We have this random variable $Y= \frac{x – μ}{\sigma}$ to convert a normal distribution $N(\mu, \sigma)$ to a $N(0, 1)$. It is quite intuitive to subtract $\mu$, since you move all the values in the $x$-axis, and thus move the mean $\mu$ to the origin of coordinates. But it does not seem intuitive to divide by the standard deviation.
This answer is basically this, however, I have not understood the answers, specifically this:
$$E[Y] = \frac{E[X] – \mu}{\sigma} = \frac{\mu-\mu}{\sigma} = 0.$$
$$\text{Var}(Y) = \frac{1}{\sigma^2}\text{Var}(X) = \frac{1}{\sigma^2}\sigma^2 = 1.$$
And I would also like to get an intuitive answer.
Best Answer
The first of the formulas uses the linearity of the expected value. To be more specific: For $a,b \in \mathbb{R}$ and a random variable $X$ we have $$ \mathbb{E}[aX] = a \mathbb{E}[X] \qquad \text{and} \qquad \mathbb{E}[X + b] = \mathbb{E}[X] + b $$ This especially implies $\mathbb{E}[a] = a$ for every constant $a \in \mathbb{R}$. Since $\mathbb{E}[X]$ is a constant, this implies $$\mathbb{E}[\mathbb{E}[X]] = \mathbb{E}[X].$$
In your case (loosely speaking) $a = \frac{1}{\sigma}$ and $b = - \mu$.
For the second. Using the definition and the result from the first formula we obtain \begin{align} \text{Var}(Y) & \overset{\textrm{Def.}}{\underset{(\star)}{=}} \mathbb{E}[(Y - \mathbb{E}[Y])^2] = \mathbb{E}[Y^2] = \mathbb{E}\left[ \left(\frac{X - \mu}{\sigma}\right)^2 \right] \overset{\textrm{L}}{=} \frac{1}{\sigma^2}\mathbb{E}\left[ \left(X - \mu\right)^2 \right] \\ & = \frac{1}{\sigma^2}\mathbb{E}\left[X^2 - 2 \mu X + \mu^2 \right] \overset{\textrm{L}}{=} \frac{1}{\sigma^2} \left(\mathbb{E}[X^2]- 2 \mu \mathbb{E}[X] + \mu^2 \right) \\ & = \frac{1}{\sigma^2} \left(\mathbb{E}[X^2]- \mu^2 \right) = \frac{1}{\sigma^2} \left(\mu^2 + \sigma^2 - \mu^2 \right) = 1, \end{align} where in the last step we use $\mathbb{E}[X^2] = \mu^2 + \sigma^2$.
The alternative definition of the variance $(\star)$ can be obtained like this: \begin{align} \mathbb{E}[(X - \mathbb{E}[X])^2] & \overset{\textrm{(L)}}{=} \mathbb{E}[X]^2 - 2 \mathbb{E}[X \mathbb{E}[X]] + \mathbb{E}[(\mathbb{E}[X])^2] \overset{\textrm{(L)}}{=} \mathbb{E}[X^2] - 2 \mathbb{E}[X] \cdot \mathbb{E}[X] + \mathbb{E}[X]^2 \\ & = \mathbb{E}[X^2] - 2 \mathbb{E}[X]^2 + \mathbb{E}[X]^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2. \end{align}