int_{-\infty}^{c} \frac{1}{\sqrt{2\pi} \alpha\sigma}\exp \left\{ \frac{-(y-(\alpha\mu + \beta))^{2}}{2\alpha^{2}\sigma^{2}} \right\}dy$$

and conclude the desired result.

Questions:

1) How does the limit of integration change from $\frac{c-\beta}{\alpha}$ to $c$?

2) When approaching similar problems, is there a good way of thinking about how the equation needs to be transformed in order to come up with a change of variables that accomplishes the desired purpose?

If $y = \alpha x + \beta$, then when $x=\frac{c-\beta}{\alpha}$ we have $$y = \alpha\cdot \frac{c-\beta}{\alpha} + \beta = (c-\beta) + \beta = c$$ The lower limit is still $-\infty$ because $y\to-\infty$ as $x\to-\infty$.
As for (2) I'm not sure there's a simple answer. You want to simplify the integral. If the substitution is linear like this one that's pretty straightforward. If $y$ is a nonlinear function of $x$, then you have to make sure the derivative of that function is part of the integrand so you can make a substitution of $dy$ in terms of $x$ and $dx$.