Prove $Y~\sim N(\mu_Y,\sigma_Y^2)$ when $Y= aX+b$ and $X \sim N(\mu,\sigma^2)$. 
Let $X$ be continuous random variable - $X \sim N(\mu,\sigma^2)$.
Let $Y$ be continuous random variable when $Y = a X + b$.
Prove that $Y$ distributes normal distribution also, and find $\mu_Y,\sigma_Y^2$.

My Attempt-
$F_Y(y) = P(Y \leq y) = P(aX+b \leq y) = P(X\leq \frac{y-b}{a}) = F_X(\frac{y-b}{a}).$
And then- $f_Y = \frac{dF_Y(y)}{dy} = \frac{dF_X(\frac{y-b}{a})}{dy} = \frac{f_X(\frac{y-b}{a})}{\frac{y}{a}}$.
But, I don't know how to continue. Can someone help me? Thanks.
 A: Given $X \sim N(\mu, \sigma^2)$ then as you have derived, one obtains
$$F_y(y) = F_X\left(\frac{y-b}{a}\right)$$
Therefore, the density of $Y$ is given by
\begin{align}
f_Y(y) &= \frac{1}{a} f_X\left(\frac{y-b}{a}\right) \\
&= \frac{1}{\sqrt{2\pi a^2 \sigma^2}} \exp\left[\frac{-1}{2\sigma^2} \left(\frac{y-b}{a}-\mu\right)^2 \right] \\
&= \ldots 
\end{align} 
Can you proceed?  
A: Other way
$$M_{Y}(t)=\mathbb{E}[e^{tY}]=\mathbb{E}[e^{t(aX+b)}]=e^{bt}\mathbb{E}[e^{atX}]=\large e^{bt}\, e^{a\mu\,t+\frac12\sigma^2a^2t^2}=e^{(a\mu+b)t+\frac12(a\sigma)^2t^2}$$
therefore
$$Y\sim N(a\mu+b,a^2\sigma^2)$$
A: Start with $U\sim Norm\left(0,1\right)$ and let $\sigma>0$.
If $X=\sigma U+\mu$ then we find $F_{X}\left(x\right)=P\left(U\leq\frac{x-\mu}{\sigma}\right)=\Phi\left(\frac{x-\mu}{\sigma}\right)$
and $f_{X}\left(x\right)=\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)$
wich can be recognized as the PDF connected with distribution $Norm\left(\mu,\sigma^2\right)$.
If $a>0$ and $Y=aX+b$ then $Y=a\sigma U+\left(a\mu+b\right)=\sigma'U+\mu'$
and the same procedure can be applied.
If $a<0$ and $Y=aX+b$ then $Y=\left(-a\right)\sigma V+\left(a\mu+b\right)=\sigma''V+\mu'$
where $V=-U$. Then $V\sim Norm\left(0,1\right)$ and the same procedure
can be applied.
