Deriving the variance from an incomplete normal distribution PDF

Question

In this Wikipedia page it is said:

$f_Z(z) = \frac{e^{-z^2}}{\sqrt{\pi}}$ ~ $N (0,(\frac{1}{\sqrt2})^2)$

My question is, why the mentioned fact in the picture is correct? can someone please show me why if:

$f_Z(z) = \frac{e^{-z^2}}{\sqrt{\pi}}$ then the distribution has a variance of 1/2?

more specifically, how can I derive the variance and expectation of a non-complete normal distribution function ?

the method I know to make a PDF into a normal distribution PDF is to do manipulations on the function like multiplying both numerator and denominator by the same number to make if fit the template of normal distribution but I cant seem to make it happen in this case. Specifically what is hard for me is the fact that the $-x^2$ is supposed to be divided by 2 regardless of the variance.

Thank you

Answer 1

The pdf of a normally distributed random variable with mean zero and variance $\sigma^2$ is

$$f(z)=\frac{1}{\sqrt{\color{green}2\cdot \pi\cdot \color{green}{\sigma^2}}} \cdot \large{e^{-\frac{1}{\color{red}{2}}\frac{z^2}{\color{red}{\sigma^2}}}} \normalsize, \, z\in \mathbb R$$

1. To get the same exponents at both functions we need that $2\cdot \sigma^2=1$, which means that the red term vanishes. Thus $\sigma^2=\frac12\Rightarrow\sigma=\frac{1}{\sqrt 2}$.
2. At the denomniator $2\cdot \sigma^2$ has to disappear as well. This leads to the same condition and result and we have no contradiction to 1.