Pove that $Y=μ+σX$ if $X\sim N(0,1)$. I want to prove that if $$X\sim N(0,1),$$ then$$Y=μ+σX$$has normal distribution with mean $μ$ and variance $σ^2$.
I searched it before, but I don't understand why I have to calculate the probability density of $X$ and also the Jacobian.
Can someone explain me clearly? Thank you in advance.
 A: Hint
$$\mathbb P\{Y\leq y\}=\mathbb P\left\{X\leq \frac{y-\mu}{\sigma }\right\}.$$
A: Characteristics of normal distribution:


*

*If random variable $X$ has normal distribution and $Y=a+bX$ where $a,b$ are constants and $b\neq0$ then also $Y$ has normal distribution.

*The distribution is completely determined by mean and variance.


According to first bullet $Y=\mu+\sigma X$ has normal distribution if $X\sim\text{Norm}(0,1)$ with: $$\mu_Y=\mathbb E(\mu+\sigma X)=\mu+\sigma\mathbb EX=\mu\text{ and }\sigma_Y^2=\mathsf{Var}(Y)=\mathsf{Var}(\mu+\sigma X)=\sigma^2\mathsf{Var}(X)=\sigma^2$$This justifies the conclusion that $Y\sim\text{Norm}(\mu,\sigma^2)$.

It might be that actually you want a proof of the first bullet. 
If $\Phi$ denotes the CDF of $X$ and $\phi$ the PDF of $X$ then: $$\phi(x)=\frac1{\sqrt{2\pi}}e^{-\frac12x^2}=\Phi'(x)$$
We can find CDF $F_{Y}(y)=P(\mu+\sigma X\leq y)=P(X\leq\frac{y-\mu}{\sigma})=\Phi(\frac{y-\mu}{\sigma})$ and - taking the derivative - PDF: $$f_Y(y)=\frac1{\sigma}\phi\left(\frac{y-\mu}{\sigma}\right)=\frac1{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{y-\mu}{\sigma}\right)^2}$$
which is the PDF associated with $\text{Norm}(\mu,\sigma^2)$.
