Proof of linear transformation of random variable

Question

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This is a part of proof of linear transformation of normal random variable, can anyone tell me how to get procedure 4 from procedure 3？

Answer 1

Let $x = \frac{v-\beta}{\alpha}$.

• $dx = \frac{1}{\alpha}dv.$
• If the bound of integration is some $X = \frac{a-\beta}{\alpha}$, then after transformation $V = a$.
• Finally, the algebra:

\begin{align} \frac{-(x-\mu)^2}{2\sigma^2} &= \frac{-\left(\frac{v-\beta}{\alpha}-\mu\right)^2}{2\sigma^2}\\ &= \frac{-\left(\frac{v^2-2\beta v + \beta^2}{\alpha^2}-2\mu\frac{v-\beta}{\alpha} + \mu^2\right)}{2\sigma^2}\\ &= \frac{-\left(\frac{v^2-2\beta v + \beta^2}{\alpha^2}-2\frac{\mu\alpha v-\mu \alpha\beta}{\alpha^2} + \frac{\alpha^2\mu^2}{\alpha^2}\right)}{2\sigma^2}\\ &= \frac{-\left(v^2-2\beta v + \color{blue}{\beta^2}-\color{red}{2\mu\alpha v}+2\mu \alpha\beta + \alpha^2\mu^2\right)}{2\alpha^2\sigma^2}\\ &= \frac{-\left(v^2-2\beta v -\color{red}{2\mu\alpha v} + \underbrace{\color{blue}{\beta^2}+2\mu\alpha\beta + \alpha^2\mu^2 }\right)}{2\alpha^2\sigma^2}\\ &= \frac{-\left(v^2\underbrace{-2\beta v -2\mu\alpha v} + (\beta + \alpha \mu)^2 \right)}{2\alpha^2\sigma^2}\\ &= \frac{-\left(v^2-2v(\beta + \alpha\mu) + (\beta + \alpha \mu)^2 \right)}{2\alpha^2\sigma^2}\\ &= \frac{-\left(v-(\beta+\alpha\mu)\right)^2}{2\alpha^2\sigma^2}\\ \square \end{align}