# proof that $Y=μ+σX$ if X∼N(0,1),

I want to proof that If$$X∼N(0,1)$$, then$$Y=μ+σX$$has the 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 Jacobian.

Can someone explain me clearly? Thank you in advance.

• Considering CDFs or CFs suffice. – Nap D. Lover Apr 17 at 19:03

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 $$\text{Norm}(\mu,\sigma^2)$$.

• why 𝜎𝔼𝑋 is disappearing in 𝜇+𝜎𝔼𝑋=𝜇 ? – RecklessSerenade Apr 19 at 2:46
• Because $\mathbb EX=0$ if $X\sim\text{Norm}(0,1)$. – drhab Apr 19 at 7:25

Hint

$$\mathbb P\{Y\leq y\}=\mathbb P\left\{X\leq \frac{y-\mu}{\sigma }\right\}.$$