Why the density of the normal law $\mathcal N(0,\sigma ^2)$ is $\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{x^2}{2\sigma ^2}}$ and not

Question

Why the density function of a random variable $X$ that follow the normal law $\mathcal N(0,\sigma ^2)$ is $\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{x^2}{2\sigma ^2}}$ and not $\frac{1}{\sigma \sqrt{\pi}}e^{-\frac{x^2}{\sigma ^2}}$ ? Indeed, this could work since $$\int_{\mathbb R}\frac{1}{\sigma \sqrt\pi}e^{-\frac{x^2}{\sigma ^2}}dx=1.$$

So why adding the $2$ in $$\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{x^2}{2\sigma ^2}}\ \ ?$$

Answer 1

The function $\frac1{\sigma\sqrt\pi}e^{-\frac{x^2}{\sigma^2}}$ is a PDF for positive $\sigma$, and this for random variables that have normal distribution with mean $0$.

However if we calculate the corresponding variance on base of this PDF then the answer is not $\sigma^2$.

So we are not dealing with $\mathcal N(0,\sigma^2)$.

Answer 2

Indeed, for any $a>0$ the function $$ f(x;a) = \sqrt{\frac{a}{\pi}}e^{-a x^2}$$

is a valid probability density, and it's actually a zero mean normal. We choose to write it by doing a change of parameter $\sigma^2=1/(2a)$ (so we get the standard zero mean normal formula) and not $\sigma^2=1/a$ merely for convenience, so that the parameter $\sigma^2$ corresponds to the variance.

Answer 3

The density

$$b\,e^{-ax^2}$$ where $b$ is the normalization constant has expectation $0$ and variance $$\frac{\displaystyle\int_{-\infty}^\infty x^2 e^{-ax^2}dx}{\displaystyle\int_{-\infty}^\infty e^{-ax^2}dx}=\frac1{2a}.$$

Answer 4

The change of variable: $$\int_{-\infty}^{\infty}\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}dx\stackrel{\frac{x}{\sqrt2}=t}=\int_{-\infty}^{\infty}\frac{1}{\sigma \sqrt{\pi}}e^{-\frac{t^2}{\sigma^2}}dt=1,\\ Var\left(\frac{x}{\sqrt{2}}\right)=\frac12Var(x)=\frac12\sigma^2.$$

Answer 5

Short answer:

Integrating by parts,

$$\frac{\displaystyle\int_{-\infty}^\infty x^2 e^{-x/2}dx}{\displaystyle\int_{-\infty}^\infty e^{-x^2/2}dx}=\frac{-\left.xe^{-x^2/2}\right|_{-\infty}^\infty+\displaystyle\int_{-\infty}^\infty x^2 e^{-x^2/2}dx}{\displaystyle\int_{-\infty}^\infty e^{-x^2/2}dx}=1$$ explains the divisor $2$. (Without it, the variance would be $\frac12$.)