if.~~X \sim N(a,~b)_. \kern{5pt} Y = aX + b
\begin{align*}
F_X(x)&=P(Y \le y) = P(aX+b \le Y)
\\[15pt]
&=P\Big(X \le \small\frac{y-b}{a}\Big)
~=~F_X\Big(\small\frac{y-b}{a}\Big)
\end{align*}
\begin{align*}
f_X(x)&= \dfrac{d}{dx}F_X(x) =
\dfrac{d}{dx}
F_X\Big(\small\frac{y-b}{a}\Big)
\\[15pt]
&=f_X\Big(\small\frac{y-b}{a}\Big)
\cdot\frac{1}{a}
\\[15pt]
&=\frac{1}{a} \cdot
\frac{1}{\sqrt{2\pi\sigma}} \cdot
\exp\bigg[
\small\frac{-1}{2\sigma^2} \cdot
\Big( \frac{y-b}{a}-\mu \Big)^2
\bigg]
\\[15pt]
&= \frac{1}{\sqrt{2\pi}(a\sigma)} \cdot
\exp\bigg[
\small\frac{-1}{2\sigma^2} \cdot
\Big( \frac{y-b-a\mu}{a}\Big)^2
\bigg]
\\[15pt]
&= \frac{1}{\sqrt{2\pi}(a\sigma)} \cdot
\exp \bigg[ - \dfrac
{\{y - (a\mu+b)\}^2}
{2a^2\sigma^2} \bigg]
\end{align*}
\therefore~~Y ~\sim~ N(aX + b,~~ a^2\sigma^2)
- 정규분포는 선형변환해도 정규분포의 성질이 유지된다.
응용 : 회귀분석
Y = \beta_0 + \beta_1x_1 + \cdots
+ \epsilon_i. \kern{20pt}
\epsilon_i \sim N(0, \sigma^2) = \text{오차항}
\begin{align*}
\text{Then, }~&~ \epsilon_i + k \kern{43.5pt} \sim N(0,1)
\\
Y_i =&~ \epsilon_i +
(\beta_0 + \beta_1x_1) \sim N(0,1)
\\[20pt]
\text{So,}~&~Y_i \sim
N(\beta_0 + \beta_1x_1,~\sigma^2)
\end{align*}