已知$X\sim N(\mu,\sigma^2)$，如何推导$(\frac{X-\mu}{\sigma})\sim N(0,1)$？

Hey there! Let's break down this proof step by step, sticking strictly to the definition of a normal distribution like you asked.

First, let's recap the definition: If a random variable $X$ follows a normal distribution with mean $\mu$ and variance $\sigma^2$ (written as $X \sim N(\mu, \sigma^2)$), its probability density function (PDF) is:
$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{{-\frac{(x-\mu)}2}{2\sigma^2}}, \quad -\infty < x < \infty$$

Our goal is to show that $Z = \frac{X - \mu}{\sigma}$ follows the standard normal distribution $N(0,1)$, whose PDF is:
$$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z2}{2}}, \quad -\infty < z < \infty$$

Here's how we get there:

1. Rewrite $X$ in terms of $Z$
   Rearrange the definition of $Z$ to solve for $X$:
   $$X = \sigma Z + \mu$$
   This is a linear, strictly increasing transformation (since $\sigma > 0$, as it's a standard deviation), which makes calculating $Z$'s PDF straightforward.

2. Calculate the derivative of the transformation
   For continuous random variables, when we apply a monotonic transformation, we need the absolute value of the derivative of $X$ with respect to $Z$. Let's compute that:
   $$\frac{dX}{dZ} = \sigma$$
   Since $\sigma$ is positive, its absolute value is just $\sigma$.

3. Apply the PDF transformation rule
   For a continuous random variable $X$ with PDF $f_X(x)$, if $X = g(Z)$ is a strictly monotonic function, the PDF of $Z$ is given by:
   $$f_Z(z) = f_X(g(z)) \times \left| \frac{dg(z)}{dz} \right|$$
   Substitute $g(z) = \sigma z + \mu$ into $f_X(x)$:
   $$f_X(\sigma z + \mu) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(\sigma z + \mu - \mu)^2}{2\sigma2}}$$
   Simplify the exponent inside the exponential function:
   $$\frac{(\sigma z)^2}{2\sigma2} = \frac{\sigma^2 z^2}{2\sigma2} = \frac{z^2}{2}$$
   So now $f_X(\sigma z + \mu)$ simplifies to:
   $$\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{z2}{2}}$$

4. Final simplification to get $Z$'s PDF
   Multiply the above result by the absolute derivative we found earlier ($\sigma$):
   $$f_Z(z) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{z2}{2}} \times \sigma$$
   The $\sigma$ terms cancel out perfectly, leaving us with:
   $$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z2}{2}}$$

That's exactly the PDF of a standard normal distribution! So we've proven that $Z = \frac{X - \mu}{\sigma} \sim N(0,1)$.

内容的提问来源于stack exchange，提问作者user122424