If $X$ follows a normal distribution, $X/\sigma$ follows a normal distribution??

Question

If $X$ follows a normal distribution, $X/\sigma$ ($\sigma$ is the standard deviation) follows a normal distribution? I have tried to prove using the transformation formula but I can't. Can anybody help me?

Answer 1

$P(X \leq x)=\frac 1 {\sqrt {2\pi} \sigma} \int_{-\infty} ^{x} e^{-(t-m)^{2}/2\sigma^{2}} dt$ where $m=EX$. Hence $P(\frac X {\sigma} \leq x)=P(X \leq \sigma x)= \frac 1 {\sqrt {2\pi} \sigma} \int_{-\infty} ^{\sigma x} e^{-(t-m)^{2}/2\sigma^{2}} dt$. Put $y=\frac t \sigma$. We get $P(\frac X {\sigma}\leq x) =\frac 1 {\sqrt {2\pi} } \int_{-\infty} ^{x} e^{-(y-\frac m {\sigma})^{2}/2} dy$. Hence $\frac X \sigma \sim N(\frac m \sigma,1)$.

Answer 2

Using the characteristic function.

Suppose $X\sim N(\mu,\sigma)$. Let $Z = \frac{X}{\sigma}$. Then, for every $t\in\mathbb{R}$, $$ \mathbb{E}[e^{it Z}] = \mathbb{E}[e^{i \frac{t}{\sigma} X}] = e^{i \frac{t}{\sigma} \mu - \frac{1}{2} \sigma^2\left(\frac{t}{\sigma}\right)^2} = e^{it \frac{\mu}{\sigma} - \frac{1}{2}t^2} $$ which is the characteristic function of a random variable distributed as $N(\frac{\mu}{\sigma},1)$, Therefore, $Z\sim N(\frac{\mu}{\sigma},1)$.